Match the following DNA-related terms to their descriptions or defin.pdf
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Match the following DNA-related terms to their descriptions or definitions: a. A cell containing only ONE copy of each chromosome. b. An individual carrying 2 of the SAME allele for a given trait. c. Pieces of DNA that contain genes. d. The OBSERVABLE characteristics of a trait. e. The process in meiosis that forms tetrads. f. Unwound, readable DNA. g. A pair of chromosomes carrying the same GENES, but not necessarily the same alleles. h. Alternate forms of a trait (Blond hair, Brown hair) i. A problem in meiosis when chromatids DO NOT separate appropriately during anaphase. j. An IDENTICAL copy of a chromosome. k. A cell containing TWO copies of each chromosome. l. The alleles present in an organism; NOT visible m. A segment of DNA that codes for a single trait. n. An individual carrying 2 DIFFERENT alleles for a given trait. o. Any situation of incorrect chromosome number. ______ Chromosome ______ Chromatin ______ Chromatid ______ Gene ______ Allele ______ Diploid ______ Haploid ______ Non-disjunction ______ Synapsis ______ Aneuploidy ______ Homologous ______ Homozygous ______ Heterozygous ______ Genotype ______ Phenotype Solution a. A cell containing only ONE copy of each chromosome. b. An individual carrying 2 of the SAME allele for a given trait. c. Pieces of DNA that contain genes. d. The OBSERVABLE characteristics of a trait. e. The process in meiosis that forms tetrads. f. Unwound, readable DNA. g. A pair of chromosomes carrying the same GENES, but not necessarily the same alleles. h. Alternate forms of a trait (Blond hair, Brown hair) i. A problem in meiosis when chromatids DO NOT separate appropriately during anaphase. j. An IDENTICAL copy of a chromosome. k. A cell containing TWO copies of each chromosome. l. The alleles present in an organism; NOT visible m. A segment of DNA that codes for a single trait. n. An individual carrying 2 DIFFERENT alleles for a given trait. o. Any situation of incorrect chromosome number. Matched below (compare a,b,c,d above with below ) ___c___ Chromosome ___f___ Chromatin ___g___ Chromatid ___m___ Gene ___h___ Allele ____k__ Diploid ___a___ Haploid ___i___ Non-disjunction ___e___ Synapsis ___o___ Aneuploidy ___j___ Homologous ___b___ Homozygous __n____ Heterozygous ___l___ Genotype ___d___ Phenotype.
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12). Choose the letter designation that represent homozygous recessi.pdf
12). Choose the letter designation that represent homozygous recessive genotypte?
c. ee
As there are always two letters in the genotype because one code for the trait come from materal
and other code for the trait comes from paternal oraganism, so every organism has two codes (
letters ). The two lowercase letters in the genotype referred as homozygous recessive genotype(
ee). The term pure may also used in place of homozygous.
13). Daughter nuclei form from each parent cell during
a. Telophase I and telophase II
In both at each pole, a daughter nucleus with nuclear membrane( from ER) and nucleolus( from
SAT chromosome) is formedand also there is disappearance of astral rays and spindle fibres.
14). The diploid human cell has _ chromosomes.
- 46
The diploid number is abbreviated as 2n because of the presence of twice the amount of
chromosomes as a haploid cell. The haploid cell has a total of 23 chromosomes while the diploid
cell has 46 chromosomes.
15). Choose all that are true about genes
- genes are located on chromosomes
- genes are units of inheritence about heritable trait
- genes are arranged in a linear sequence on a chromosome
- each gene has its own location,in one type of chromosome
16).Mendal theory of independent assortment states that
a. genes for different traits are randomly distributed to gametes during meiosis.
17). Pairs of homologus chromosomes exist
a. during meiosis I
It is a special type of division in which chromosomes duplicate only once, but cell divides twice.
So one parental cell produces 4 daughter cells.
18). Alternate forms of genes are called_
- alleles
19). The physical appearance of an individual is referrred to as its
b. Phenotype
20). The genetic make- up of an individual is referred to as its
a.Genotype
21). Why the sexual reproduction benefit for the organism?
all of these are true.
22). The various forms of particular gene are called
b. alleles
Solution
12). Choose the letter designation that represent homozygous recessive genotypte?
c. ee
As there are always two letters in the genotype because one code for the trait come from materal
and other code for the trait comes from paternal oraganism, so every organism has two codes (
letters ). The two lowercase letters in the genotype referred as homozygous recessive genotype(
ee). The term pure may also used in place of homozygous.
13). Daughter nuclei form from each parent cell during
a. Telophase I and telophase II
In both at each pole, a daughter nucleus with nuclear membrane( from ER) and nucleolus( from
SAT chromosome) is formedand also there is disappearance of astral rays and spindle fibres.
14). The diploid human cell has _ chromosomes.
- 46
The diploid number is abbreviated as 2n because of the presence of twice the amount of
chromosomes as a haploid cell. The haploid cell has a total of 23 chromosomes while the diploid
cell has 46 chromosomes.
15). Choose all that are true about genes
- genes are located on chromosomes
- genes are units of inheriten.
3.2 Notes
Guided notes covering material from Topic 3.2 of the updated IB Biology syllabus for 2016 exams. Notes sequence and prompts are based on the Oxford IB Biology textbook by Allott and Mindorff.
please! 1. Match and pair the following basic genetic conce.pdf
please!
1. Match and pair the following basic genetic concepts with their respective
descriptions/definitions: 1. Gene a. Two different alleles 2. Allele b. Observable heritable feature
that may vary among organisms. 3. Phenotype c. Two identical alleles 4. Genotype d. Describes
the specific alleles present in the offspring. 5. Homozygous e. Segments of DNA responsible for
producing a particular trait. 6. Heterozygous f. Variants of a gene present for a particular
character. 2. From the figure presented here, determine the following: a. Which letter(s) show a
phenotype? b. Which letter(s) show a genotype? c. In letter (c), are we looking at a homozygous
recessive, a homozygous dominant, or a heterozygous individual? d. In letter (e), are we looking
at a homozygous recessive, a homozygous dominant, or a heterozygous individual?
3. A and a are dominant and recessive alleles, respectively, of the same gene. Which genotype(s)
would result in an individual with the dominant trait? A) AA and aa B) AA and aa C) only AA
D) AA and Aa E) only Aa 4. For Labrador retrievers, black fur color (B) is dominant to yellow
fur color (b). If a heterozygous black male mated with a homozygous yellow female, what would
the puppies look like? Determine the phenotypic ratios. - Step 1: Determine the genotype of the
female and male Labradors. Heterozygous black male: Homozygous yellow female: - Step 2:
Write/Type the cross between the two Labradors. X -Step 3: Do a Punnett Square to determine
the probable genotypes for the offspring and to determine the phenotypic ratio.
-Step 4: Determine the Phenotype ratios of the progeny: s. There are many humans traits that are
fairly simple and that follow the Mendelian pattern of inheritance. However, most of our traits
are much more complex, involving many genes or interactions between genes. For example, hair
color is determined by at least four genes, each one coding for the productions of melanin, a
brown pigment. Because the effect of these genes is cumulative, hair color can range from
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following table..
Chapter 2 problems.1. (Problem 1) Of the following, whi.pdf
Chapter 2 problems.
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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
.
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this slide will help you in IGCSE examination in subject biology. Hopefully it can be helpful in understanding biology syllabus.
Firefly luciferase is a commonly used reporter gene. The gene origina.pdf
Firefly luciferase is a commonly used reporter gene. The gene originates from fireflies (a type of
beetle) and when it combines with its substrate, luciferin, light is produced. Luciferin is very
expensive to purchase. You would therefore like to figure out the biosynthetic pathway for
making luciferin so that you can get bacteria to make luciferin. This will enable you pay off all
your college loans! You have obtained 5 mutant firefly stocks that each fail to glow at night.
The flies in each stock have the wild type luciferase gene but ultimately fail to make the luciferin
substrate. Each stock has a unique mutation in one of the genes required to make luciferin. You
know that the biosynthesis of luciferin involves 5 intermediate compounds: A, B, C, D, and E.
However, you do not know the order that these compounds appear in the pathway. When you
feed these intermediate compounds to the mutant fireflies, they sometimes are able glow (+) and
sometimes still cannot glow (-). However, when you feed the mutant fireflies luciferin, they
always glow. Based on the results in the table above, write the biosynthetic pathway for
luciferin, showing the order of the intermediates (A-E) and the genes (M1-M5) required to
convert each intermediate to the next in the pathway.
Solution
This is a probem, where in you can get the answer by the means of trial and error.
Now that we have all pieces to this puzzle, lets join them.
E>B>D>C>A this is the pathway in the production of Luciferase..
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Solution
1 University of Wisconsin - Madison ECE/Comp Sci 352 Digital Systems Fundamentals Charles
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& Thomas Kaminski © 2001 Prentice Hall, Inc Logic and Computer Design Fundamentals Logic
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represent NOT or invert. 2 Logic and Computer Design Fundamentals © 2001 Prentice Hall, Inc
Chapter 2-Part 7 3 NAND Gates The basic positive logic NAND gate is denoted by the
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function with a NOT applied. We call this symbol for a NAND gate an AND-Invert. The small
circle represents the invert function. If we apply DeMorgan\'s Law we get: X Y Z X YZ = X +
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(NAND) We call this symbol for a NAND gate the Invert - OR since all inputs are inverted,
followed by the OR function. Both symbols represent the NAND gate - it is sometimes more
logically descriptive to use one form over the other. A NAND gate with one input degenerates to
an inverter. X Y Z F(X, Y,Z) = X + Y + Z 3 Logic and Computer Design Fundamentals © 2001
Prentice Hall, Inc Chapter 2-Part 7 5 NAND Function Implementation NAND gates can
implement a simplified Sum-ofProducts form. Constructing two level NAND-NAND gate
circuit: The first level is two 2-input NAND gates using ANDInvert. The second level is one 2-
input NAND gate using Invert-OR. Using the NAND relationship, we have: G(A,B,C,D) = AB
CD = AB + CD = A B + CD A B C D G(A,B,C,D) = AB + CD Logic and Computer Design
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be combined and deleted: A B C D G(A,B,C,D) This form of the implementation is the Sum-of-
Products form. 4 Logic and Computer Design Fundamentals © 2001 Prentice Hall, Inc Chapter
2-Part 7 7 NAND Implementation (Cont.) In the implementation, the bubbles are on opposite
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results To implement an equation like: F(A,B,C) = A + BC, the NAND for A degenerates t.
More Related Content
Similar to Match the following DNA-related terms to their descriptions or defin.pdf
Which of the following correctly matches a phase of the cell cycle wit.pdf
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12). Choose the letter designation that represent homozygous recessive genotypte?
c. ee
As there are always two letters in the genotype because one code for the trait come from materal
and other code for the trait comes from paternal oraganism, so every organism has two codes (
letters ). The two lowercase letters in the genotype referred as homozygous recessive genotype(
ee). The term pure may also used in place of homozygous.
13). Daughter nuclei form from each parent cell during
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In both at each pole, a daughter nucleus with nuclear membrane( from ER) and nucleolus( from
SAT chromosome) is formedand also there is disappearance of astral rays and spindle fibres.
14). The diploid human cell has _ chromosomes.
- 46
The diploid number is abbreviated as 2n because of the presence of twice the amount of
chromosomes as a haploid cell. The haploid cell has a total of 23 chromosomes while the diploid
cell has 46 chromosomes.
15). Choose all that are true about genes
- genes are located on chromosomes
- genes are units of inheritence about heritable trait
- genes are arranged in a linear sequence on a chromosome
- each gene has its own location,in one type of chromosome
16).Mendal theory of independent assortment states that
a. genes for different traits are randomly distributed to gametes during meiosis.
17). Pairs of homologus chromosomes exist
a. during meiosis I
It is a special type of division in which chromosomes duplicate only once, but cell divides twice.
So one parental cell produces 4 daughter cells.
18). Alternate forms of genes are called_
- alleles
19). The physical appearance of an individual is referrred to as its
b. Phenotype
20). The genetic make- up of an individual is referred to as its
a.Genotype
21). Why the sexual reproduction benefit for the organism?
all of these are true.
22). The various forms of particular gene are called
b. alleles
Solution
12). Choose the letter designation that represent homozygous recessive genotypte?
c. ee
As there are always two letters in the genotype because one code for the trait come from materal
and other code for the trait comes from paternal oraganism, so every organism has two codes (
letters ). The two lowercase letters in the genotype referred as homozygous recessive genotype(
ee). The term pure may also used in place of homozygous.
13). Daughter nuclei form from each parent cell during
a. Telophase I and telophase II
In both at each pole, a daughter nucleus with nuclear membrane( from ER) and nucleolus( from
SAT chromosome) is formedand also there is disappearance of astral rays and spindle fibres.
14). The diploid human cell has _ chromosomes.
- 46
The diploid number is abbreviated as 2n because of the presence of twice the amount of
chromosomes as a haploid cell. The haploid cell has a total of 23 chromosomes while the diploid
cell has 46 chromosomes.
15). Choose all that are true about genes
- genes are located on chromosomes
- genes are units of inheriten.
3.2 Notes
Guided notes covering material from Topic 3.2 of the updated IB Biology syllabus for 2016 exams. Notes sequence and prompts are based on the Oxford IB Biology textbook by Allott and Mindorff.
please! 1. Match and pair the following basic genetic conce.pdf
please!
1. Match and pair the following basic genetic concepts with their respective
descriptions/definitions: 1. Gene a. Two different alleles 2. Allele b. Observable heritable feature
that may vary among organisms. 3. Phenotype c. Two identical alleles 4. Genotype d. Describes
the specific alleles present in the offspring. 5. Homozygous e. Segments of DNA responsible for
producing a particular trait. 6. Heterozygous f. Variants of a gene present for a particular
character. 2. From the figure presented here, determine the following: a. Which letter(s) show a
phenotype? b. Which letter(s) show a genotype? c. In letter (c), are we looking at a homozygous
recessive, a homozygous dominant, or a heterozygous individual? d. In letter (e), are we looking
at a homozygous recessive, a homozygous dominant, or a heterozygous individual?
3. A and a are dominant and recessive alleles, respectively, of the same gene. Which genotype(s)
would result in an individual with the dominant trait? A) AA and aa B) AA and aa C) only AA
D) AA and Aa E) only Aa 4. For Labrador retrievers, black fur color (B) is dominant to yellow
fur color (b). If a heterozygous black male mated with a homozygous yellow female, what would
the puppies look like? Determine the phenotypic ratios. - Step 1: Determine the genotype of the
female and male Labradors. Heterozygous black male: Homozygous yellow female: - Step 2:
Write/Type the cross between the two Labradors. X -Step 3: Do a Punnett Square to determine
the probable genotypes for the offspring and to determine the phenotypic ratio.
-Step 4: Determine the Phenotype ratios of the progeny: s. There are many humans traits that are
fairly simple and that follow the Mendelian pattern of inheritance. However, most of our traits
are much more complex, involving many genes or interactions between genes. For example, hair
color is determined by at least four genes, each one coding for the productions of melanin, a
brown pigment. Because the effect of these genes is cumulative, hair color can range from
blonde (little melanin) to very dark brown (much melanin). 2 Human traits are of great interest to
us and there are few traits that seem to exhibit Mendelian inheritance, such as the ones shown
below (this can be found in your lab manual). Figure 134 some madiy absenutle human
Menoulae tak. Mid-digital hair b. Tongue rolling c. Widow's peak d. Earlobe attachment e.
Hitchhiker's thumb f. Relative finger length With the information provided in your lab manual,
Section 13-3 (pages 172-174), choose one out the six traits presented here and complete the
following table..
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
.
INHERITANCE PART 1.pptx
this slide will help you in IGCSE examination in subject biology. Hopefully it can be helpful in understanding biology syllabus.
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feed these intermediate compounds to the mutant fireflies, they sometimes are able glow (+) and
sometimes still cannot glow (-). However, when you feed the mutant fireflies luciferin, they
always glow. Based on the results in the table above, write the biosynthetic pathway for
luciferin, showing the order of the intermediates (A-E) and the genes (M1-M5) required to
convert each intermediate to the next in the pathway.
Solution
This is a probem, where in you can get the answer by the means of trial and error.
Now that we have all pieces to this puzzle, lets join them.
E>B>D>C>A this is the pathway in the production of Luciferase..
Find all hazards in this circuit. Redesign the circuit as a three-le.pdf
Find all hazards in this circuit. Redesign the circuit as a three-level NOR circuit that is free of
all hazards Show how four 2-to-1 and one 4-to-1 multiplexers could be connected to form an 8-
to-1 MUX with three control inputs
Solution
1 University of Wisconsin - Madison ECE/Comp Sci 352 Digital Systems Fundamentals Charles
R. Kime Section 2 – Fall 2001 Chapter 2 – Combinational Logic Circuits – Part 7 Charles Kime
& Thomas Kaminski © 2001 Prentice Hall, Inc Logic and Computer Design Fundamentals Logic
and Computer Design Fundamentals © 2001 Prentice Hall, Inc Chapter 2-Part 7 2 NAND and
NOR Implementation We found that we could implement general Boolean equations with these
three primitives: • AND • OR • NOT In this section we will find that either of two gates, the
NAND gate or the NOR gate can be used to implement arbitrary logic functions. We use the
Positive Logic Convention (where all signals are active high) and a small circle to on a symbol to
represent NOT or invert. 2 Logic and Computer Design Fundamentals © 2001 Prentice Hall, Inc
Chapter 2-Part 7 3 NAND Gates The basic positive logic NAND gate is denoted by the
following symbol: • AND-Invert (NAND) NAND comes from NOT AND, I. e., the AND
function with a NOT applied. We call this symbol for a NAND gate an AND-Invert. The small
circle represents the invert function. If we apply DeMorgan\'s Law we get: X Y Z X YZ = X +
Y + Z F(X, Y,Z) = X Y Z Logic and Computer Design Fundamentals © 2001 Prentice Hall, Inc
Chapter 2-Part 7 4 NAND Gates (Cont.) Applying DeMorgan\'s Law gives: • Invert-OR
(NAND) We call this symbol for a NAND gate the Invert - OR since all inputs are inverted,
followed by the OR function. Both symbols represent the NAND gate - it is sometimes more
logically descriptive to use one form over the other. A NAND gate with one input degenerates to
an inverter. X Y Z F(X, Y,Z) = X + Y + Z 3 Logic and Computer Design Fundamentals © 2001
Prentice Hall, Inc Chapter 2-Part 7 5 NAND Function Implementation NAND gates can
implement a simplified Sum-ofProducts form. Constructing two level NAND-NAND gate
circuit: The first level is two 2-input NAND gates using ANDInvert. The second level is one 2-
input NAND gate using Invert-OR. Using the NAND relationship, we have: G(A,B,C,D) = AB
CD = AB + CD = A B + CD A B C D G(A,B,C,D) = AB + CD Logic and Computer Design
Fundamentals © 2001 Prentice Hall, Inc Chapter 2-Part 7 6 NAND Implementation (Cont.) In
the implementation, note that the bubbles are on opposite ends of the same line. Thus, they can
be combined and deleted: A B C D G(A,B,C,D) This form of the implementation is the Sum-of-
Products form. 4 Logic and Computer Design Fundamentals © 2001 Prentice Hall, Inc Chapter
2-Part 7 7 NAND Implementation (Cont.) In the implementation, the bubbles are on opposite
ends of the same line. By , they can be combined and deleted: A sum-of-products (SOP) form
results To implement an equation like: F(A,B,C) = A + BC, the NAND for A degenerates t.
Fergie has the choice between investing in State of New York bond at.pdf
Fergie has the choice between investing in State of New York bond at 10.2 percent and a
Surething bond at 11.5 percent. Assuming that both bonds have the same nontax characteristics
and that Fergie has a 30 percent marginal tax rate, in which bond should she invest?
Solution
After tax rate of return for Surething bond =11.5(1-0.3)=8.05%
Hence since State of New York bond has a higher rate of return;State of New York bond bond
must be invested upon..
Explain the “life” of a secreted protein molecule - trace the pathwa.pdf
Explain the “life” of a secreted protein molecule - trace the pathway in a cell by which a secreted
protein, such as an antibody, would be produced in an animal cell, starting in the nucleus.
Solution
Cells secrete proteins for their own cytosol or for release into surrounding extracellular fluid.
These secretions may include various hormones, digestive enzymes, antibodies, mucus etc.
The transcription process (conversion of DNA to mRNA) occurs inside the nucleus. This newly
formed mRNA is translated into proteins in the cytosol at cellular structures called ribosomes.the
ribosome remains in the cytosol if the protein has to be used within the cell, whereas if it is
destined for secretion outside the cell, the signal sequence (first few amino acids) is synthesized
on the ribosome and then the ribosome docks at the rough endoplasmic reticulum (RER) for
completion of protein synthesis. Proteins enter and cross endoplasmic recticulum (ER)
membrane co-translationally (i.e., they cross ER during synthesis of polypeptide). In the lumen
of ER, proteins undergo glycosylation and molecular chaperones aid in the process of protein
folding. Misfolded proteins are retrotranslocated to the cytosol, where they are degraded by a
proteasome action.
The vesicles with correctly folded proteins then enter the golgi appratus where further
posttranslational modifications occur including cleavage and functionalization. When the protein
moves through the entire Golgi apparatus, it buds off as secretion vesicles. Now, with the help of
cell\'s cytoskeleton, it moves towards the edge of the cell and attaches itself to the membrane.
Eventually, vesicle fuses with the cell membrane releasing its contents out of the cell through
the process of exocytosis.
This secretion pathway is followed for all protein molecules including antibodies..
7. Using to Table 1 to assist you, explain the rational for the diver.pdf
7. Using to Table 1 to assist you, explain the rational for the diversity of the mycobiome, as
compared to the number of bacterial species identified in the lungs of CF patients. This should be
a full paragraph, not one or two sentences.
Solution
From the tabulated data given, it can be clearly noticed that majority of the population suffers
from ascomycete infections. Particularly, the most infectious species can be identified as
Candida albicans followed by C. dubliniensis. Saccharomyces cerevisiae infections are also very
high followed by C. parapsilosis and C. glabrata. Other types of infections are also evident in
patients but these organisms remain the most frequent infective agents for patients with cystic
fibrosis.
Since all of these microbes spread the disease from one organism to another, they are highly
communicable in nature. Also, since very high frequency of some of the species can be noticed
from the data, it is also suggested that the infections are opportunistic in nature i.e. they will
infect the individual as soon as his immunity against these microbes is minimized. Importantly,
since a huge population is suffering form nearly similar kind of infections, it can be suggested
that fixation of these microbial infections is possible in this population which can be clinically
deleterious in advanced stages because eradication of these microbes using multiple antibiotic
treatment would be difficult.
Hence, it can be evaluated from the data that the statisitical figures are clinically alarming and
demand immediate attention..
A. list two vertebrae that spinal nerve t12 travels between .pdf
A. list two vertebrae that spinal nerve t12 travels between
B. Define the function of ground substance in connective tissue proper.
A. list two vertebrae that spinal nerve t12 travels between
B. Define the function of ground substance in connective tissue proper.
B. Define the function of ground substance in connective tissue proper.
Solution
A. Two vertebrae that spinal nerve t12 travels between are-cervical spine vertebrae and lumbar
spine vertebrae
B. Groud substance is an amorphous gel-like substance present in the connective tissue proper,
which surrounds the cells. Glycoproteins and Hydrophilic muco-polysacharides present in the
ground substance attract water to induce variation in viscosity. Groud substance of all connective
tissue proper make intimate contact with the cells. Hence, these cells doesnot occupy distinct
space within the matrix..
Do people from different cultures experience emotions differently O.pdf
Do people from different cultures experience emotions differently? One study designed to
examine this question collected data from 411 college students from five different cultures. The
participants were asked to record, on a 1 (never) to 7 (always) scale, how much of the time they
typically felt eight specific emotions. These were averaged to produce the global emotion score
for each participant. Here is a summary of this measure:
Note that the convention of giving the standard deviations in parentheses after the means saves a
great deal of space in a table such as this. (a) The ANOVA F statistic was reported as 5.57. Give
the degrees of freedom and either an approximate (from Table E) or an exact (from software) P-
value. Sketch a picture of the F distribution that illustrates the P-value. (Round your
P-value to four decimal places.)
What do you conclude?
We have strong evidence at the 5% level that the means are not all the same.
We have no evidence at the 5% level that the means are different.
(b) Without doing any additional formal analysis, describe the pattern in the means that appears
to be responsible for your conclusion in part (a). Are there pairs of means that are quite similar?
European Americans have the highest emotion scores, Indians are in the middle, and the other
three cultures are lowest.
Hispanic Americans have the highest emotion scores, Japanese are in the middle, and the other
three cultures are lowest.
Japanese have the highest emotion scores, Hispanic Americans are in the middle, and the other
three cultures are lowest.
Asian Americans have the highest emotion scores, Japanese are in the middle, and the other
three cultures are lowest.
European Americans have the highest emotion scores, Hispanic Americans are in the middle,
and the other three cultures are lowest. Culture n Mean (s)European American
Asian American
Japanese
Indian
Hispanic American48
32
91
160
804.39 (1.01)
4.35 (1.18)
4.71 (1.13)
4.34 (1.26)
5.04 (1.16)
Solution
(a) numerator df= 5-1=4
denominator df = 48+32+91+160+80-5 =406
P-value = P(F with df=4 and df=406 >5.57 ) =0.0002 (from F table)
We have strong evidence at the 5% level that the means are not all the same.
---------------------------------------------------------------------------------------------------------
(b)Hispanic Americans have the highest emotion scores, Japanese are in the middle, and the
other three cultures are lowest..
Describe the basic elements found in robot controllers. (Choose all t.pdf
Describe the basic elements found in robot controllers. (Choose all that apply.) central
processing unit (CPU) input/output devices nanobots running on treadmills memory teach
station wide-mouth frog
Solution
SOlution :
The components listed of robotic controllers are
CPU
Input/output devices
Memory.
Consider the following probability distribution The Normal distribut.pdf
Consider the following probability distribution The Normal distribution with mean 22.1 &
variance 5.0367 The Binomial distribution with p = 0.5 The x2 distribution on 14 degrees of
freedom which of the above are symmetric distributions? (i), (ii), (iii) (i), (ii) only (i), (iii) only
only (i) only (iii)
Solution
Answer is b.
The chi square distribution is symmetric for df >=30..
1. Please explain 1. What is cosolidation of a soi 2. What are the .pdf
1. Please explain: 1. What is cosolidation of a soi? 2. What are the assumptions for 1-D
consoildation theory proposed by Kar Terzaghi? 3. Compared to compaction, what are the
difference? 4. How do you define the process of a soil consolidation? 2. For settlements, what
are the major components of a soil settlment? What is the difference between between settlment
in sands and clays, please describe the details.
Solution
1) Consolidation is the reduciton of soil volume by expulsion of pore water.
2) assumptions:
3) Compaction is expulsion of air and consolidation is expulsion of water.
4) Process of soil consolidation is not instataneous. It ia time taking process..
What type of gametes did Sutton predict Mendel (parental, recombin.pdf
What type of gametes did Sutton predict? Mendel? (parental, recombinant, both)
Solution
1. According to Mendel’s law of segregation :
2 . The Chromosomal Theory of Inheritance (Theodor Boveri and Walter Sutton --1903) was
consistent with Mendel\'s laws and the basic postulates were:
-----------------------These lead to the conclusion that both Sutton and Mendel predicted parental
types of gametes because in case of parental type , gametes are produced in equal proportion but
recombinant type produces gametes in unequal proportions.
Besides this , In 1911, after observing that linked traits could on occasion be inherited separately,
Thomas Hunt Morgan suggested that \"crossovers\" can occur between linked genes and
crossover is the prime reason for recombination.
With reference to Barth (2014), discuss how the author assessed the .pdf
With reference to Barth (2014), discuss how the author assessed the extent to which unmodified
historical cost, modified historical cost, and fair value measurement possess the qualitative
characteristics specified in the IFRS Conceptual Framework.
Solution
The conceptual framework in IFRS states the qualitative characteristics of the financial
statements to be useful which are as follows:
1. Fundamental : Fundamental qualitative characteristics are relevance and faithful
representation.
2. Enhancing: Enhancing qualitative characteristics are comparability, completeness, verifiablity
and understandability.
According to Barth(2014), Historical cost does not possess the qualitative characteristics of the
IFRS conceptual framework as it does not facilitates comparability and relevance but yes it is
sometimes relevant. The historical cost should be modified to use in financial statements, it
should be indexed for Inflation.
Modified historical cost possess the qualitative characteristics of the IFRS.
It is understandable, complete, accurate, but not considered as faithful representation. This is
widely used but both modified historical cost and unmodified historical cost lack meaning.
Whereas the Fair value presentation is more consistent with the new framework and hence it
should be widely used. This paper concludes that assets and liabilities at large should be
measured at fair value to provide the financial statements with the qualitative characteristics..
1. Peter Grant - the evolutionary biologist who has studied finches .pdf
1. Peter Grant - the evolutionary biologist who has studied finches on Daphne Major (an island
in the Galapagos) for several decades - collected data on birds\' beak sizes, weight, diet and
mating behavior to test his hypotheses. This kind of data is an example of:
2.
In the video \"How Does Evolution Really Work?\", biologists described an example where the
length of individual hummingbird beaks grew longer in response to changes in their food source
(flowers).
True
False
3
The observation that the individuals possessing favorable traits are more likely to survive and
reproduce than those possessing less favorable traits is the basis for:
4.
Which one of the following descriptions, pertaining to a scientific hypothesis or theory, is NOT
true?
5.
Which of the following is involved in the scientific method (MARK ALL THAT APPLY)?
making observations
6.
Due to ___________________, \"bad science” (e.g. fake data or fake results) is eventually
discovered and revealed.
powerful police forces
7.
Darwin\'s Hypothesis of Evolution by Natural Selection included the observation that
populations often grow faster than the food supply and therefore individuals need to compete for
resources. Darwin borrowed this idea from the work of:
8.
Lamarck\'s \"Inheritance of Acquired Characteristics\" can best be described as:
disease and starvation are required to keep population size in check
9.
Aristotle and other early philosophers believed that species were \"immutable\" or \"fixed\",
which means:
species are always changing
species don\'t change over timeA.taxonomy
B.artificial selectionC.empirical evidenceD.non-factual informationE.a belief
Solution
1)B)Artificial Selection as it tells about the way birds survive and their mating.
2)True yes it is rightly stated.
3)D)Natural selection which says traits move from one generation to other.
and the ones with more traits survive more than other
here.
4)E)it must deal with evidence that is empirical (i.e. the kind that can be observed).
COMMUTER PASSES Five different types of monthly commuter passes are o.pdf
COMMUTER PASSES Five different types of monthly commuter passes are offered by a city\'s
local transit authority for each of three different groups of passengers: youths, adults, and senior
citizens. How many different kinds of passes must be printed each month?
Solution
five different types passes are offeres by authority
categories are three
So, diffenerent kinds of passes would be 5 *3 = 15.
If a researcher rejects a null hypothesis when that hypothesis is ac.pdf
If a researcher rejects a null hypothesis when that hypothesis is actually true, he has committed:
a type I error
a type II error
a type III error
no error
a type I error
a type II error
a type III error
no error
Solution
If a researcher rejects a null hypothesis when that hypothesis is actually true, he has committed:
a type I error
a type II error
a type III error
no error
A type I error is the incorrectly rejecting a true null hypothesis.
a type I error
a type II error
a type III error
no error.
Write an algorithm that reads a list of integers from the keyboard, .pdf
Write an algorithm that reads a list of integers from the keyboard, creates
a list of them using linked list implementation, and prints the result
Solution
/*
Algorithm:
Implementation of linked list.
Create a node type class which will hold an integer and next pointer to point to another node.
Create a constructor which will construct an object of type node with the integer type data as the
only parameter
and initializing next pointer to NULL.
Create LinkedList class which will only have head pointer.
Create a constructor which will initialize this head pointer to NULL.
Create a append method which will recieve an input of type integer from the user and append it
to LinkedList
using below steps:
1.Create a Node type object.
1.IF head is NULL assign a Node type object created in 1st step.
2.IF head is not NULL then traverse to the end of the linked list till you found a Node which has
next pointer
set to NULL.
3.Add Node type object created in first step to the next pointer of last node.
Create a print method which will traverse the list from the head of the linked list to the last
node(Node
with next pointer NULL) and while traversing it will keep printing the node data.
Get the userinput:
Ask user for integer continously till user input -1 and append all the integers to the linked list.
*/
PROGRAM:
#include
using namespace std;
class Node
{
public:
int data;
Node *next;
Node(int data)
{
this->data = data;
this->next = NULL;
}
};
class LinkedList
{
private:
Node *head;
public:
LinkedList()
{
head = NULL;
}
void addToFront(int data)
{
Node *temp = new Node(data);
if(head == NULL)
head = temp;
else
{
temp->next = head;
head = temp;
}
}
void append(int data)
{
Node *temp = new Node(data);
Node *current = head;
if(current == NULL)
{
head = temp;
return;
}
while(current->next != NULL)
{
current = current->next;
}
current->next = temp;
}
void print()
{
Node *current = head;
if(current == NULL)
cout << \"No element in list. \ \";
while(current != NULL)
{
cout << current->data << \" \";
current = current->next;
}
cout << endl;
}
void deleteLinkedList()
{
Node *prev;
Node *current = head;
while(current!=NULL)
{
prev = current;
current = current->next;
delete prev;
}
}
void deleteLastNode()
{
Node *current = head;
Node *prev = NULL;
if(current == NULL)
return;
while(current->next != NULL)
{
prev = current;
current = current->next;
}
if(prev == NULL)
{
delete current;
head = NULL;
}
else
{
prev->next = NULL;
delete current;
}
}
~LinkedList()
{
deleteLinkedList();
}
};
int main()
{
LinkedList list;
int data;
while(1)
{
cout << \"Enter the integer(-1 to Quit) : \";
cin >> data;
if(data==-1)
break;
list.append(data);
}
cout << \"Content of LinkedList is : \";
list.print();
return 0;
}
OUTPUT:
Enter the integer(-1 to Quit) : 1
Enter the integer(-1 to Quit) : 2
Enter the integer(-1 to Quit) : 3
Enter the integer(-1 to Quit) : 4
Enter the integer(-1 to Quit) : -1
Content of LinkedList is : 1 2 3 4.
Which of the following values of r allows a perfect prediction of sc.pdf
Which of the following values of r allows a perfect prediction of scores on one variable from
knowledge of scores on the other variable?
Question 13 options:
0
0.50
–1.00
2.00
0
0.50
–1.00
2.00
Solution
ANS: D) –1.00
Reason: Value of second variable is dependden on the result of the first variable. We don\'t see a
realtion between them prior to the result fof first.
Which of the following is a way in which meiosis differs from mitosi.pdf
Which of the following is a way in which meiosis differs from mitosis?
In meiosis, a spindle of microtubules does not form.
Meiosis produces only diploid daughter cells.
Meiosis is found in asexual reproduction.
In meiosis, the parent cell divides twice.
Meiosis never produces gametes (sex cells). Question Part
Solution
Answer is 4
In meiosis parent cell divides twice whereas in mitosis parent cell divides only once.
Meiosis is a reductional division which occurs in diploid cell to produce four tabloid cells
whereas mitosis is a educational division which either occur in haploid or diploid cell to produce
haploid or diploid daughter cells.
Meiosis produces balloon gametes.
Also, meiosis is not found in asexual reproduction.
Meiosis and mitosis both involves formation of spindle of microtubules..
What is the inheritance of the following pedigree X-linked recessiv.pdf
What is the inheritance of the following pedigree? X-linked recessive X-linked dominant
autosomal dominant autosomal recessive Y-link ed
Solution
d. autosomal recessive
In autosomal recessive inheritance, two copies of a disease allele are required for an individual to
be susceptible to expressing the phenotype..
What is the profitability index for an investment with the following.pdf
What is the profitability index for an investment with the following cash flows given a 6 percent
required return?
1.38
0.72
1.06
0.81
1.23
What is the profitability index for an investment with the following cash flows given a 6 percent
required return?
Solution
Present value of inflows=cash inflow*Present value of discounting factor(rate%,time period)
15700/1.06+18000/1.06^2+17400/1.06^3
=$45440.6322
PI=Present value of inflows/Present value of outflows
=$45440.6322/$37000
which is equal to
=1.23(Approx)..
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Firefly luciferase is a commonly used reporter gene. The gene origina.pdf
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- 1. Match the following DNA-related terms to their descriptions or definitions: a. A cell containing only ONE copy of each chromosome. b. An individual carrying 2 of the SAME allele for a given trait. c. Pieces of DNA that contain genes. d. The OBSERVABLE characteristics of a trait. e. The process in meiosis that forms tetrads. f. Unwound, readable DNA. g. A pair of chromosomes carrying the same GENES, but not necessarily the same alleles. h. Alternate forms of a trait (Blond hair, Brown hair) i. A problem in meiosis when chromatids DO NOT separate appropriately during anaphase. j. An IDENTICAL copy of a chromosome. k. A cell containing TWO copies of each chromosome. l. The alleles present in an organism; NOT visible m. A segment of DNA that codes for a single trait. n. An individual carrying 2 DIFFERENT alleles for a given trait. o. Any situation of incorrect chromosome number. ______ Chromosome ______ Chromatin ______ Chromatid ______ Gene ______ Allele ______ Diploid ______ Haploid ______ Non-disjunction ______ Synapsis ______ Aneuploidy ______ Homologous ______ Homozygous ______ Heterozygous ______ Genotype ______ Phenotype Solution a. A cell containing only ONE copy of each chromosome.
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