E. None of the above.E. Actually, they differ in all of these ways.pdf
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E. None of the above. E. Actually, they differ in all of these ways. E. Actually, all of these are accurate. E. None of the above.1.Bacteria are good experimental organisms for genetic studies for all of the following reasons EXCEPT:A. they have large numbers of progeny.B. they reproduce rapidly.C. they are diploid.D. they reproduce asexually.E. Actually, all of these are reasons bacteria are useful in genetic studies. Solution 1. Bacteria are good experimental organisms for genetic studies for all of the following reasons except they have a relatively small, diploid genome that is easier to characterize. So, the correct answer is c. 2. A plasmid is a small, circular, double-stranded DNA molecule that is distinct from a cell\'s chromosomal DNA. So, the correct answer is plasmid DNA is circular like bacterial DNA. Option A. 3. Option B (Transference). Transference is the unconscious transfer to others of feelings, and attitudes which were originally associated with important figures in one\'s early life. 4. Option C (minutes ) is the correct answer. It is a unit of meaure in baterial conjugation experiment..
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Question 1 Random changes in gene frequency that occur in small .docx
Question 1
Random changes in gene frequency that occur in small populations describes the concept of
A.
convergent evolution.
B.
adaptive radiation.
C.
sympatric speciation.
D.
genetic drift.
E.
parapatric speciation.
5 points
Question 2
On a molecular level, all organisms
A.
have the same bases in DNA and RNA, the same amino acids in proteins, but use different genetic codes.
B.
have the same bases in DNA and RNA, but different amino acids in proteins.
C.
have the same bases in DNA and RNA, the same amino acids in proteins, and use the same genetic code.
D.
show no similarities.
E.
use different means of storing genetic information.
5 points
Question 3
Which of the following is the best description of an adaptation?
A.
a trait that will be beneficial in the future
B.
a trait that was beneficial to survival in the past
C.
an environmental change that selects for certain traits
D.
a trait that is of no advantage to survival
E.
a trait that is currently beneficial to survival
5 points
Question 4
Which of the following statements about natural selection is true?
A.
All individuals within a population have an equal likelihood of survival. Survival is based on chance.
B.
Natural selection results in those individuals within a population who are best-adapted surviving and producing more offspring.
C.
Natural selection leads to extinction.
D.
Natural selection causes variations to arise within a population.
E.
Natural selection leads to increased likelihood of survival for certain individuals based on variation. The variation must come from outside the population.
5 points
Question 5
Embryonic development includes all of the following except
A.
cleavage.
B.
growth.
C.
organogenesis.
D.
gastrulation.
E.
fertilization.
5 points
Question 6
Vestigial organs
A.
include the human appendix and wisdom teeth and a snake’s hipbones.
B.
did not have a function when they first developed, but do so today.
C.
are not found in present-day organisms.
D.
are similar in many organisms.
E.
must have evolved individually in different organisms.
5 points
Question 7
Most mutations lead to a decreased chance of survival for the mutant organism. Rarely, a mutation can provide an increased chance of survival. These rare mutations provide a(n) _____________________ for the mutant.
A.
directional shift
B.
selective disadvantage
C.
adaptive
D.
selective advantage
E.
environmental susceptibility
5 points
Question 8
Which of the following is not true of the punctuated equilibrium theory?
A.
It suggests that most species undergo little phenotypic change during most of their evolutionary duration.
B.
It proposes that most evolutionary change is concentrated in relatively brief episodes of change.
C.
It was inspired in part by patterns observed in the fossil record.
D.
It appears to be well supported by the fossil record.
E.
It suggests that subgroups diverge and evolve at about equal and constant rates.
5 poin ...
In-class introduction to basic Punnett square set-up and problem s.docx
In-class introduction to basic Punnett square set-up and problem solving, Part 1
Problem-solving tips:
· A Punnett square allows you to predict the possible genetic outcome of children based on the genetic make-up of the parents.
· First, read the problem and figure out whether the trait of interest or genetic disorder is found on the dominant allele or the recessive allele because that will have an impact on how you interpret the results of the Punnett square.
· Select a letter to represent the trait or disorder and define the dominant and recessive alleles. For example: For eye color, B (dominant) = brown eyes and b (recessive) = blue eyes. For achondroplasia (dwarfism), A (dominant) = achondroplasia and a (recessive) = normal allele.
· If it is a sex-linked question, remember to include the sexual genotypes of the parents (XX for mom and XY for dad).
· Write down all possible genotypes & phenotypes and use this information to help you set up the Punnett square.
1. Practice question on a human trait. In reality, eye color is controlled by multiple genes and is a complex trait. For simplicity, we’ll assume that brown eyes are dominant to blue eyes. Answer the questions below.
a) Select a letter for this trait and define the dominant and recessive alleles.
B (dominant) =
b (recessive) =
b) Write down all possible genotypes and phenotypes for individuals in the population
Possible genotypes
(the 2 alleles an individual has)
Possible phenotypes (the physical appearance of a trait)
Homozygous dominant individuals
Homozygous recessive individuals
Heterozygous individuals
c) Set up the Punnett square and solve this problem. Kristy is heterozygous and Mark has blue eyes. What percentage of their offspring will have blue eyes?
Kristy's genotype
Mark's genotype
a) Select a letter for this genetic condition and define the dominant and recessive alleles.
F (dominant) =
f (recessive) =
b) Write down all possible genotypes and phenotypes for individuals in the population
Possible genotypes
(the 2 alleles an individual has)
Possible phenotypes (the physical appearance of a trait)
Homozygous dominant individuals
Homozygous recessive individuals
Heterozygous individuals
c) Set up the Punnett square and solve this problem. Kristy and Mark are carriers for cystic fibrosis. The term carrier is only used when a condition is on the recessive allele. Carriers are heterozygous individuals who are normal and show no symptoms of the disorder, but they have the ability to pass on the mutated recessive allele to their offspring. What percentage of their children will be normal? What percentage of their children will be carriers?
Kristy's genotype
Mark's genotype
2. Practice question on a genetic condition. Cystic fibrosis (CF) is an autosomal, recessive condition that results in mucus buildup in the lungs and digestive system organs. As a result, CF .
Cbse class 10 science chapter 9 heredity and evolution2015 11-09
Cbse class 10 science chapter 9 heredity and evolution
1. The case of albinism among the Hopi Indians of Black M.pdf
1. The case of albinism among the Hopi Indians of Black Mesa is different from other
cases of albinism because of all of the following EXCEPT:
A. the incidence is much higher among the Hopis.
B. the albinism seems to have provided a reproductive advantage.
C. the albinism seems to have conferred social advantages.
D. the albinism did not involve vision defects.
E. Actually, all of the above are accurate.
2. Genetics has NOT played a key role in which of the following areas?
A. agriculture
B. medicine
C. pharmaceutical production
D. urban design and layout
E. Actually, it has played a key role in all of these areas.
3. Genetic variation in populations:
A. demonstrates descent from a common ancestor.
B. provides the basis of adaptation.
C. explains the use of similar genetic systems in all organisms.
D. is a consequence of origin of species.
E. All of the above.
4. Genetics as a discipline can be subdivided into all of the following EXCEPT:
A. transmission genetics.
B. acquisition genetics.
C. molecular genetics.
D. population genetics.
E. Actually, all of these are found in the discipline.
5. While modern genetics starts with the experiments of Gregor Mendel in 1866, it wasn\'t
until the start of the 20th century that those studies became well known, in several areas, NOT
including:
A. Sutton, who proposed that genes were on chromosomes.
B. Morgan, who identified fruit fly mutations.
C. Haldane and others who used populations to connect genetics to evolution.
D. Watson and Crick\'s determination of the structure of DNA.
E. Actually, Mendel\'s work was critical in each of these areas.
6. Prokaryotes and eukaryotes differ in that:
A. prokaryotes have a nucleus.
B. they decode DNA to different amino acids.
C. prokaryotes tend to be larger.
D. prokaryotes tend to be less complex.
E. None of the above.
7. Genes:
A. are carried on chromosomes.
B. come in pairs in eukaryotic cells.
C. come in alternative forms, known as alleles.
D. are encoded by DNA.
E. All of the above.
8. Phenotypes are:
A. measurable traits of an organism.
B. not inherited directly.
C. not changeable during the life of an organism.
D. A and B.
E. All of the above.
9. Evolution:
A. is genetic change.
B. starts with variation.
C. does not require change of DNA.
D. A and B.
E. All of the above.
10. The flow of information is as follows:
A. protein to DNA to RNA.
B. protein to RNA to DNA.
C. DNA to RNA to protein.
D. RNA to protein to DNA.
E. None of the above.
11. Mutations are changes of DNA:
A. that can affect a single gene.
B. that can affect large regions of chromosomes.
C. that can be passed on to daughter cells.
D. that can change amino acid sequences.
E. All of the above.
Solution
1). C. the albinism seems to have conferred social advantages.
2). D. urban design and layout
3). D. is a consequence of origin of species.
4). B. acquisition genetics.
5). C. Haldane and others who used populations to connect genetics to evolution.
6). E. None of the above.
7). A. are carried on ch.
Teaching with Clickers for Deep Learning
Presentation given to the Vanderbilt ophthalmology department, January 17, 2014
1. Of the following, which is a difference between prokaryotic and e.pdf
1. Of the following, which is a difference between prokaryotic and eukaryotic cells?
A. Eukaryotic cells have a nuclear envelope surrounding the genetic material, whereas
prokaryotes have no nucleus.
B. Prokaryotic cells have multiple paired chromosomes, but eukaryotic cells have
only one.
C. Prokaryotic cells have linear DNA, whereas eukaryotes have circular DNA.
D. Eukaryotes have histone proteins, whereas prokaryotes do not.
E. Plasmids contain some genes for eukaryotes, whereas organelles contain some
genes for prokaryotes.
2. Which of the following are fundamental events that must take place in cell reproduction?
A. A cell\'s genetic information must be copied.
B. The copies of the genetic information must be separated from one another.
C. The cell must divide.
D. All of the above are fundamental events that must take place in cell
reproduction.
3. What are three essential structural elements of a functional eukaryotic chromosome?
A. Histone proteins, spindle fibers, and a pair of telomeres
B. Origins of replication, chromatids, and kinetochores
C. A centromere, a pair of telomeres, and origins of replication
D. A centromere, chromatids, and kinetochores
E. Spindle fibers, kinetochores, and chromatids
4. The chromosomal structures that serve as the attachment point for spindle fibers are the
_____________________.
A. telomeres
B. origins of replication
C. centromeres
D. kinetochore proteins
E. histone proteins
5. The natural ends of eukaryotic chromosomes that serve to stabilize the ends of the
chromosomes are the ______________________.
A. telomeres
B. origins of replication
C. centromeres
D. kinetochore proteins
E. histone proteins
6. _______________ is the part of interphase within the cell cycle where the cell growth and
protein synthesis necessary for cell division and DNA replication take place.
A. G1
B. S
C. G2
D. M
E. Telophase
7. __________________ is the part of interphase within the cell cycle where additional
biochemical reactions take place preparing the cell for mitosis.
A. G1
B. S
C. G2
D. M
E. Telophase
8. __________________ is the part of interphase within the cell cycle where DNA replication
takes place.
A. G1
B. S
C. G2
D. M
E. Telophase
9. What are checkpoints?
A. Key transition points in the cell cycle that lead to cancer
B. Periods during the cell cycle that ensure all cellular components are in good
working order
C. Transition points during the process of replication that allow DNA repair
D. The last process that occurs before a cell divides
E. Critical points that ensure correct and complete cytokinesis
10. What is the stage in mitosis when sister chromatids separate at the centromere and move to
opposite poles of the cell.
A. Prophase
B. Prometaphase
C. Metaphase
D. Anaphase
E. Telophase
11. What is the stage of mitosis where the chromosomes are arranged on the equatorial plane of
the cell.
A. prophase
B. prometaphase
C. metaphase
D. anaphase
E. telophase
12. What is the stage of mitosis where the chromosomes ini.
Chapter2...22. (Problem 3) Which of the following are.pdf
Chapter2...>2>
2. (Problem 3) Which of the following are fundamental events that must take place in cell
reproduction?
A. A cell\'s genetic information must be copied.
B. The copies of the genetic information must be separated from one another.
C. The cell must divide.
D. All of the above are fundamental events that must take place in cell reproduction.
3. (Problem 4 part 1): What are three essential structural elements of a functional
eukaryotic chromosome?
A. Histone proteins, spindle fibers, and a pair of telomeres
B. Origins of replication, chromatids, and kinetochores
C. A centromere, a pair of telomeres, and origins of replication
D. A centromere, chromatids, and kinetochores
E. Spindle fibers, kinetochores, and chromatids
4. (Problem 4 part 2 ) The chromosomal structures that serve as the attachment point for
spindle fibers are the _____________________.
A. telomeres
B. origins of replication
C. centromeres
D. kinetochore proteins
E. histone proteins
5. (Problem 4 part 3) The natural ends of eukaryotic chromosomes that serve to stabilize
the ends of the chromosomes are the ______________________.
A. telomeres
B. origins of replication
C. centromeres
D. kinetochore proteins
E. histone proteins
6. (Problem 6, part 1) _______________ is the part of interphase within the cell cycle
where the cell growth and protein synthesis necessary for cell division and DNA replication take
place.
A. G1
B. S
C. G2
D. M
E. Telophase
15. (Problem 10) Why are the two cells produced by the cell cycle genetically identical?
A. An exact copy of each DNA molecule was created in S phase giving rise to the two
identical sister chromatids.
B. Chromosomal replication in prophase ensures two identical copies of each
chromosome for the two cells.
C. Mitosis ensures that each cell receives one copy of the two identical sister
chromatids created during S phase.
D. Both A and B are reasons that the two cells produced by the cell cycle are identical.
E. Both A and C are reasons that the two cells produced by the cell cycle are identical.
16. (Problem 11, part 1) For the stages of meiosis I below, which is incorrectly matched to a
major event that takes place during that stage?
A. Prophase I—synapsis
B. Metaphase I—homologous pairs of chromosomes line up on the equatorial plane of
the metaphase plate
C. Anaphase I—separation of sister chromatids
D. Telophase I—separated chromosomes reach spindle poles at opposite ends of the
cell
E. Prophase I—genetic recombination through crossing over
17. (Problem 11, part 2) For the stages of meiosis II below, which is incorrectly matched to
the major event that takes place during that stage?
A. Prophase II—spindle fibers form
B. Metaphase II—homologous pairs of chromosomes line up along the equatorial
plane of the metaphase II plate
C. Anaphase II—sister chromatids separate
D. Telophase II—spindle fibers breakdown
E. Prophase II—nuclear envelope breaks down
18. (Problem 12) What are the major results of meiosis?
A. Production of hapl.
Recommended
Question 1 Random changes in gene frequency that occur in small .docx
Question 1
Random changes in gene frequency that occur in small populations describes the concept of
A.
convergent evolution.
B.
adaptive radiation.
C.
sympatric speciation.
D.
genetic drift.
E.
parapatric speciation.
5 points
Question 2
On a molecular level, all organisms
A.
have the same bases in DNA and RNA, the same amino acids in proteins, but use different genetic codes.
B.
have the same bases in DNA and RNA, but different amino acids in proteins.
C.
have the same bases in DNA and RNA, the same amino acids in proteins, and use the same genetic code.
D.
show no similarities.
E.
use different means of storing genetic information.
5 points
Question 3
Which of the following is the best description of an adaptation?
A.
a trait that will be beneficial in the future
B.
a trait that was beneficial to survival in the past
C.
an environmental change that selects for certain traits
D.
a trait that is of no advantage to survival
E.
a trait that is currently beneficial to survival
5 points
Question 4
Which of the following statements about natural selection is true?
A.
All individuals within a population have an equal likelihood of survival. Survival is based on chance.
B.
Natural selection results in those individuals within a population who are best-adapted surviving and producing more offspring.
C.
Natural selection leads to extinction.
D.
Natural selection causes variations to arise within a population.
E.
Natural selection leads to increased likelihood of survival for certain individuals based on variation. The variation must come from outside the population.
5 points
Question 5
Embryonic development includes all of the following except
A.
cleavage.
B.
growth.
C.
organogenesis.
D.
gastrulation.
E.
fertilization.
5 points
Question 6
Vestigial organs
A.
include the human appendix and wisdom teeth and a snake’s hipbones.
B.
did not have a function when they first developed, but do so today.
C.
are not found in present-day organisms.
D.
are similar in many organisms.
E.
must have evolved individually in different organisms.
5 points
Question 7
Most mutations lead to a decreased chance of survival for the mutant organism. Rarely, a mutation can provide an increased chance of survival. These rare mutations provide a(n) _____________________ for the mutant.
A.
directional shift
B.
selective disadvantage
C.
adaptive
D.
selective advantage
E.
environmental susceptibility
5 points
Question 8
Which of the following is not true of the punctuated equilibrium theory?
A.
It suggests that most species undergo little phenotypic change during most of their evolutionary duration.
B.
It proposes that most evolutionary change is concentrated in relatively brief episodes of change.
C.
It was inspired in part by patterns observed in the fossil record.
D.
It appears to be well supported by the fossil record.
E.
It suggests that subgroups diverge and evolve at about equal and constant rates.
5 poin ...
In-class introduction to basic Punnett square set-up and problem s.docx
In-class introduction to basic Punnett square set-up and problem solving, Part 1
Problem-solving tips:
· A Punnett square allows you to predict the possible genetic outcome of children based on the genetic make-up of the parents.
· First, read the problem and figure out whether the trait of interest or genetic disorder is found on the dominant allele or the recessive allele because that will have an impact on how you interpret the results of the Punnett square.
· Select a letter to represent the trait or disorder and define the dominant and recessive alleles. For example: For eye color, B (dominant) = brown eyes and b (recessive) = blue eyes. For achondroplasia (dwarfism), A (dominant) = achondroplasia and a (recessive) = normal allele.
· If it is a sex-linked question, remember to include the sexual genotypes of the parents (XX for mom and XY for dad).
· Write down all possible genotypes & phenotypes and use this information to help you set up the Punnett square.
1. Practice question on a human trait. In reality, eye color is controlled by multiple genes and is a complex trait. For simplicity, we’ll assume that brown eyes are dominant to blue eyes. Answer the questions below.
a) Select a letter for this trait and define the dominant and recessive alleles.
B (dominant) =
b (recessive) =
b) Write down all possible genotypes and phenotypes for individuals in the population
Possible genotypes
(the 2 alleles an individual has)
Possible phenotypes (the physical appearance of a trait)
Homozygous dominant individuals
Homozygous recessive individuals
Heterozygous individuals
c) Set up the Punnett square and solve this problem. Kristy is heterozygous and Mark has blue eyes. What percentage of their offspring will have blue eyes?
Kristy's genotype
Mark's genotype
a) Select a letter for this genetic condition and define the dominant and recessive alleles.
F (dominant) =
f (recessive) =
b) Write down all possible genotypes and phenotypes for individuals in the population
Possible genotypes
(the 2 alleles an individual has)
Possible phenotypes (the physical appearance of a trait)
Homozygous dominant individuals
Homozygous recessive individuals
Heterozygous individuals
c) Set up the Punnett square and solve this problem. Kristy and Mark are carriers for cystic fibrosis. The term carrier is only used when a condition is on the recessive allele. Carriers are heterozygous individuals who are normal and show no symptoms of the disorder, but they have the ability to pass on the mutated recessive allele to their offspring. What percentage of their children will be normal? What percentage of their children will be carriers?
Kristy's genotype
Mark's genotype
2. Practice question on a genetic condition. Cystic fibrosis (CF) is an autosomal, recessive condition that results in mucus buildup in the lungs and digestive system organs. As a result, CF .
Cbse class 10 science chapter 9 heredity and evolution2015 11-09
Cbse class 10 science chapter 9 heredity and evolution
1. The case of albinism among the Hopi Indians of Black M.pdf
1. The case of albinism among the Hopi Indians of Black Mesa is different from other
cases of albinism because of all of the following EXCEPT:
A. the incidence is much higher among the Hopis.
B. the albinism seems to have provided a reproductive advantage.
C. the albinism seems to have conferred social advantages.
D. the albinism did not involve vision defects.
E. Actually, all of the above are accurate.
2. Genetics has NOT played a key role in which of the following areas?
A. agriculture
B. medicine
C. pharmaceutical production
D. urban design and layout
E. Actually, it has played a key role in all of these areas.
3. Genetic variation in populations:
A. demonstrates descent from a common ancestor.
B. provides the basis of adaptation.
C. explains the use of similar genetic systems in all organisms.
D. is a consequence of origin of species.
E. All of the above.
4. Genetics as a discipline can be subdivided into all of the following EXCEPT:
A. transmission genetics.
B. acquisition genetics.
C. molecular genetics.
D. population genetics.
E. Actually, all of these are found in the discipline.
5. While modern genetics starts with the experiments of Gregor Mendel in 1866, it wasn\'t
until the start of the 20th century that those studies became well known, in several areas, NOT
including:
A. Sutton, who proposed that genes were on chromosomes.
B. Morgan, who identified fruit fly mutations.
C. Haldane and others who used populations to connect genetics to evolution.
D. Watson and Crick\'s determination of the structure of DNA.
E. Actually, Mendel\'s work was critical in each of these areas.
6. Prokaryotes and eukaryotes differ in that:
A. prokaryotes have a nucleus.
B. they decode DNA to different amino acids.
C. prokaryotes tend to be larger.
D. prokaryotes tend to be less complex.
E. None of the above.
7. Genes:
A. are carried on chromosomes.
B. come in pairs in eukaryotic cells.
C. come in alternative forms, known as alleles.
D. are encoded by DNA.
E. All of the above.
8. Phenotypes are:
A. measurable traits of an organism.
B. not inherited directly.
C. not changeable during the life of an organism.
D. A and B.
E. All of the above.
9. Evolution:
A. is genetic change.
B. starts with variation.
C. does not require change of DNA.
D. A and B.
E. All of the above.
10. The flow of information is as follows:
A. protein to DNA to RNA.
B. protein to RNA to DNA.
C. DNA to RNA to protein.
D. RNA to protein to DNA.
E. None of the above.
11. Mutations are changes of DNA:
A. that can affect a single gene.
B. that can affect large regions of chromosomes.
C. that can be passed on to daughter cells.
D. that can change amino acid sequences.
E. All of the above.
Solution
1). C. the albinism seems to have conferred social advantages.
2). D. urban design and layout
3). D. is a consequence of origin of species.
4). B. acquisition genetics.
5). C. Haldane and others who used populations to connect genetics to evolution.
6). E. None of the above.
7). A. are carried on ch.
Teaching with Clickers for Deep Learning
Presentation given to the Vanderbilt ophthalmology department, January 17, 2014
1. Of the following, which is a difference between prokaryotic and e.pdf
1. Of the following, which is a difference between prokaryotic and eukaryotic cells?
A. Eukaryotic cells have a nuclear envelope surrounding the genetic material, whereas
prokaryotes have no nucleus.
B. Prokaryotic cells have multiple paired chromosomes, but eukaryotic cells have
only one.
C. Prokaryotic cells have linear DNA, whereas eukaryotes have circular DNA.
D. Eukaryotes have histone proteins, whereas prokaryotes do not.
E. Plasmids contain some genes for eukaryotes, whereas organelles contain some
genes for prokaryotes.
2. Which of the following are fundamental events that must take place in cell reproduction?
A. A cell\'s genetic information must be copied.
B. The copies of the genetic information must be separated from one another.
C. The cell must divide.
D. All of the above are fundamental events that must take place in cell
reproduction.
3. What are three essential structural elements of a functional eukaryotic chromosome?
A. Histone proteins, spindle fibers, and a pair of telomeres
B. Origins of replication, chromatids, and kinetochores
C. A centromere, a pair of telomeres, and origins of replication
D. A centromere, chromatids, and kinetochores
E. Spindle fibers, kinetochores, and chromatids
4. The chromosomal structures that serve as the attachment point for spindle fibers are the
_____________________.
A. telomeres
B. origins of replication
C. centromeres
D. kinetochore proteins
E. histone proteins
5. The natural ends of eukaryotic chromosomes that serve to stabilize the ends of the
chromosomes are the ______________________.
A. telomeres
B. origins of replication
C. centromeres
D. kinetochore proteins
E. histone proteins
6. _______________ is the part of interphase within the cell cycle where the cell growth and
protein synthesis necessary for cell division and DNA replication take place.
A. G1
B. S
C. G2
D. M
E. Telophase
7. __________________ is the part of interphase within the cell cycle where additional
biochemical reactions take place preparing the cell for mitosis.
A. G1
B. S
C. G2
D. M
E. Telophase
8. __________________ is the part of interphase within the cell cycle where DNA replication
takes place.
A. G1
B. S
C. G2
D. M
E. Telophase
9. What are checkpoints?
A. Key transition points in the cell cycle that lead to cancer
B. Periods during the cell cycle that ensure all cellular components are in good
working order
C. Transition points during the process of replication that allow DNA repair
D. The last process that occurs before a cell divides
E. Critical points that ensure correct and complete cytokinesis
10. What is the stage in mitosis when sister chromatids separate at the centromere and move to
opposite poles of the cell.
A. Prophase
B. Prometaphase
C. Metaphase
D. Anaphase
E. Telophase
11. What is the stage of mitosis where the chromosomes are arranged on the equatorial plane of
the cell.
A. prophase
B. prometaphase
C. metaphase
D. anaphase
E. telophase
12. What is the stage of mitosis where the chromosomes ini.
Chapter2...22. (Problem 3) Which of the following are.pdf
Chapter2...>2>
2. (Problem 3) Which of the following are fundamental events that must take place in cell
reproduction?
A. A cell\'s genetic information must be copied.
B. The copies of the genetic information must be separated from one another.
C. The cell must divide.
D. All of the above are fundamental events that must take place in cell reproduction.
3. (Problem 4 part 1): What are three essential structural elements of a functional
eukaryotic chromosome?
A. Histone proteins, spindle fibers, and a pair of telomeres
B. Origins of replication, chromatids, and kinetochores
C. A centromere, a pair of telomeres, and origins of replication
D. A centromere, chromatids, and kinetochores
E. Spindle fibers, kinetochores, and chromatids
4. (Problem 4 part 2 ) The chromosomal structures that serve as the attachment point for
spindle fibers are the _____________________.
A. telomeres
B. origins of replication
C. centromeres
D. kinetochore proteins
E. histone proteins
5. (Problem 4 part 3) The natural ends of eukaryotic chromosomes that serve to stabilize
the ends of the chromosomes are the ______________________.
A. telomeres
B. origins of replication
C. centromeres
D. kinetochore proteins
E. histone proteins
6. (Problem 6, part 1) _______________ is the part of interphase within the cell cycle
where the cell growth and protein synthesis necessary for cell division and DNA replication take
place.
A. G1
B. S
C. G2
D. M
E. Telophase
15. (Problem 10) Why are the two cells produced by the cell cycle genetically identical?
A. An exact copy of each DNA molecule was created in S phase giving rise to the two
identical sister chromatids.
B. Chromosomal replication in prophase ensures two identical copies of each
chromosome for the two cells.
C. Mitosis ensures that each cell receives one copy of the two identical sister
chromatids created during S phase.
D. Both A and B are reasons that the two cells produced by the cell cycle are identical.
E. Both A and C are reasons that the two cells produced by the cell cycle are identical.
16. (Problem 11, part 1) For the stages of meiosis I below, which is incorrectly matched to a
major event that takes place during that stage?
A. Prophase I—synapsis
B. Metaphase I—homologous pairs of chromosomes line up on the equatorial plane of
the metaphase plate
C. Anaphase I—separation of sister chromatids
D. Telophase I—separated chromosomes reach spindle poles at opposite ends of the
cell
E. Prophase I—genetic recombination through crossing over
17. (Problem 11, part 2) For the stages of meiosis II below, which is incorrectly matched to
the major event that takes place during that stage?
A. Prophase II—spindle fibers form
B. Metaphase II—homologous pairs of chromosomes line up along the equatorial
plane of the metaphase II plate
C. Anaphase II—sister chromatids separate
D. Telophase II—spindle fibers breakdown
E. Prophase II—nuclear envelope breaks down
18. (Problem 12) What are the major results of meiosis?
A. Production of hapl.
1. One condition that seems to lead to senescence and not apoptosi.docx
1. One condition that seems to lead to senescence and not apoptosis in stem cells is
a. shortening of telomeres below a critical limit
b. base damage to DNA
c. unequal division of chromosomes at anaphase
d. none of the above
2. Knowing that even in the oldest people's stem cells seem to have many cell divisions left to them when cultured in vitro (in a dish), the you might contend that a possible reason for that was
a. Absence of inhibitory factors found in the old bodies in the new culture medium
b. factors in the cellular environment of old people's bodies
c.
The presence of high levels of growth factors in the culture serum
d. All of the above
3. In 1962, Leonard Hayflick conducted an experiment that supported Weismann's Theory of programmed death. He proposed The Hayflick Limit which can be reached faster by:
a. Increasing the rate of cell division
b. Underfeeding cells
c. Decreasing the rate of mutations
d. Synthetically elongating telomeres.
4. The antagonistic pleiotropy theory of aging assumes that each gene has only one effect on the phenotype of an organism at different stages of its life. True or False
5. In Conboy's study on heterochronic parabiotic pairs, which phrase described what future studies should focus on according to the results of this study?
a. adult stem cells
b. changes in aging cells
c. muscle cell regeneration
d. changes in aging plasma
6.- 7. Briefly explain what happened to old stem and progenitor cells in a young systemic environment.
8. Which substance is the most common in cells?
a. carbohydrates
b. salts and minerals
c. proteins
d. fats
e. water
9.
- 11. Describe how an enzyme may change the transcription patterns of a cell by adding chemical groups to histone proteins. What is this called?
12. Acetyl L-Carnitine/Alpha lipoic acid are ____________ that are produced naturally by the body in _______
a. Amino acids, large amounts
b. Antioxidants, small amounts
c. Toxins, abundance
d. ROS, trace amounts
13. Most genes that determine lifespan in C. elegans are directly in DNA repair. True or False
14.- 16. What is the preferable study group in mammalian testing of aging interventions Mice or monkeys? Why?
17 - 19. If you were to run a retirement community what would you do to improve the health/lives of the people who depend on you?
20. The ____________ ____________ theory , states that there exists a biological clock and a time dependent plan or program that directs a limitation to a lifespan?
a. DNA damage
b. Evolveabity Theory
c. Mutation-accumulation
d. free radical
21. - 24. Unlike inorganic objects like rocks or even computers (at present) , living organisms can repair damage or rebuild or even recycle damages at the sub-cellular level. Give three
instances cellular systems involved in repairing cellular damages.
25. . An often given example ...
Chapter 2 problems.1. (Problem 1) Of the following, whi.pdf
Chapter 2 problems.
1. (Problem 1) Of the following, which is a difference between prokaryotic and eukaryotic
cells?
A. Eukaryotic cells have a nuclear envelope surrounding the genetic material, whereas
prokaryotes have no nucleus.
B. Prokaryotic cells have multiple paired chromosomes, but eukaryotic cells have only
one.
C. Prokaryotic cells have linear DNA, whereas eukaryotes have circular DNA.
D. Eukaryotes have histone proteins, whereas prokaryotes do not.
E. Plasmids contain some genes for eukaryotes, whereas organelles contain some
genes for prokaryotes.
2. (Problem 3) Which of the following are fundamental events that must take place in cell
reproduction?
A. A cell\'s genetic information must be copied.
B. The copies of the genetic information must be separated from one another.
C. The cell must divide.
D. All of the above are fundamental events that must take place in cell reproduction.
3. (Problem 4 part 1): What are three essential structural elements of a functional
eukaryotic chromosome?
A. Histone proteins, spindle fibers, and a pair of telomeres
B. Origins of replication, chromatids, and kinetochores
C. A centromere, a pair of telomeres, and origins of replication
D. A centromere, chromatids, and kinetochores
E. Spindle fibers, kinetochores, and chromatids
4. (Problem 4 part 2 ) The chromosomal structures that serve as the attachment point for
spindle fibers are the _____________________.
A. telomeres
B. origins of replication
C. centromeres
D. kinetochore proteins
E. histone proteins
5. (Problem 4 part 3) The natural ends of eukaryotic chromosomes that serve to stabilize
the ends of the chromosomes are the ______________________.
A. telomeres
B. origins of replication
C. centromeres
D. kinetochore proteins
E. histone proteins
6. (Problem 6, part 1) _______________ is the part of interphase within the cell cycle
where the cell growth and protein synthesis necessary for cell division and DNA replication take
place.
A. G1
B. S
C. G2
D. M
E. Telophase
15. (Problem 10) Why are the two cells produced by the cell cycle genetically identical?
A. An exact copy of each DNA molecule was created in S phase giving rise to the two
identical sister chromatids.
B. Chromosomal replication in prophase ensures two identical copies of each
chromosome for the two cells.
C. Mitosis ensures that each cell receives one copy of the two identical sister
chromatids created during S phase.
D. Both A and B are reasons that the two cells produced by the cell cycle are identical.
E. Both A and C are reasons that the two cells produced by the cell cycle are identical.
16. (Problem 11, part 1) For the stages of meiosis I below, which is incorrectly matched to a
major event that takes place during that stage?
A. Prophase I—synapsis
B. Metaphase I—homologous pairs of chromosomes line up on the equatorial plane of
the metaphase plate
C. Anaphase I—separation of sister chromatids
D. Telophase I—separated chromosomes reach spindle poles at opposite ends of the
cell
.
Biology 204 Principles of Biology I Assignment 2C
Biology 204 Principles of Biology I Assignment 2C
For students with first names starting with the letters O to Z.
This assignment is graded out of 110 points, and is worth 10% of your final mark. Please submit this assignment after you have completed Chapter 16 and before you write the final exam
Question 1 This type of selection favors both extremes of the .docx
Question 1
This type of selection favors both extremes of the phenotype spectrum within a population.
A.
disruptive
B.
natural
C.
neutral
D.
stabilizing
E.
directional
5 points
Question 2
On a molecular level, all organisms
A.
have the same bases in DNA and RNA, the same amino acids in proteins, but use different genetic codes.
B.
have the same bases in DNA and RNA, the same amino acids in proteins, and use the same genetic code.
C.
use different means of storing genetic information.
D.
have the same bases in DNA and RNA, but different amino acids in proteins.
E.
show no similarities.
5 points
Question 3
Most mutations lead to a decreased chance of survival for the mutant organism. Rarely, a mutation can provide an increased chance of survival. These rare mutations provide a(n) _____________________ for the mutant.
A.
selective disadvantage
B.
environmental susceptibility
C.
adaptive
D.
directional shift
E.
selective advantage
5 points
Question 4
After being produced in the seminiferous tubules, sperm mature in the
A.
seminal vesicle.
B.
vas deferens.
C.
epididymis.
D.
ejaculatory duct.
E.
urethra.
5 points
Question 5
Random changes in gene frequency that occur in small populations describes the concept of
A.
genetic drift.
B.
adaptive radiation.
C.
parapatric speciation.
D.
sympatric speciation.
E.
convergent evolution.
5 points
Question 6
The primary male hormone is
A.
produced throughout a male’s entire lifetime.
B.
produced in the brain.
C.
testosterone.
D.
estrogen.
E.
only involved in sperm production.
5 points
Question 7
Which of the following statements about natural selection is true?
A.
Natural selection leads to increased likelihood of survival for certain individuals based on variation. The variation must come from outside the population.
B.
Natural selection results in those individuals within a population who are best-adapted surviving and producing more offspring.
C.
Natural selection causes variations to arise within a population.
D.
All individuals within a population have an equal likelihood of survival. Survival is based on chance.
E.
Natural selection leads to extinction.
5 points
Question 8
Which of the following is not true of the punctuated equilibrium theory?
A.
It was inspired in part by patterns observed in the fossil record.
B.
It suggests that most species undergo little phenotypic change during most of their evolutionary duration.
C.
It suggests that subgroups diverge and evolve at about equal and constant rates.
D.
It appears to be well supported by the fossil record.
E.
It proposes that most evolutionary change is concentrated in relatively brief episodes of change.
5 points
Question 9
Animals that possess both male and female sexual organs
A.
are hermaphroditic and typically cross-fertilize.
B.
are dioecious and typically self-fertilize.
C.
are parthenogenetic and typically self-fertilize.
D.
are hermaphroditic and typically ...
Shape cocci bacill spirilii Gram positive t negative.pdf
Shape (cocci) (bacill) (spirilii) Gram positive (t) negative (-1) Stain (Violet) [pink) 0. 2 Shape (cocci)
(bacilli) (spirilli) Gram positive ( t) negative (1) stain (Vviolet) (pink] Q. 3 Eubacteria are/have the
following except: (A) Membrane-bound organelles B. Capable of methanogenesis C. Unicellular D.
Peptidoglycan in cell wall E. A and B Q. 4 Cyanobacteria (blue-green algae) are/have the following
except: A. chloroplasts B. capable of nitrogenation C. Unicellular Eukaryotes D. Capable of
photosynthesis E. Chlorophylls and accessory pigments Q. 5 Nostoc sp. and Anaboena sp. have
special cells calledheterocysts which function to fix nitrogen.Q. 6 Which kingdom do the protists
belonc? A. Animalia B. Plantac C. Fungi D. Bacteria E. None of the above Q.7. Poramedum
coudotum is an alveolate that is in the phplume it is a(n) treph. Q.8. Plosmodium ivax is a(n) it it a
sore forting paraute that causes. Q. 9 Areolates are/have: A. Photosynthetic and heterotrophic
species 8. Flattened cavities under plasms membrancs of the cells C. Some have cilla or fagells
D. Parasitic and symbiotic species E. All of the above Q. 10 The ceil walls of diatoms and
radiolarlans are made of while the foraminferans are made of Q. 11. Which is the definitive host for
Ptaunodilum sp, humans or mosevitas? Q. 12 fuglena ishas eacept: A. Autotroptic and/or
heterotrophic 8. Mitochondria C. Contractile vacuoles 9. Cilis E. Eyespot Q.13 Thyonosomes are:
A. Saprophytic 0. Photoautotrophic C. Paratitic D. Heterotrophicdentify the following: : 4 19.
EuglenaTrue or False Two 'normally diatibuted' popubtions wath ditterent meass can hwe the ame
sandard deviation. Mode is the most common measure of cental tendency. False varabality is a
measure of central tendeney. What is a better measure of variabley or dasenyon than the sample
or population varance? A) degrees of liredom (an) of confidence interval (C) (C) Zhmple randard
deviation (s) D) range The (eactis Lerevne) finches had a ngher ate of wnival durine the 18 monts
of dreveht in the Galapagos while the (sactus / ucaund) finches had a higher rate cf survikal during
Nigh predplation/hooding period due to tI wino..
Can you please help with the questions. 1. Which type of signalin.pdf
Can you please help with the questions.
1. Which type of signaling DOES NOT involve a soluble signal molecule? a) endocrine b)
paracrine c) neuronal d) contact dependent 2. Which statement conceming signal transduction is
true? a) Cells can only respond to one signal at a time. b) Each cell has exactly one type of
receptor, present in multiple copies. c) Signal transduction usually occurs in a single step. d) The
biological response to a given ligand may not be the same in each cell. e) None of these answers
are true. 3. Receptors involved in signaling can be found a) in the plasma membrane b) in the
nucleus c) in the cytosol or the plasma membrane d) in the cytosol, nucleus or plasma membrane
4. Which of the following applies to second messengers? a) usually activate kinases b) amplify
the signal from a membrane receptor c) can be cAMP or Ca++ d) are small molecules or ions
that are soluble in the cytosol e) All of these answers. 5. Which second messenger is NOT
soluble? a) calcium b) IP3 c) DAG d) CAMP 6. This type of cell surface receptor alters the
membrane potential. a) ion channel receptor b) G protein-coupled receptors c) enzyme-coupled
receptors d) none of these answers e) all of these answers.
DNA replicates conservatively, Semiconservatively, liberally, by hyb.pdf
DNA replicates conservatively, Semiconservatively, liberally, by hybridization by congression.
A bacterial artificial chromosome is a vector with centromeres, telameres, and an origin of
replication. with centromeres and an origin of replication derived from a plasmid with an origin
of replication. derived from a plasmid with a centromere, derived from a complete bacterial
chromosome. The central dogma of molecular biology is____________ rightarrow
___________ rightarrow ___________ Which of the following correctly represents the
distinction between the terms \"genotype\" and \"phenotype\"? Phenotype is the genetic
composition of an organism, while genotype is the physical appearance. There is no difference
between a genotype and a phenotype. A phenotype is dominant and a genotype is recessive.
The genotype is the genetic composition of an organism, whereas the phenotype is its physical
appearance.
Solution
5. b) semiconservatively
The DNA is doube stranded with two strans in antipolar nature. During replication the two
strands sepearate. A new strand is added to each one of them and thus we get two new double
stranded DNA strands with one old strand and one new strand. Hence it is called Semi-
conservative.
6. c) Derived from plasmid with an origin of replication.
The Bacterial artificial chromosome is derived from fully functional F plasmid. It is used to
clone large DNA inserts and express inside E. coli. It has its own origin of replication and genes
for homologous recombination. This Ori, helps it to replicate on its own inside the bacterial cell
and maintain itself. However unlike actual chromosome it does not contain centromers or
telomeres. Hence answer is option C.
7. DNA --> RNA --> Protein
It is the central dogma of Biology, in which the DNA acts as genetic material passes on its
information to RNA (mRNA-messenger RNA) by a process called as Transcription. The mRNA
is later Translated with the help of ribosomes into proteins. Athough there is another transition
exists in which RNA is converted back to DNA, which is called reverse transcription ( Not
mentioned in the question ).
8. d) The genotype is the genetic compostion of an organism, whereas the phenotype is its
physical appearance.
The genotype is the genetic content, what genes the organism possesses. These genes gets
transcribed and translated into proteins, which make up the appearance of the organism, in other
words Phenotype. Phenotype depends largely on the genotype and genotype is hereditory..
LOS At the time of her death on September 4, 2015, Alicia held the .pdf
LOS At the time of her death on September 4, 2015, Alicia held the following assets. Bonds of
Emerald Tool Corporation Stock in Drab Corporation Insurance policy (face amount of
$400,000) Fair Market Value $ 900,000 1,100,000 on the life of her father, Mitch Traditional
IRAs *Cash surrender value. 80,000* 300,000 Alicia was also the life tenant of a trust (fair
market value of $2 million) created by her late husband Bert. (The executor of Bert\'s estate had
made a QTIP election.) In October, Alicia\'s estate received an interest payment of $11,500
($6,000 accrued before September 4, 2015) paid by Emerald and a cash dividend of $9,000 from
Drab. The Drab dividend was declared on August 19 and was payable to date of re- cord
shareholders on September 3, 2015. Although Mitch survives Alicia, she is the designated
beneficiary of the policy. The IRAs are distributed to Alicia\'s children. What amount is included
in Alicia\'s gross estate?
Solution
Alicia’s Gross Estate:-
Bonds of Emrald Tool Corporation - 900,000
Interest on Emrald Tool Bonds – 11,500…………..(accrued interest is to be included )
Stock in Drab Corporation - 1,100,000
Cash Dividend – 9,000
Insurance Policy – 80,000 ……….( Cash surrender value)
Traditional IRA – 300,000
Value of Trust – 2,000,000
Thus, Total asset – 4,400,500.
Match the following terms with the definitions given belowTerms Gi.pdf
Match the following terms with the definitions given below
Terms Given
a. Heap
b. Memory leak
c. Inaccessible object
d. Dangling pointer
e. Destructor
f. Dynamic data
Definitions given
i. The loss of available space that occurs when dynamic data is not deallocation properly.
ii. The area of memory used for allocation and deallocation of dynamic data.
iii. An object that has been allocated but has no pointer pointing to it.
iv. Variables created with the new operation.
v. A pointer that points to a dellocated object.
vi. A member function invoked when an object goes out of scope.
Solution
Heap - The area of memory used for allocation and deallocation of dynamic data.
Memory leak - The loss of available space that occurs when dynamic data is not deallocation
properly.
Inaccessible object - An object that has been allocated but has no pointer pointing to it.
Dangling pointer - A pointer that points to a dellocated object.
Destructor - vi. A member function invoked when an object goes out of scope.
Dynamic data - Variables created with the new operation..
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Similar to E. None of the above.E. Actually, they differ in all of these ways.pdf
1. One condition that seems to lead to senescence and not apoptosi.docx
1. One condition that seems to lead to senescence and not apoptosis in stem cells is
a. shortening of telomeres below a critical limit
b. base damage to DNA
c. unequal division of chromosomes at anaphase
d. none of the above
2. Knowing that even in the oldest people's stem cells seem to have many cell divisions left to them when cultured in vitro (in a dish), the you might contend that a possible reason for that was
a. Absence of inhibitory factors found in the old bodies in the new culture medium
b. factors in the cellular environment of old people's bodies
c.
The presence of high levels of growth factors in the culture serum
d. All of the above
3. In 1962, Leonard Hayflick conducted an experiment that supported Weismann's Theory of programmed death. He proposed The Hayflick Limit which can be reached faster by:
a. Increasing the rate of cell division
b. Underfeeding cells
c. Decreasing the rate of mutations
d. Synthetically elongating telomeres.
4. The antagonistic pleiotropy theory of aging assumes that each gene has only one effect on the phenotype of an organism at different stages of its life. True or False
5. In Conboy's study on heterochronic parabiotic pairs, which phrase described what future studies should focus on according to the results of this study?
a. adult stem cells
b. changes in aging cells
c. muscle cell regeneration
d. changes in aging plasma
6.- 7. Briefly explain what happened to old stem and progenitor cells in a young systemic environment.
8. Which substance is the most common in cells?
a. carbohydrates
b. salts and minerals
c. proteins
d. fats
e. water
9.
- 11. Describe how an enzyme may change the transcription patterns of a cell by adding chemical groups to histone proteins. What is this called?
12. Acetyl L-Carnitine/Alpha lipoic acid are ____________ that are produced naturally by the body in _______
a. Amino acids, large amounts
b. Antioxidants, small amounts
c. Toxins, abundance
d. ROS, trace amounts
13. Most genes that determine lifespan in C. elegans are directly in DNA repair. True or False
14.- 16. What is the preferable study group in mammalian testing of aging interventions Mice or monkeys? Why?
17 - 19. If you were to run a retirement community what would you do to improve the health/lives of the people who depend on you?
20. The ____________ ____________ theory , states that there exists a biological clock and a time dependent plan or program that directs a limitation to a lifespan?
a. DNA damage
b. Evolveabity Theory
c. Mutation-accumulation
d. free radical
21. - 24. Unlike inorganic objects like rocks or even computers (at present) , living organisms can repair damage or rebuild or even recycle damages at the sub-cellular level. Give three
instances cellular systems involved in repairing cellular damages.
25. . An often given example ...
Chapter 2 problems.1. (Problem 1) Of the following, whi.pdf
Chapter 2 problems.
1. (Problem 1) Of the following, which is a difference between prokaryotic and eukaryotic
cells?
A. Eukaryotic cells have a nuclear envelope surrounding the genetic material, whereas
prokaryotes have no nucleus.
B. Prokaryotic cells have multiple paired chromosomes, but eukaryotic cells have only
one.
C. Prokaryotic cells have linear DNA, whereas eukaryotes have circular DNA.
D. Eukaryotes have histone proteins, whereas prokaryotes do not.
E. Plasmids contain some genes for eukaryotes, whereas organelles contain some
genes for prokaryotes.
2. (Problem 3) Which of the following are fundamental events that must take place in cell
reproduction?
A. A cell\'s genetic information must be copied.
B. The copies of the genetic information must be separated from one another.
C. The cell must divide.
D. All of the above are fundamental events that must take place in cell reproduction.
3. (Problem 4 part 1): What are three essential structural elements of a functional
eukaryotic chromosome?
A. Histone proteins, spindle fibers, and a pair of telomeres
B. Origins of replication, chromatids, and kinetochores
C. A centromere, a pair of telomeres, and origins of replication
D. A centromere, chromatids, and kinetochores
E. Spindle fibers, kinetochores, and chromatids
4. (Problem 4 part 2 ) The chromosomal structures that serve as the attachment point for
spindle fibers are the _____________________.
A. telomeres
B. origins of replication
C. centromeres
D. kinetochore proteins
E. histone proteins
5. (Problem 4 part 3) The natural ends of eukaryotic chromosomes that serve to stabilize
the ends of the chromosomes are the ______________________.
A. telomeres
B. origins of replication
C. centromeres
D. kinetochore proteins
E. histone proteins
6. (Problem 6, part 1) _______________ is the part of interphase within the cell cycle
where the cell growth and protein synthesis necessary for cell division and DNA replication take
place.
A. G1
B. S
C. G2
D. M
E. Telophase
15. (Problem 10) Why are the two cells produced by the cell cycle genetically identical?
A. An exact copy of each DNA molecule was created in S phase giving rise to the two
identical sister chromatids.
B. Chromosomal replication in prophase ensures two identical copies of each
chromosome for the two cells.
C. Mitosis ensures that each cell receives one copy of the two identical sister
chromatids created during S phase.
D. Both A and B are reasons that the two cells produced by the cell cycle are identical.
E. Both A and C are reasons that the two cells produced by the cell cycle are identical.
16. (Problem 11, part 1) For the stages of meiosis I below, which is incorrectly matched to a
major event that takes place during that stage?
A. Prophase I—synapsis
B. Metaphase I—homologous pairs of chromosomes line up on the equatorial plane of
the metaphase plate
C. Anaphase I—separation of sister chromatids
D. Telophase I—separated chromosomes reach spindle poles at opposite ends of the
cell
.
Biology 204 Principles of Biology I Assignment 2C
Biology 204 Principles of Biology I Assignment 2C
For students with first names starting with the letters O to Z.
This assignment is graded out of 110 points, and is worth 10% of your final mark. Please submit this assignment after you have completed Chapter 16 and before you write the final exam
Question 1 This type of selection favors both extremes of the .docx
Question 1
This type of selection favors both extremes of the phenotype spectrum within a population.
A.
disruptive
B.
natural
C.
neutral
D.
stabilizing
E.
directional
5 points
Question 2
On a molecular level, all organisms
A.
have the same bases in DNA and RNA, the same amino acids in proteins, but use different genetic codes.
B.
have the same bases in DNA and RNA, the same amino acids in proteins, and use the same genetic code.
C.
use different means of storing genetic information.
D.
have the same bases in DNA and RNA, but different amino acids in proteins.
E.
show no similarities.
5 points
Question 3
Most mutations lead to a decreased chance of survival for the mutant organism. Rarely, a mutation can provide an increased chance of survival. These rare mutations provide a(n) _____________________ for the mutant.
A.
selective disadvantage
B.
environmental susceptibility
C.
adaptive
D.
directional shift
E.
selective advantage
5 points
Question 4
After being produced in the seminiferous tubules, sperm mature in the
A.
seminal vesicle.
B.
vas deferens.
C.
epididymis.
D.
ejaculatory duct.
E.
urethra.
5 points
Question 5
Random changes in gene frequency that occur in small populations describes the concept of
A.
genetic drift.
B.
adaptive radiation.
C.
parapatric speciation.
D.
sympatric speciation.
E.
convergent evolution.
5 points
Question 6
The primary male hormone is
A.
produced throughout a male’s entire lifetime.
B.
produced in the brain.
C.
testosterone.
D.
estrogen.
E.
only involved in sperm production.
5 points
Question 7
Which of the following statements about natural selection is true?
A.
Natural selection leads to increased likelihood of survival for certain individuals based on variation. The variation must come from outside the population.
B.
Natural selection results in those individuals within a population who are best-adapted surviving and producing more offspring.
C.
Natural selection causes variations to arise within a population.
D.
All individuals within a population have an equal likelihood of survival. Survival is based on chance.
E.
Natural selection leads to extinction.
5 points
Question 8
Which of the following is not true of the punctuated equilibrium theory?
A.
It was inspired in part by patterns observed in the fossil record.
B.
It suggests that most species undergo little phenotypic change during most of their evolutionary duration.
C.
It suggests that subgroups diverge and evolve at about equal and constant rates.
D.
It appears to be well supported by the fossil record.
E.
It proposes that most evolutionary change is concentrated in relatively brief episodes of change.
5 points
Question 9
Animals that possess both male and female sexual organs
A.
are hermaphroditic and typically cross-fertilize.
B.
are dioecious and typically self-fertilize.
C.
are parthenogenetic and typically self-fertilize.
D.
are hermaphroditic and typically ...
Shape cocci bacill spirilii Gram positive t negative.pdf
Shape (cocci) (bacill) (spirilii) Gram positive (t) negative (-1) Stain (Violet) [pink) 0. 2 Shape (cocci)
(bacilli) (spirilli) Gram positive ( t) negative (1) stain (Vviolet) (pink] Q. 3 Eubacteria are/have the
following except: (A) Membrane-bound organelles B. Capable of methanogenesis C. Unicellular D.
Peptidoglycan in cell wall E. A and B Q. 4 Cyanobacteria (blue-green algae) are/have the following
except: A. chloroplasts B. capable of nitrogenation C. Unicellular Eukaryotes D. Capable of
photosynthesis E. Chlorophylls and accessory pigments Q. 5 Nostoc sp. and Anaboena sp. have
special cells calledheterocysts which function to fix nitrogen.Q. 6 Which kingdom do the protists
belonc? A. Animalia B. Plantac C. Fungi D. Bacteria E. None of the above Q.7. Poramedum
coudotum is an alveolate that is in the phplume it is a(n) treph. Q.8. Plosmodium ivax is a(n) it it a
sore forting paraute that causes. Q. 9 Areolates are/have: A. Photosynthetic and heterotrophic
species 8. Flattened cavities under plasms membrancs of the cells C. Some have cilla or fagells
D. Parasitic and symbiotic species E. All of the above Q. 10 The ceil walls of diatoms and
radiolarlans are made of while the foraminferans are made of Q. 11. Which is the definitive host for
Ptaunodilum sp, humans or mosevitas? Q. 12 fuglena ishas eacept: A. Autotroptic and/or
heterotrophic 8. Mitochondria C. Contractile vacuoles 9. Cilis E. Eyespot Q.13 Thyonosomes are:
A. Saprophytic 0. Photoautotrophic C. Paratitic D. Heterotrophicdentify the following: : 4 19.
EuglenaTrue or False Two 'normally diatibuted' popubtions wath ditterent meass can hwe the ame
sandard deviation. Mode is the most common measure of cental tendency. False varabality is a
measure of central tendeney. What is a better measure of variabley or dasenyon than the sample
or population varance? A) degrees of liredom (an) of confidence interval (C) (C) Zhmple randard
deviation (s) D) range The (eactis Lerevne) finches had a ngher ate of wnival durine the 18 monts
of dreveht in the Galapagos while the (sactus / ucaund) finches had a higher rate cf survikal during
Nigh predplation/hooding period due to tI wino..
Can you please help with the questions. 1. Which type of signalin.pdf
Can you please help with the questions.
1. Which type of signaling DOES NOT involve a soluble signal molecule? a) endocrine b)
paracrine c) neuronal d) contact dependent 2. Which statement conceming signal transduction is
true? a) Cells can only respond to one signal at a time. b) Each cell has exactly one type of
receptor, present in multiple copies. c) Signal transduction usually occurs in a single step. d) The
biological response to a given ligand may not be the same in each cell. e) None of these answers
are true. 3. Receptors involved in signaling can be found a) in the plasma membrane b) in the
nucleus c) in the cytosol or the plasma membrane d) in the cytosol, nucleus or plasma membrane
4. Which of the following applies to second messengers? a) usually activate kinases b) amplify
the signal from a membrane receptor c) can be cAMP or Ca++ d) are small molecules or ions
that are soluble in the cytosol e) All of these answers. 5. Which second messenger is NOT
soluble? a) calcium b) IP3 c) DAG d) CAMP 6. This type of cell surface receptor alters the
membrane potential. a) ion channel receptor b) G protein-coupled receptors c) enzyme-coupled
receptors d) none of these answers e) all of these answers.
DNA replicates conservatively, Semiconservatively, liberally, by hyb.pdf
DNA replicates conservatively, Semiconservatively, liberally, by hybridization by congression.
A bacterial artificial chromosome is a vector with centromeres, telameres, and an origin of
replication. with centromeres and an origin of replication derived from a plasmid with an origin
of replication. derived from a plasmid with a centromere, derived from a complete bacterial
chromosome. The central dogma of molecular biology is____________ rightarrow
___________ rightarrow ___________ Which of the following correctly represents the
distinction between the terms \"genotype\" and \"phenotype\"? Phenotype is the genetic
composition of an organism, while genotype is the physical appearance. There is no difference
between a genotype and a phenotype. A phenotype is dominant and a genotype is recessive.
The genotype is the genetic composition of an organism, whereas the phenotype is its physical
appearance.
Solution
5. b) semiconservatively
The DNA is doube stranded with two strans in antipolar nature. During replication the two
strands sepearate. A new strand is added to each one of them and thus we get two new double
stranded DNA strands with one old strand and one new strand. Hence it is called Semi-
conservative.
6. c) Derived from plasmid with an origin of replication.
The Bacterial artificial chromosome is derived from fully functional F plasmid. It is used to
clone large DNA inserts and express inside E. coli. It has its own origin of replication and genes
for homologous recombination. This Ori, helps it to replicate on its own inside the bacterial cell
and maintain itself. However unlike actual chromosome it does not contain centromers or
telomeres. Hence answer is option C.
7. DNA --> RNA --> Protein
It is the central dogma of Biology, in which the DNA acts as genetic material passes on its
information to RNA (mRNA-messenger RNA) by a process called as Transcription. The mRNA
is later Translated with the help of ribosomes into proteins. Athough there is another transition
exists in which RNA is converted back to DNA, which is called reverse transcription ( Not
mentioned in the question ).
8. d) The genotype is the genetic compostion of an organism, whereas the phenotype is its
physical appearance.
The genotype is the genetic content, what genes the organism possesses. These genes gets
transcribed and translated into proteins, which make up the appearance of the organism, in other
words Phenotype. Phenotype depends largely on the genotype and genotype is hereditory..
Similar to E. None of the above.E. Actually, they differ in all of these ways.pdf (20)
1. One condition that seems to lead to senescence and not apoptosi.docx
1. One condition that seems to lead to senescence and not apoptosi.docx
Chapter 2 problems.1. (Problem 1) Of the following, whi.pdf
Chapter 2 problems.1. (Problem 1) Of the following, whi.pdf
Question 1 This type of selection favors both extremes of the .docx
Question 1 This type of selection favors both extremes of the .docx
Shape cocci bacill spirilii Gram positive t negative.pdf
Shape cocci bacill spirilii Gram positive t negative.pdf
Biological Transformation Of Bacteria And Pglo Plasmid Dna
Biological Transformation Of Bacteria And Pglo Plasmid Dna
Can you please help with the questions. 1. Which type of signalin.pdf
Can you please help with the questions. 1. Which type of signalin.pdf
DNA replicates conservatively, Semiconservatively, liberally, by hyb.pdf
DNA replicates conservatively, Semiconservatively, liberally, by hyb.pdf
More from calderoncasto9163
LOS At the time of her death on September 4, 2015, Alicia held the .pdf
LOS At the time of her death on September 4, 2015, Alicia held the following assets. Bonds of
Emerald Tool Corporation Stock in Drab Corporation Insurance policy (face amount of
$400,000) Fair Market Value $ 900,000 1,100,000 on the life of her father, Mitch Traditional
IRAs *Cash surrender value. 80,000* 300,000 Alicia was also the life tenant of a trust (fair
market value of $2 million) created by her late husband Bert. (The executor of Bert\'s estate had
made a QTIP election.) In October, Alicia\'s estate received an interest payment of $11,500
($6,000 accrued before September 4, 2015) paid by Emerald and a cash dividend of $9,000 from
Drab. The Drab dividend was declared on August 19 and was payable to date of re- cord
shareholders on September 3, 2015. Although Mitch survives Alicia, she is the designated
beneficiary of the policy. The IRAs are distributed to Alicia\'s children. What amount is included
in Alicia\'s gross estate?
Solution
Alicia’s Gross Estate:-
Bonds of Emrald Tool Corporation - 900,000
Interest on Emrald Tool Bonds – 11,500…………..(accrued interest is to be included )
Stock in Drab Corporation - 1,100,000
Cash Dividend – 9,000
Insurance Policy – 80,000 ……….( Cash surrender value)
Traditional IRA – 300,000
Value of Trust – 2,000,000
Thus, Total asset – 4,400,500.
Match the following terms with the definitions given belowTerms Gi.pdf
Match the following terms with the definitions given below
Terms Given
a. Heap
b. Memory leak
c. Inaccessible object
d. Dangling pointer
e. Destructor
f. Dynamic data
Definitions given
i. The loss of available space that occurs when dynamic data is not deallocation properly.
ii. The area of memory used for allocation and deallocation of dynamic data.
iii. An object that has been allocated but has no pointer pointing to it.
iv. Variables created with the new operation.
v. A pointer that points to a dellocated object.
vi. A member function invoked when an object goes out of scope.
Solution
Heap - The area of memory used for allocation and deallocation of dynamic data.
Memory leak - The loss of available space that occurs when dynamic data is not deallocation
properly.
Inaccessible object - An object that has been allocated but has no pointer pointing to it.
Dangling pointer - A pointer that points to a dellocated object.
Destructor - vi. A member function invoked when an object goes out of scope.
Dynamic data - Variables created with the new operation..
JAVA...With N.E.T_B.E.A.N.S___________________________________.pdf
JAVA...W\'i\'t\'h
N.E.T_B.E.A.N.S_______________________________________________________________
_____________________________________________________________________________
_____________________________________________________________________________
____________________________________________________
***********Product*********************
package rooster.nest;
import java.text.NumberFormat;
public class Product {
private long id;
private String code;
private String description;
private double price;
public long getId() {
return id;
}
public void setId(long id) {
this.id = id;
}
public String getCode() {
return code;
}
public void setCode(String code) {
this.code = code;
}
public String getDescription() {
return description;
}
public void setDescription(String name) {
this.description = name;
}
public double getPrice() {
return price;
}
public void setPrice(double price) {
this.price = price;
}
public String getPriceFormatted() {
NumberFormat currencyFormatter =
NumberFormat.getCurrencyInstance();
return currencyFormatter.format(getPrice());
}
}
********DBException************
package crows.foot;
/*
* This is just a wrapper class so we can throw a common exception for
* the UI to catch without tightly coupling the UI to the database layer.
*/
public class DBException extends Exception {
DBException() {}
DBException(Exception e) {
super(e);
}
}
*******DBUtil************
package crows.foot;
import java.sql.Connection;
import java.sql.DriverManager;
import java.sql.SQLException;
public class DBUtil {
private static Connection connection;
private DBUtil() {}
public static synchronized Connection getConnection() throws DBException {
if (connection != null) {
return connection;
}
else {
try {
// set the db url, username, and password
String url = \"jdbc:mysql://localhost:3306/mma\";
String username = \"mma_user\";
String password = \"sesame\";
// get and return connection
connection = DriverManager.getConnection(
url, username, password);
return connection;
} catch (SQLException e) {
throw new DBException(e);
}
}
}
public static synchronized void closeConnection() throws DBException {
if (connection != null) {
try {
connection.close();
} catch (SQLException e) {
throw new DBException(e);
} finally {
connection = null;
}
}
}
}
*************ProductDB**************
package crows.foot;
import java.sql.Connection;
import java.sql.PreparedStatement;
import java.sql.ResultSet;
import java.sql.SQLException;
import java.util.ArrayList;
import java.util.List;
import rooster.nest.Product;
public class ProductDB {
private static Product getProductFromRow(ResultSet rs) throws SQLException {
int productID = rs.getInt(1);
String code = rs.getString(2);
String description = rs.getString(3);
double price = rs.getDouble(4);
String note = rs.getString(5);
Product p = new Product();
p.setId(productID);
p.setCode(code);
p.setDescription(description);
p.setPrice(price);
return p;
}
public static List getAll() throws DBException {
String sql = \"SELECT * FROM Product.
Let us roll a die twice. The number from the first throw is denoted .pdf
Let us roll a die twice. The number from the first throw is denoted by X1 and the number from
the second throw is denoted by X2. Let us define a new variable Y by Y(x1+x2-5)^2.
* Make a table that shows all cases of Y with their probabilities.
* What is the expectation value of Y, i.e. E[Y]?
* Obtain the variance of Y, V[Y].
Solution
Let us roll a die twice. The number from the first throw is denoted by X1 and the number from
the second throw is denoted by X2. Let us define a new variable Y by Y(x1+x2-5)^2.
* Make a table that shows all cases of Y with their probabilities.
* What is the expectation value of Y, i.e. E[Y]?
* Obtain the variance of Y, V[Y].
We can use the following:
Sum of two die
Probability
2
1/36
3
2/36 = 1/18
4
3/36 = 1/12
5
4/36 = 1/9
6
5/36
7
6/36 = 1/6
8
5/36
9
4/36 = 1/9
10
3/36 = 1/12
11
2/36 = 1/18
12
1/36
Total
36/36
sum of dice
y
p
y^2
y*p
y^2*p
2
9
1/36
81
0.25
2.25
3
4
1/18
16
0.222222
0.888889
4
1
1/12
1
0.083333
0.083333
5
0
1/9
0
0
0
6
1
5/36
1
0.138889
0.138889
7
4
1/6
16
0.666667
2.666667
8
9
5/36
81
1.25
11.25
9
16
1/9
256
1.777778
28.44444
10
25
1/12
625
2.083333
52.08333
11
36
1/18
1296
2
72
12
49
1/36
2401
1.361111
66.69444
sums:
1
9.833333
236.5
To get the y column, subtract 5 from the sum column and then square.
The second column and the third column of this table show the possible values of y and their
probabilities.
E(Y) and V(Y) are calculated using the sums in the last row of this table.
E(Y) = 9.8333
Sum of two die
Probability
2
1/36
3
2/36 = 1/18
4
3/36 = 1/12
5
4/36 = 1/9
6
5/36
7
6/36 = 1/6
8
5/36
9
4/36 = 1/9
10
3/36 = 1/12
11
2/36 = 1/18
12
1/36
Total
36/36.
Is there relationship between having a high school education and kno.pdf
Is there relationship between having a high school education and knowledge of the orbit of the
earth, which one is independent variable and which one is dependent variable?
Solution
There is relationship between having a high school education and knowledge of the orbit of the
earth. While hving high school education is an independent variable, knowledge of the orbit of
earth is dependent upon the former information. If one has high school education then only he
can know about the orbit of the earth..
In a survey, 600 mothers and fathers were asked about the importance.pdf
In a survey, 600 mothers and fathers were asked about the importance of sports for boys and
girls. Of the parents interviewed, 70% said the genders are equal and should have equal
opportunities to participate in sports.
1. find the probability of getting at least 67% but no more than 73%. Use the exact binomial
calculation.
Solution
The probability is
P(0.67.
in html 1 A Header with font size ranging from h4 to h1 with at l.pdf
in html
1: A Header with font size ranging from h4 to h1 with at least your name and class
included. Follow the example on the printout I gave in class.
2. Two horizontal lines separating the head and body – spanning the screen and of size 5.
3. A background image chosen judiciously so that fonts will be displayed clearly. If you
wish not to use an image, you could use a different background color.
4. An image to the left and one to the right, aligned with each other and enough space
in between that you could add some written comments, description, etc. It should be
very similar to the example shown.
5. Keep adding more comments, etc. below the images that must now stretch across the
page. Somewhere in all of this writing, create a link that will take the user down
somewhere near the bottom of the page. After reading at the bottom of the page, there
must be another link that will take the user back to the point where he/she left.
6. Make all links blue and visited links red.
7. (a) Now create a bulleted list (unordered with square as the bulleted symbol).
(b) Inside of the bulleted list, you must have a sublist that is ordered (using numbers).
The font face of this sublist must be different than that of the rest of the
document.
(c) Inside of this sublist, create an unordered sublist with disc as the bulleted
symbols. Look at the example given.
8. Create a table with the following properties:
(a) A width of 80% and centered.
(b) 10 rows and 6 columns
(c) The background of the table must be different from rest of document.
(d) The first row must consist of an image not more than 2 inches in height.
(e) The label for the first column can be for example NAMES, centered, bold and take up two
rows
(f) The second row must span 5 columns and can have as its header TEST SCORES or WEEK
DAYS or ..., centered and bold.
(g) The third row must consist of 5 individual columns each with a header that is centered and
bold as shown in the example.
(h) Fill in the table with appropriate data as shown in the example making sure that the names
are left aligned and the scores are centered but not bold.
9. After the table, follow it up with some written material then include some sort of
quote as shown in the example. Make sure you use italics for the quote, different
color font and for the author of the quote also use different color font.
10. Use a horizontal line to separate this part of the document from what is to follow.
11. Create a link of some web site that you like. Make sure that the link is live i.e. I can
activate the link and be able to go to that web site.
12. Create a link to another page that you have created as demonstrated in the example.
(A) Only the written words are active.
(B) Both the image and the written words are active.
(C) Only the image is active.
13. Add your email address below. Again, make sure that it is active so that I can email
you. Also experiment by using an image for the email rather than words, as shown in
the example
Solut.
Identify these monomers (general classification). Are the monomers h.pdf
Identify these monomers (general classification). Are the monomers hydrophilic, or
amphipathic? Write your answer in the space above each monomer. How many functional
groups can you identify? Circle and name them. Show how you would make these monomers
into a polymer.
Solution
A and B. a.Cysteine (Cys, C) - Hydrophobic and aliphatic R group
b. Phenylalanine (Phe, F) - Hydrophobic and aromatic R group
C. Functional groups
Cysteine
- COOH - Carboxyl
- NH2 - Amino
- SH - Sulphydryl
Phenylalanine
- COOH - Carboxyl
- NH2- Amino
- Aromatic.
How is the Internet come about How was the World Wide Web ( WWW .pdf
How is the Internet come about ? How was the World Wide Web ( WWW ) developed ?
How is the Internet come about ? How was the World Wide Web ( WWW ) developed ?
Solution
It was in late 1960s the initial concepts of internet came into exostence, with ARPANET(
Advanced Research Projects Agency Networ), which was originally founded by US Department
of Defence. They introduced packet switching to communicate between systems over a single
network.Later on in 1970s the TCP/IP was got introduced, so the ARPANET adopted TCP/IP in
1983, and from there the network of network model came into existencce, which later known as
Internet.
WWW (World Wide Web) is an aspect of Internet, like email. Coomonly WWW is being
misinterpreted as the so called Internet. WWW is an information space in which one can locate
any kind of information through a URL(Uniform Resource Locator). It was in 1989, Tim
Berners-Lee invented the Web. He developed a web browser which he was an employee at
CERN in Switzerland. Lter on the commercial potential in web makes the growth of web in an
exponential paced manner..
Explain the answers in sufficient details i.e show the work ace Wha.pdf
Explain the answers in sufficient details i.e show the work ace? What is the probability of
drawing a red ace? Are these events (\"ace\" and \"red\") mutually exclusive? Are they
independent? e. List two events that are mutually exclusive for a single draw from a deck of
cards. What is the probability of drawing a red f. king? What is the probability of drawing a face
card in hearts? Are these two events mutually exclusive? Are they independent? 20. The human
genome is composed of the four DNA nucleotides: A, T, G, and C. Some regions of the human
genome are extremely G-C rich (i.e., a high proportion of the DNA nucleotides there are guanine
and cytosine). Other regions are relatively A-T rich (i.e., a high proportion of the DNA
nucleotides there are adenine and thymine). Imagine that you want to compare nucleotide
sequences from two regions of the
Solution
a) Let we have two regions, Region1 and Region2.
It is given that 60% nucleotides in 1st region is A&T (30% each) and remaining 40% is of G&C
nucleotides(20% each). Probability of the nucleotides in Region1:
Similarly for region2, Probability of the nucleotides in Region2:
Probability of getting both are same nucleiods,we must multiply the individual probabilities.
Probability of getting both has type A nucleoids=P(A1)*P(A2)
Probability of getting both has type T nucleoids=P(T1)*P(T2)
Probability of getting both has type G nucleoids=P(G1)*P(G2)
Probability of getting both has type C nucleoids=P(C1)*P(C2)
Required probability=P(A1)*P(A2)+ P(T1)*P(T2)+ P(G1)*P(G2)+ P(C1)*P(C2)
=0.25
b)
It is given that two regions are independent. To find probability to select three nucleoids
sequence (triplets) from each of the two regions are same.
All combinations made with the nucleoids A, T, G and C taking three at a time 4C3.
The number of all combinations of 4 things taken 3 at a time is 4C3=4
Possible selections are ATG, TGC, GCA and CAT
Suppose we get outcome as ATG from Region1, so Probability of getting it isNucleotidesP
(N)A10.2T10.2G10.3C10.3.
Each figure is an isometry image of the figure at the right. Tell whe.pdf
Each figure is an isometry image of the figure at the right. Tell whether their orientations are the
same or opposite. Then classify the isometry.
Solution
27, 28, - Yes, they have same orientation
29 - Have opposite orientation
All images are isometric to the original image, its just that figure 29 is flipped..
Describe the difference between a. The pulmonary and systemic circula.pdf
Describe the difference between a. The pulmonary and systemic circulation b. An artery and a
vein c. An atrium and a ventricle
Solution
a. Pulmonary and systemic circulation
The cardiovascular system comprise of two circulatory pathsways namely pulmonary and
sytemic circulation.The pulmonary circulation circuits through the lungs carrying the oxygenated
blood whereas the systemic circulation circuit through the rest of the body to provide oxygenated
blood to various body parts.
In the pulmonary circulation, blood travels through capillaries on the alveoli, air sacs in the lungs
allowing the gas exchange. Here, as blood flows, the size of the vessel decreases from artery to
arteriole, from vein to venuleand finally to capillaries,which serve as vessels for gas and nutrient
exchange.In the systemic circulation,the movement of blood takes place from the heart through
the body providing oxygen and nutrients to the tissues of the body while bringing back the
deoxygenated to the heart. Systemic and pulmonary circulation transition to the opposite to each
other. Systemic circulation is a much larger and higher pressure system than pulmonary
circulation.
Thus systemic circulation is the larger path that blood takes on, routing oxygenated blood around
the tissues of the body and deoxygenated blood from them back to the heart , while pulmonary
circulation is the smaller loop that blood takes in the lungs.
b. An artery and a Vein
Blood is circulated in the body through arteries, veins and capillaries in a closed circulatory
system.
All blood vessels leading from heart carrying oxygenated blood (except pulmonary artery) to
various organs of the body are called as arteries.
Whereas, vessels which carry deoxygenated blood (except pulmonary vein) from various organs
of the body to the heart are called as veins.
Arteries have thick elastic muscular walls and valves are absent, whereas, Veins have thin non-
elastic valves and valves are present to prevent bacl flow of blood.
c. An atrium and a Ventricle
There are 4 chambers in the heart; the atria refer to the upper chambers while the ventricles refer
to the lower chambers.
Atria (atrium) refers to the upper chambers of the heart that receive the impure blood from the
veins to send it to the ventricles.
Ventricles are small cavities or chambers that are present within the left chamber of the heart that
accepts blood from the arteries (left atrium) and then contracts to force into the aorta. Both right
and the left part of our heart has an atrium and one ventricle.
The walls of the ventricles are thicker, while that of the atria are thinner. They ventricles contain
valves to pump the blood in and out of the heart..
Cross a homozygous running, heterozygous black mouse with a waltzing.pdf
Cross a homozygous running, heterozygous black mouse with a waltzing brown mouse. What is
the probable phenotype ratio? Show punnet square and all work.
Solution
Dihybrid cross
Homozygous running & heterozygous black -mouse x heterozygous black & waltzing
mouse
Parental gametes --> RRBb x rrbb
RB Rb RB Rb
rb RrBb Rrbb RrBb Rabb
rb RrBb Rrbb RrBb Rrbb
rb RrBb Rrbb RrBb Rrbb
rb RrBb Rrbb RrBb Rrbb
F1-phenotypes
Running black mouse
Running brown mouse.
Describe some effects that cybertechnology has had so far for our se.pdf
Describe some effects that cybertechnology has had so far for our sense of self. Questions to
think about as you write: In the age of information how are we beginning to conceive ourselves
in relation to nature? How is cybertechnology changing the way he view ourselves and others?
Does cybertechnology bring us closer together or create a rift? Are humans becoming more
computer-like? How?
MUST BE AT LEAST 300 WORDS IN TOTAL.
Solution
Information technologies are in a stable level of change and break-through.We can see that
information technologies have brought great social changes compared to the past years before
they appeared.In the past days of disconnect,we plugged into each other\'s nervous system by
passing our books,on discussions with others and by sharing our thoughts and dreams.But
connectedness made one become others databases snd servers learning on eachother\'s
memories.The expanded technology increased the lenghts of people\'s thoughts with this
increasing world.It brought people\'s imaginations and thoughts into existence.
The effects of technology has to do with whether humans desire greater simplicity or complexity
as they progress. Many technology creators acclaim the value of convergence. While that might
severely reduce the number of ‘devices’ that we need and use, it also adds complexity to the task
of using a device.Alternately, people have written about how we desire purity, and will therefore
look for technologies that will do one thing, and do it extremely well. The challenge before
technology innovators, clearly, is to enable technologies to communicate with each other, and
sort problems out, the way humans do.Technology will enable diverse cultures to collaborate
more efficiently, in every sphere. It will bring people and organizations together, closer.
Technology will allow people to live multiple lives, assume and live out multiple
identities.Every day, as millions of people interact over the internet, they navigate worlds that are
simulated, they create virtual personalities, and forge online relationships. Fantasy and role-play,
both natural and important functions of child development, are fostered by adult chatrooms and
virtual reality. In a distributed, yet isolated world, people desire connections. In doing so, they
think not of identity but identity crises.
Technological world is a brilliant opportunity for global connections and it gives the exprience of
leading different lives and cultures.We must believe that internet brings us together as without
internet it is not so easy to communicate with eachother.But on the other hand, it does have some
negative effects on people, tearing them apart in way of face to face communication making such
relations undervalued.
Computers, or operating systems to be exact, are lacking in the the fundamental capabilities
possessed by all sentient systems: the ability to reason, to learn from observation and experience,
to grow intelectually over time. As of today, computers .
Choose a deployment platform that allows for the implementation o.pdf
Choose a deployment platform that allows for the implementation of SOA. Describe the
architectural layers that are included in the platform and analyze the flexibility of the platform
for supporting different types of services. Determine whether this platform supports loose
coupling or if it is strongly implementation dependent.
Choose a deployment platform that allows for the implementation of SOA. Describe the
architectural layers that are included in the platform and analyze the flexibility of the platform
for supporting different types of services. Determine whether this platform supports loose
coupling or if it is strongly implementation dependent.
Solution
Service-oriented architectures (SOA) are based on the software services, which are high-level
software components that include web services like (Rest , Soap). Implementation of an SOA
requires tools and run-time infrastructure software. This is collectively referred to as a service-
oriented architecture implementation framework or (SOAIF).
Application development and deployment :
In the traditional software development process, translating the requirements into working
distributed systems is time-consuming and difficult, requiring several stages of manual
development and then its deployment. This complex, error-friendlytask can be effectively done
using a higher-level, component-based SOAIF. The SOAIF incorporates tools that let processes
that are developed, using standards such as Business Process Execution Language (BPEL), can
be easily translated into distributed, high-level services, which are easy to develop, manipulate,
and debug and hence more effecient . These services are easily composed of implementation-
level data flows without the user or developer having to track complex middleware concepts,
such as topics or queues. Further, the implementation-level services can run on any machine
across the network due to the built-in dynamic deployment support SOAIF provides.
Architectural Layers
Enterprise service bus :core infrastructure of an SOAIF is typically provided by an enterprise
service bus (ESB), which points out the challenges in composing, deploying, and managing
distributed, service-based enterprise applications. The ESB incorporates systems connectivity
using web services, Java EE, .NET Framework, and other standards also
ESB also do not provide any guaranty of loose coupling to meet evolving Service-oriented needs
. Implementing ESB to meet SOA requirements require the addition of extra functionality to
compose for business Services and provide secure Service interactions.
Business process management :
Business process management (BPM) considers systems and IT assets as activities or tasks that
participate in well-coordinated and centrally organised Business processes. Traditionally, the
challenge of BPM is that while it is possible to construct any processes that achieve integration
goals, enterprises typically use BPM tools only at the time of designing , modeling.
Abstract Base Class (C++ Program)Create an abstract base class cal.pdf
Abstract Base Class (C++ Program)
Create an abstract base class called Animal which has the following pure virtual member
functions: talk( ), move( ) and the following attribute: animalType (which is a pointer to a
dynamically allocated string.)
Then create the following derived classes (from the animal class): reptile, bird, and mammal.
Here are the concrete classes:
From the reptile class derive lizard and snake.
From bird derive eagle and chicken.
From mammal derive bear, hyena and lion.
Each concrete class needs to override the talk() and move() functions. Also each concrete class
will have the operator<<( ) defined. The output of cout << smokey; (assuming smokey is a bear)
will be the animalType , the sound he makes (talk( ) ) and how he moves ( move( )). Here is an
example:
int main()
{
bear smokey; // using the concrete class
cout << smokey // prints out: bear, growl, walk
animal *aptr = new bear; // I could have used an animal pointer
cout << *aptr; // prints the same results: bear, growl, walk
return 0;
}
Create an array of animal pointers called zoo. Make it 10 in size. Here is what the menu should
look like:
----Zoo Cages----
1 [empty]
2 [empty]
3 [empty]
4 [snake]
5 [eagle]
6 [bear]
7 [empty]
8 [empty]
9 [lion]
10 [hyena]
----Menu----
1) Add
2) Remove
3) Display all animals
4) Display animal count
5) Exit
If the user selects 1) then ask for a cage number and the animal he wishes to put in that cage. Be
sure to check the cage for an existing animal first. The user cannot put an animal in a cage that is
already occupied.
If the user selects 2) then ask for the cage number and remove the animal from that cage (use the
destructor). If there is no animal in the cage, issue an error to the user.
If the user selects 3) then display the contents of each cage. The display function will show each
cage number as well as the animalType of the animal, the sound it makes and how it moves, in
other words, use the operator<<( ). All this information must be on a single screen. The output
screen is up to you.
If the user selects 4) then display the count of the number of animals in your zoo. In the above
example, you will have 5. You should use a static data member in the animal class to keep track
of the number of animals. You should increase that count each time the animal constructor is
called and decrease the count when the animal destructor is called.
***Use Virtual Functions and the animal base class MUST be abstract
Solution
main.cpp
#include //input output
#include //strings
#include //Randoms
#include //Randoms
#include //For input
#include \"lizard.h\"
#include \"snake.h\"
#include \"chicken.h\"
#include \"eagle.h\"
#include \"hyena.h\"
#include \"bear.h\"
#include \"lion.h\"
using namespace std;
//declare our static variable from the animal class
int animal::numAnimals;
int main()
{
//array of animal pointers
animal *zoo[10] =
{ NULL };
//initialize our static variable to zero
animal::numAnimals = 0;
//To test if cage is empty, it is nu.
A range of values estimated to have a high probably of containing th.pdf
A range of values estimated to have a high probably of containing the population value is called:
Point estimate
Standard error of the mean
Confidence interval
Standard deviation
Point estimate
Standard error of the mean
Confidence interval
Standard deviation
Solution
It is
OPTION C: CONFIDENCE INTERVAL. [ANSWER]
***********
The confidence interval is the range of values in which the true population parameter value is
there..
A thrown lump of clay hits a block in mid air and sticks to it. what.pdf
A thrown lump of clay hits a block in mid air and sticks to it. what happens to the total kinetic
energy of the ball+clay system during this inetraction, ignoring gravity?
Solution
As the lump of clay sticks to th block, it is an example of inelasti collision. During the inelastic
collision, the momentum will be conserved for ball+clay system but Kinetic energy will not.
Total Kinetic Energy of ball+clay system will decrease during the interaction..
Answer die following questions with short answers Explain the d.pdf
Answer die following questions with short answers: Explain the difference between stream
ciphers and block cipher. Explain the difference between passive and active attacks. Describe
the difference between authentication and non-repudiation Explain the difference between a
MAC and message digest. Give one advantage and one disadvantage of public-key cryptography
over symmetric-key cryptogra Why is CBC mode considered preferable to ECB mode?
Solution
a)
stream cipher and block cipher differ in whitch the plain text is encoded and decoded.
the basic difference is how the logic used for the encryption and decryption. the idea of block
cipher is partition the plain text into relatively larger blocks and further each block will be
encoded.
and same key is used for both encryption and decryption.example DES,IDEA
the idea of stream cipher is plain text is encrypted and transfered bit by bit.but key property of
stream cipher is different key is generated for each block or bit.example -ISAAC,FISH
b)
passive attack is indirect attack.the attacked host is completely unaware about the attack.
hence it is called passive attack like some body listens the packet transformation or observing
the host in network.
active attack is direct attack.its nothing but the attacked one gets aware of the attack.
like somebody listens the id and password and stop the services(denial of service) or logic
bomb.
both are harmful.
c)
authentication and non repudiation are two different things.
authentication verifies the who you are(user-id).where the non-repudiation verifies the what you
did(messages).
authentication is done through a central instance where non repudiation can done by anyone
d)
message digest is a hash of a message.its output of a cryptographic hash function appilied to
input data it refered to as message.
mac is a piece of information that proves the integrity of a message.
message digest is take only single input and produces a message digest(aka hash) if the message
change it produce different hash code.
mac (message authentication code) is take input and security code to check the integrity of
message. HMAC is one of this kind.
e)
advantage of public key cryptography over the symmetric key cryptography is increased
security and convenience.
and no need of key distribution(transmitted to anyone).and it provide digital signatures that
cannot be repudiated.
the disadvantage of public key cryptography over symmetric cryptography is speed.they are
many symmetic key methods runs faster than the public key cryptography.
f)
the reason why cbc is considered better than ecb has nothing to do with situations involving
an attacker with partial cipher text.
problem with ECB is that it leaks the information like when two blocks have same plain text
then the corresponding cipher blocks also be same where in case
of the CBC it randomly independent of plain text to produce the cipher text..
A single breeding pair of rabbits is introduced to Australia in 1900.pdf
A single breeding pair of rabbits is introduced to Australia in 1900. Assume (for simplicity) that
each pair can produce 4 offspring each year (2 male, 2 female). Also assume, that all births
happen in the last hour of the last day each year and the parents die immediately after offspring
are born. a) How many rabbits are there after 10 years? b) How many rabbits are there after 25
years? If they have 20 offspring each year, c) How many rabbits are there after 10 years? d) How
many rabbits are there after 25 years?
Solution
Each pair can produce 4 offsprings each year and parents die after giving birth to offsprings. So,
offsprings makes the population and parents don\'t add to population.
After first year total population is 4. Next year these 4 rabbits make 2 pairs and give birth to 4 off
springs each. Second year population becomes 8, multiplies by 2 from last year. This continues
till 10th year and then till 25th year.
At the end of 10th year, the population of rabbits will be 2048 and at the end of 25th year the
population of rabbits will be 33554432.
Now when each year 20 offsprings are produced, after the end of first year total population is 20.
At the end of second year, 20 rabbits make 10 pairs and give birth to 20 rabbits each taking total
population to 10*20=200. At the end of 3rd year 200 rabbits make 100 pairs and give birth to 20
each taking total population to 100*20=2000.
In this way at the end of each year population gets multiplied by 10.Therefore, at the end of 10
year total population will be 2*1010 and at the end of 25 years total rabbit population is will be
2*1025..
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