1.) The normal eye color of Drosophila is red, but strains in which .pdf
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1.) The normal eye color of Drosophila is red, but strains in which all flies have brown eyes are available. Similarly, wings are normally long, but there are strains with short wings. A female from a pure line with brown eyes and short wings is crossed with a male from a normal pure line. The F1 consists of normal females and short-winged males. An F2 is then produced by intercrossing the F1. Both sexes of F2 flies show phenotypes as follows: 3/8 red eyes, long wings 3/8 red eyes, short wings 1/8 brown eyes, long wings 1/8 brown eyes, short wings Deduce the inheritance of these phenotypes; use clearly defined genetic symbols of your own invention. State the genotypes of all three generations and the genotypic proportions of the F1 and F2. 2.) When a cell of genotype A/a ; B/b ; C/c having all the genes on separate chromosome pairs divides mitotically, what are the genotypes of the daughter cells? 3.) A presumed dihybrid in Drosophila, B/b ; F/f, is test-crossed with b/b ; f/f. (B = black body ; b = brown body; F = forked bristles; f = unforked bristles.) The results are black, forked 230 brown, forked 240 black, unforked 210 brown, unforked 250 Use the 2 test to determine if these results fit the results expected from testcrossing the hypothesized dihybrid. 4.) In mice, dwarfism is caused by an X-linked recessive allele, and pink coat is caused by an autosomal dominant allele (coats are normally brownish). If a dwarf female from a pure line is crossed with a pink male from a pure line, what will be the phenotypic ratios in the F1 and F2 in each sex? (Invent and define your own gene symbols.) Solution Answer 2. If a cell divides mitotically (equational division), the genetic constitution of the daughter cells remains the same as the parents. So a parent with the genotype A/a; B/b; C/c will give rise to daughter cells with the genotype A/a; B/b; C/c. However, if the cell was to divide meiotically (reductional division), the genetic constitution would be different as only one half of the parental chromosomes goes to the daughter cell. So the genotypes would be as follows: ABC or AbC or ABc or Abc or aBC or abC or aBc or abc Answer 3. For the given cross BbFf X bbff For such a test cross the resultant phenotypic ration obtained is 1 (BF):1 (Bf):1 (bF):1 (bf). For dihybrid cross, the degree of freedom will be 4 – 1 = 3 Chi square test is as follows BF Bf bF bf Totals Observed phenotype 230 210 240 250 930 Expected phenotype 930x ¼= 232.5 930x ¼= 232.5 930x ¼= 232.5 930x ¼= 232.5 930 Observed – expected -2.5 -22.5 7.5 17.5 (Obs – Exp)2 Expected 6.25/232.5= 0.0268 506.25/232.5= 2.1774 56.25/232.5= 0.2419 306.25/232.5= 1.3172 3.7633 Looking at the probability value table for degree of freedom = 3, df .90 .70 .50 .30 .20 .10 .05 .01 3 0.58 1.42 2.37 3.67 4.64 6.25 7.82 11.35 Hence we can deduce that there is more than 20% probability for the occurrence of this event. Answer 4. Let XD = Normal phenotype (tall) Xd = Dwarf phenotype P = Pink coat p = Br.
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Drosophila - Laboratory Report 2
- The student conducted an experiment crossing Drosophila F1 phenotypes to obtain F2 phenotypes.
- They counted 53 wild type flies and 19 sepia flies in the F2 generation.
- A chi-square test found the observed ratio of phenotypes (53:19) did not significantly deviate from the expected Mendelian ratio of 3:1, with a chi-square value of 0.074 and p-value above 0.05.
- Therefore, the results supported the hypothesis that the sepia mutation behaves as an autosomal recessive trait.
this is all of the information that I have please help Lab 5 In.pdf
this is all of the information that I have please help Lab 5 Introduction - Genetic mapping See
figure 5.1 for a schematic of the fly the cross you initially started with, you'll either crosses you
have been working on. Two labs ago, map the distance between the w gene and the m you set up
a pair of reciprocal parental crosses, gene, or between the w gene and the y gene. between mutant
and wild type flies (fig. 5.1a). You had one of two different mutant strains, each with two mutant
phenotypes - either white eyes all F2 individuals will receive only recessive (w) and miniature
wings (m), or white eyes (w) alleles from this parent (fig. 5.1d, orange and and yellow body (y).
The phenotypes of the F1 yellow chromosomes). Because of this, the flies should have indicated
to you that all mutant phenotype of each F2 fly will tell you which phenotypes in question are x-
linked recessive alleles (mutant or WT) were inherited from the (fig. 5.1b). heterozygous female
F1 parent (fig. 5.1d, dark and light blue chromosomes). The first F2 fly Last lab, you used the F1
flies from one of shown in figure 5.1d inherited 'a B' from the your parental crosses to set up an
F1 cross (fig. heterozygous parent and will end up with the consequence. After crossing two pure
breeding this F2, you observe a ABphenotype and parents, F1 offspring will be heterozygous for
therefore know that this fly received 'A B' from nearly all genes in question - the exception is X -
the heterozygous parent. linked genes in the male offspring. Since the Y chromosome is
equivalent to recessive alleles for The goal of genetic mapping is to X-linked genes, these F1
males are recessive for determine the likelihood of cross over between all X-linked genes and act
as a test cross. The two loci/genes. If we score the phenotypes of a heterozygous F1 females and
recessive F1 males large F2 population from our crosses, we can (test cross) can be used to map
the distance determine the recombination frequency of your between the genes causing the two
phenotypes two genes. of your parental mutant female. Depending on
e) F2 phenotype scoring: f) Recombination frequency: Eigure 5.1: Schematic of your Drosophila
crosses, See text of lab 5 intro for description.
For a given F1 gamete for the F2 individual it number of flies, it is easy to calculate the creates),
if no cross over occurs between the two recombinant frequency between your two genes genes in
question, the F2 phenotype will be the (fig. 5.1ef ). same as one of the original parents - either
fully WT or double mutant in this case. In figure 5.1d Recall from last time that a lower
recombinant these have blue chromosomes of a single colour. frequency is observed when
genetic map If a crossover does occur between the two genes, distances are small. When genes
are close to the F2 fly will have a phenotype unlike either of each other, there's a narrow range
on the the parents - a recombinant phenotype (shown chromosome for a random crossover to
land.
Linkage and recombination of gene
- A three-point test cross was conducted in Drosophila to map the locations of three genes - black body (b), purple eyes (pr), and curved wings (c) - on the same chromosome.
- Offspring were classified into eight phenotypic classes. Reciprocal classes occurred in roughly equal numbers, indicating recombination between loci.
- Map distances were calculated between loci based on percentages of recombinants. A discrepancy in the direct vs. indirect distance between b and c was resolved by accounting for double crossovers.
- The study demonstrated how three-point test crosses can order loci and more accurately measure map distances compared to two-point crosses by detecting double crossovers.
Genetics chapter 5 part 1
This document provides an overview of genetic linkage and mapping. It discusses how Thomas Morgan established the chromosome theory of inheritance by discovering genetic linkage between genes located on the same chromosome. Genes located close together on a chromosome assort together more often than genes farther apart due to less recombination between linked genes. The document describes Morgan's early experiments in Drosophila that demonstrated genetic linkage between eye color and wing size genes. It also explains how Alfred Sturtevant created the first genetic linkage map by quantifying recombination frequencies between linked genes on the X chromosome. The concept of map units (centiMorgans) is introduced as a way to represent the physical distance between genes based on their recombination rate.
Bft1033 6 linkage genes_print
Morgan mapped three genes (yellow body (y), white eyes (w), miniature wings (m)) in Drosophila using genetic crosses. He found that the genes exhibited different recombination frequencies: w-y was 0.5%, w-m was 34.5%, and m-y was 35.4%. These recombination frequencies were used to construct the first genetic map, with map units (mu) representing the distance between genes. Morgan's work demonstrated that genes located farther apart on a chromosome recombine more frequently during meiosis than genes located closer together, establishing the basis for using recombination frequencies to determine gene order and distance.
Figure 3- Ventral view of the female and male abdomen Virgin females-.docx
Figure 3: Ventral view of the female and male abdomen Virgin females: Difference between a virgin and a mated female fruit fly. Virgin (left): pale in colour, black spot in abdomen (meconium) and often bloated. Mated (right): more pigmented, no meconium. (Source: Twitter) Crosses: For this lab we are observing and crossing the F 1 offspring. The parents and some individuals from the offspring will show the phenotypic traits for the mutations, some will show the wildtype phenotype. The parental cross and F 1 are showed below: Dihybrid cross 1: P l = apterous (vestigial wings showed) x sepia Dihybrid cross 2 : P l = wildtype x white Cross 1: sepia male x apterous virgin female 9 Cross 2: white virgin female 9 x wildtype male So, what is your observed ratio? To determine the proportion each class represents of the total number of offspring, divide \# observed/total offspring: Proportion of wildtype 171/220 = 0.78 , which can also be shown as 78% (multiply by 100 ) Proportion of apterous 49/220 = 0.22 , which can also be shown as 22% To calculate the observed ratio in the offspring among the phenotypic classes: 1. Write the classes with the corresponding numbers as a ratio: 171 wildtype: 49 apterous. 2. Simplify the ratio by dividing each number by the smallest number in the ratio. So, 171/49 = 3.5 and 49/49 = 1 . So, the simplified ratio is 3.5 wildtype: 1 apterous. This is obviously close to a 3:1 ratio, but is it close enough to support your hypothesis? This is the point at which you would do your Chi-Square calculations. Review questions (5 pts) 1) What is the Parental cross genotype and phenotype? What would you expect for the genotype and phenotype about the F 1 and F 2 ? 2) What are the expected rations for a monohybrid and for a dihybrid cross?
.
Concepts of Genetics 10th Edition Klug Test Bank
This document contains 36 multiple choice questions about basic concepts in Mendelian genetics. The questions cover topics such as: influential scientists in early genetics; Mendel's work with pea plants; monohybrid and dihybrid crosses; phenotypic and genotypic ratios; independent assortment; Chi-square tests; and probabilities in human genetics. Sample questions are provided along with their answers to illustrate the scope and level of detail assessed.
Problems2. This is a map for a diploid plantR--------35-------.pdf
Problems:
2. This is a map for a diploid plant:
R--------35--------E--------17---------F
One parental plant is R e f/r E F. It is mated with a homozygous recessive plant: r e f/r e f. Give
the proportion of the offspring assuming there is both no interference, and complete interference.
3. In a newly discovered insect, the three genes R, A, and P located on the same chromosome
(R= red body, A= spotted coloring, P=purple eyes, r=white body, a=solid coloring, p=white
eyes). A heterozygous parent(R A P/r a p) is mated with a homozygous recessive parent (r a p/ r
a p).
The following offspring are shown below:
(210) dominant for all traits
(16) red body, solid coloring, purple eyes
(70) red body, spotted coloring, white eyes
(50) red body, solid coloring, white eyes
(200) recessive for all traits
(55) white body, spotted coloring, purple eyes
(13) white body, spotted coloring, white eyes
(72) white body, solid coloring, purple eyes
Perform a three-point linkage to determine the gene order and relative distance between the
genes. How did the distance between the 2 genes affect the frequency of crossing-over events?
4. Here are four recombinant sequences that came from four Hfr matings during an interrupted
mating experiment. Determine the order and relative distance by setting A to 0/100 minutes.
Hfr1) A T1 H T2
8 26 33 53
Hfr2) E1 T1 E2 E3
4 12 24 54
Hfr3) T2 S T3 A
7 37 52 62
Hfr4) T3 C E1 E3
2 17 22 72
5. In a u-tube experiment, a researcher placed a kanamycin resistant strain of E. coli on one side
of a membrane impervious to the cells, and a wild-type strain on the other side. After a day, he
then plated a sample from both sides of the tube on plates containing kanamycin and observed
growth on both plates. He then repeated this experiment, but this time he added DNase to the
tube and observed the same results when he plated both cell types. What happened inside the
tube?
6. Pretend the following is a chromosome that underwent some changes. What were they? (the
dot is the centromere).
A B C D E F . G H I J A B C G . F E H I J
a. pericentric inversion, deletion
b. paracentric inversion, deletion
c. deletion, translocation
d. Robertsonian translocation,
pericntric inversion
7. Given:
Type of mutation Cistron A or B
1. point A
2. point A
3. point B
4. point B
5. deletion A
Results from 5 crosses:
1x2 (plaque) 2x3 (plaque) 3x4 (plaque) 4x5 (plaque)
1x3 (plaque) 2x4 (plaque) 3x5 (plaque)
1x4 (plaque) 2x5 (no plaque)
1x5 (plaque)
8. The following nucleotide sequence is found in a piece of DNA
5’-TTAACA-3’
3’-AATTGT-5’
if the cytosince undergoes a tautameric shift for one replication cycle draw two round of
replication and explain what type of mutation occurred (Transition or Transversion)?
Bonus:
Explain the relationship between mosaicism and x-inactivation using an example.
9. In a diploid fly, the three loci R E and D are linked as follows
R--------------22cm------------E----------16cm-------------D
One fly is available to you and its genotype is R.
Recommended
Drosophila - Laboratory Report 2
- The student conducted an experiment crossing Drosophila F1 phenotypes to obtain F2 phenotypes.
- They counted 53 wild type flies and 19 sepia flies in the F2 generation.
- A chi-square test found the observed ratio of phenotypes (53:19) did not significantly deviate from the expected Mendelian ratio of 3:1, with a chi-square value of 0.074 and p-value above 0.05.
- Therefore, the results supported the hypothesis that the sepia mutation behaves as an autosomal recessive trait.
this is all of the information that I have please help Lab 5 In.pdf
this is all of the information that I have please help Lab 5 Introduction - Genetic mapping See
figure 5.1 for a schematic of the fly the cross you initially started with, you'll either crosses you
have been working on. Two labs ago, map the distance between the w gene and the m you set up
a pair of reciprocal parental crosses, gene, or between the w gene and the y gene. between mutant
and wild type flies (fig. 5.1a). You had one of two different mutant strains, each with two mutant
phenotypes - either white eyes all F2 individuals will receive only recessive (w) and miniature
wings (m), or white eyes (w) alleles from this parent (fig. 5.1d, orange and and yellow body (y).
The phenotypes of the F1 yellow chromosomes). Because of this, the flies should have indicated
to you that all mutant phenotype of each F2 fly will tell you which phenotypes in question are x-
linked recessive alleles (mutant or WT) were inherited from the (fig. 5.1b). heterozygous female
F1 parent (fig. 5.1d, dark and light blue chromosomes). The first F2 fly Last lab, you used the F1
flies from one of shown in figure 5.1d inherited 'a B' from the your parental crosses to set up an
F1 cross (fig. heterozygous parent and will end up with the consequence. After crossing two pure
breeding this F2, you observe a ABphenotype and parents, F1 offspring will be heterozygous for
therefore know that this fly received 'A B' from nearly all genes in question - the exception is X -
the heterozygous parent. linked genes in the male offspring. Since the Y chromosome is
equivalent to recessive alleles for The goal of genetic mapping is to X-linked genes, these F1
males are recessive for determine the likelihood of cross over between all X-linked genes and act
as a test cross. The two loci/genes. If we score the phenotypes of a heterozygous F1 females and
recessive F1 males large F2 population from our crosses, we can (test cross) can be used to map
the distance determine the recombination frequency of your between the genes causing the two
phenotypes two genes. of your parental mutant female. Depending on
e) F2 phenotype scoring: f) Recombination frequency: Eigure 5.1: Schematic of your Drosophila
crosses, See text of lab 5 intro for description.
For a given F1 gamete for the F2 individual it number of flies, it is easy to calculate the creates),
if no cross over occurs between the two recombinant frequency between your two genes genes in
question, the F2 phenotype will be the (fig. 5.1ef ). same as one of the original parents - either
fully WT or double mutant in this case. In figure 5.1d Recall from last time that a lower
recombinant these have blue chromosomes of a single colour. frequency is observed when
genetic map If a crossover does occur between the two genes, distances are small. When genes
are close to the F2 fly will have a phenotype unlike either of each other, there's a narrow range
on the the parents - a recombinant phenotype (shown chromosome for a random crossover to
land.
Linkage and recombination of gene
- A three-point test cross was conducted in Drosophila to map the locations of three genes - black body (b), purple eyes (pr), and curved wings (c) - on the same chromosome.
- Offspring were classified into eight phenotypic classes. Reciprocal classes occurred in roughly equal numbers, indicating recombination between loci.
- Map distances were calculated between loci based on percentages of recombinants. A discrepancy in the direct vs. indirect distance between b and c was resolved by accounting for double crossovers.
- The study demonstrated how three-point test crosses can order loci and more accurately measure map distances compared to two-point crosses by detecting double crossovers.
Genetics chapter 5 part 1
This document provides an overview of genetic linkage and mapping. It discusses how Thomas Morgan established the chromosome theory of inheritance by discovering genetic linkage between genes located on the same chromosome. Genes located close together on a chromosome assort together more often than genes farther apart due to less recombination between linked genes. The document describes Morgan's early experiments in Drosophila that demonstrated genetic linkage between eye color and wing size genes. It also explains how Alfred Sturtevant created the first genetic linkage map by quantifying recombination frequencies between linked genes on the X chromosome. The concept of map units (centiMorgans) is introduced as a way to represent the physical distance between genes based on their recombination rate.
Bft1033 6 linkage genes_print
Morgan mapped three genes (yellow body (y), white eyes (w), miniature wings (m)) in Drosophila using genetic crosses. He found that the genes exhibited different recombination frequencies: w-y was 0.5%, w-m was 34.5%, and m-y was 35.4%. These recombination frequencies were used to construct the first genetic map, with map units (mu) representing the distance between genes. Morgan's work demonstrated that genes located farther apart on a chromosome recombine more frequently during meiosis than genes located closer together, establishing the basis for using recombination frequencies to determine gene order and distance.
Figure 3- Ventral view of the female and male abdomen Virgin females-.docx
Figure 3: Ventral view of the female and male abdomen Virgin females: Difference between a virgin and a mated female fruit fly. Virgin (left): pale in colour, black spot in abdomen (meconium) and often bloated. Mated (right): more pigmented, no meconium. (Source: Twitter) Crosses: For this lab we are observing and crossing the F 1 offspring. The parents and some individuals from the offspring will show the phenotypic traits for the mutations, some will show the wildtype phenotype. The parental cross and F 1 are showed below: Dihybrid cross 1: P l = apterous (vestigial wings showed) x sepia Dihybrid cross 2 : P l = wildtype x white Cross 1: sepia male x apterous virgin female 9 Cross 2: white virgin female 9 x wildtype male So, what is your observed ratio? To determine the proportion each class represents of the total number of offspring, divide \# observed/total offspring: Proportion of wildtype 171/220 = 0.78 , which can also be shown as 78% (multiply by 100 ) Proportion of apterous 49/220 = 0.22 , which can also be shown as 22% To calculate the observed ratio in the offspring among the phenotypic classes: 1. Write the classes with the corresponding numbers as a ratio: 171 wildtype: 49 apterous. 2. Simplify the ratio by dividing each number by the smallest number in the ratio. So, 171/49 = 3.5 and 49/49 = 1 . So, the simplified ratio is 3.5 wildtype: 1 apterous. This is obviously close to a 3:1 ratio, but is it close enough to support your hypothesis? This is the point at which you would do your Chi-Square calculations. Review questions (5 pts) 1) What is the Parental cross genotype and phenotype? What would you expect for the genotype and phenotype about the F 1 and F 2 ? 2) What are the expected rations for a monohybrid and for a dihybrid cross?
.
Concepts of Genetics 10th Edition Klug Test Bank
This document contains 36 multiple choice questions about basic concepts in Mendelian genetics. The questions cover topics such as: influential scientists in early genetics; Mendel's work with pea plants; monohybrid and dihybrid crosses; phenotypic and genotypic ratios; independent assortment; Chi-square tests; and probabilities in human genetics. Sample questions are provided along with their answers to illustrate the scope and level of detail assessed.
Problems2. This is a map for a diploid plantR--------35-------.pdf
Problems:
2. This is a map for a diploid plant:
R--------35--------E--------17---------F
One parental plant is R e f/r E F. It is mated with a homozygous recessive plant: r e f/r e f. Give
the proportion of the offspring assuming there is both no interference, and complete interference.
3. In a newly discovered insect, the three genes R, A, and P located on the same chromosome
(R= red body, A= spotted coloring, P=purple eyes, r=white body, a=solid coloring, p=white
eyes). A heterozygous parent(R A P/r a p) is mated with a homozygous recessive parent (r a p/ r
a p).
The following offspring are shown below:
(210) dominant for all traits
(16) red body, solid coloring, purple eyes
(70) red body, spotted coloring, white eyes
(50) red body, solid coloring, white eyes
(200) recessive for all traits
(55) white body, spotted coloring, purple eyes
(13) white body, spotted coloring, white eyes
(72) white body, solid coloring, purple eyes
Perform a three-point linkage to determine the gene order and relative distance between the
genes. How did the distance between the 2 genes affect the frequency of crossing-over events?
4. Here are four recombinant sequences that came from four Hfr matings during an interrupted
mating experiment. Determine the order and relative distance by setting A to 0/100 minutes.
Hfr1) A T1 H T2
8 26 33 53
Hfr2) E1 T1 E2 E3
4 12 24 54
Hfr3) T2 S T3 A
7 37 52 62
Hfr4) T3 C E1 E3
2 17 22 72
5. In a u-tube experiment, a researcher placed a kanamycin resistant strain of E. coli on one side
of a membrane impervious to the cells, and a wild-type strain on the other side. After a day, he
then plated a sample from both sides of the tube on plates containing kanamycin and observed
growth on both plates. He then repeated this experiment, but this time he added DNase to the
tube and observed the same results when he plated both cell types. What happened inside the
tube?
6. Pretend the following is a chromosome that underwent some changes. What were they? (the
dot is the centromere).
A B C D E F . G H I J A B C G . F E H I J
a. pericentric inversion, deletion
b. paracentric inversion, deletion
c. deletion, translocation
d. Robertsonian translocation,
pericntric inversion
7. Given:
Type of mutation Cistron A or B
1. point A
2. point A
3. point B
4. point B
5. deletion A
Results from 5 crosses:
1x2 (plaque) 2x3 (plaque) 3x4 (plaque) 4x5 (plaque)
1x3 (plaque) 2x4 (plaque) 3x5 (plaque)
1x4 (plaque) 2x5 (no plaque)
1x5 (plaque)
8. The following nucleotide sequence is found in a piece of DNA
5’-TTAACA-3’
3’-AATTGT-5’
if the cytosince undergoes a tautameric shift for one replication cycle draw two round of
replication and explain what type of mutation occurred (Transition or Transversion)?
Bonus:
Explain the relationship between mosaicism and x-inactivation using an example.
9. In a diploid fly, the three loci R E and D are linked as follows
R--------------22cm------------E----------16cm-------------D
One fly is available to you and its genotype is R.
Chromosomal mutations
Variations in chromosome structure or number can arise spontaneously or be induced. There are four main types of chromosomal mutations: rearrangements, aneuploidy, polyploidy, and changes in chromosome number. Chromosomal mutations can cause genetic disorders and contribute to human miscarriages and stillbirths. Specific examples of chromosomal mutations include deletions, duplications, inversions, translocations, and changes in chromosome number like trisomy, monosomy, and polyploidy. These mutations begin with breaks in DNA that are repaired, sometimes incorrectly.
AP Biology Ch 12 gene linkage groups and chromosome maps
1. The document discusses linkage groups and chromosome mapping in genetics. It provides examples of crosses involving fruit flies that demonstrate genes located on the same chromosome tend to be inherited together, while genes on different chromosomes assort independently.
2. The document explains how observing recombinant vs parental offspring types allows geneticists to determine if genes are linked and calculate the distance between genes on a chromosome.
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Geneticschapter5part1 140222103829-phpapp02
Morgan established the chromosome theory of inheritance through his work with genetic linkage and mapping in fruit flies. His student Sturtevant created the first genetic linkage map by ordering genes based on their recombination frequencies on fruit fly chromosomes. Genetic linkage mapping is based on the principle that genes located closer together on a chromosome assort together more often than genes farther apart due to less recombination between linked genes. This technique allows genes to be ordered and distances between genes to be measured in centiMorgans based on recombination frequencies observed in test crosses.
AP BIOLOGY MENDELIAN GENETICS AND 2 Student Packet.pdf
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The genetics eye color drosohila 15 pollock
- This document outlines a genetics experiment using Drosophila melanogaster to study eye color inheritance.
- Students will set up crosses between fly strains with different eye colors and analyze the results over multiple generations.
- In week 3, chromatography will be used to separate pterin eye pigments from fly strains, which will then be identified under UV light based on their fluorescent colors.
Genetics
Genetic crosses can be used to predict the likelihood of offspring inheriting different traits from their parents. A monohybrid cross looks at inheritance of one trait, while a dihybrid cross examines two traits. Punnett squares are used to display all possible combinations of alleles from each parent. For monohybrid crosses involving homozygous parents, the offspring will all be heterozygous. Dihybrid crosses have multiple genotype and phenotype ratios depending on if the parents are homozygous or heterozygous.
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1. The document discusses genetics concepts like linkage groups, chromosome mapping, parental and recombinant types. It provides examples of genetic crosses in pea plants and fruit flies to demonstrate these concepts.
2. Recombinant types result from crossovers during meiosis, where alleles switch positions on homologous chromosomes. The frequency of recombinants allows calculation of distance between genes.
3. A sample genetics problem is presented, asking the reader to analyze a dihybrid cross in pea plants and determine the chromosome arrangement and distance between genes.
Genetics Review.pdf
1. The document provides an overview of genetics concepts including: Mendelian inheritance patterns; DNA structure and replication; gene expression and regulation; and modern applications like biotechnology.
2. Key topics covered include alleles and inheritance ratios; meiosis; linkage and crossing over; mutations and their effects; the central dogma of DNA to protein; gene regulation in prokaryotes and eukaryotes; and recombinant DNA techniques.
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Genetics Review.pdf
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2. It also includes sample multiple choice questions testing these concepts, such as questions about inheritance patterns demonstrated in genetic crosses, DNA replication, transcription and translation processes, and recombinant DNA techniques.
3. The review concludes with sample free response questions asking students to explain meiosis, chromosomal abnormalities resulting from defects in meiosis, and how the human genome demonstrates both continuity and variation.
Recombination and LinkageA Three point test cross in Drosophil.docx
A three-point test cross was conducted in Drosophila to determine the order and distance between three linked genes (y, cv, and f) on the X chromosome. F2 progeny were scored for phenotypes to determine the number of each genotype. Recombination frequencies and a genetic map were constructed based on the data. Chi-square tests compared the observed and expected values, and some deviations from expectations were observed. The results provided insight into genetic mapping and factors that influence mapping accuracy.
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The document discusses restriction fragment length polymorphism (RFLP) analysis, which refers to variations in restriction sites between individuals in a population. These variations are useful for genetic mapping as they allow researchers to distinguish between chromosomes from different individuals. The document provides examples of how RFLP analysis can be used to map genes and determine genetic distances between loci by examining patterns produced by restriction enzyme digestion and southern blotting.
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- At least one child out of three will be phenotypically normal.
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There are two main ways heritable variations can arise: independent assortment of genes during meiosis and fertilization, which introduces new combinations of parental genes, and spontaneous gene and chromosome mutations. Independent assortment occurs when parental chromosomes separate during meiosis and recombine in new configurations during fertilization. Gene mutations happen due to errors in DNA replication, while chromosome mutations result from changes like gains or losses of chromosomes. Both types of mutations can lead to new traits.
Genetics
This document contains genetics concept questions and their answers related to topics like Mendelian inheritance, sex determination, extensions of Mendelian inheritance, and blood types. Some example questions include the probability of offspring genotypes from given parental crosses, determining heterozygotes vs homozygotes, inheritance patterns like sex-influenced traits, and solving genetics problems using rules of probability. The responses provide detailed explanations and use techniques like Punnett squares, product rule, and binomial/multinomial expansions to calculate the requested probabilities.
BIO Virtual fly lab (final draft)
The document summarizes Gregor Mendel's experiments with pea plants and how they laid the foundation for genetics. It then describes how the author will simulate Mendel's experiments virtually using fruit flies. The author performs a series of crosses between fruit flies with different traits and uses a chi-square analysis to determine if the results match expected Mendelian ratios. The crosses are designed to show traits behaving as dominant, recessive, recessive sex-linked, and dihybrid traits.
CODOMINANCE.pptx
Codominance occurs when two alleles are simultaneously expressed in a heterozygote. For example, some chickens have black and white feathers that are speckled due to codominance - the alleles for black (B) and white (W) feathers are both expressed at the same time in chickens with genotype BW. If two speckled chickens were crossed, there would be a 50% probability that their offspring would be speckled and a 25% probability of being either black or white.
Pre-AP_Biology_Unit_04_-_Genetics_Practice_Set (1).pdf
This document provides genetics practice problems and examples related to monohybrid and dihybrid crosses, inheritance of sex-linked traits, blood typing, and other genetics concepts. It begins with practice problems identifying genotypes and predicting phenotypes based on genetic crosses. The document then provides step-by-step examples and explanations of monohybrid and dihybrid crosses, inheritance patterns such as incomplete dominance and codominance, sex-linked traits, and multiple alleles as seen in human blood typing. Students are to complete related practice problems and predict outcomes of genetic crosses based on provided genotypes.
Pre-AP_Biology_Unit_04_-_Genetics_Practice_Set.pdf
This document provides genetics practice problems and instructions for solving monohybrid and dihybrid cross problems. It includes examples of setting up Punnett squares and predicting offspring genotypes and phenotypes for crosses involving one or two gene traits. Students are asked to solve sample genetics problems involving traits such as pea pod shape, eye color in mice, and fur color/texture in guinea pigs. The document emphasizes following a five step process: determining parent genotypes, setting up a Punnett square, predicting offspring genotypes and phenotypes, and working backwards from given offspring data.
It was noted that excess water can kill plants. If water is not toxi.pdf
It was noted that excess water can kill plants. If water is not toxic, and if all plants require water,
then how can excess water kill plants? What is actually killing the plants?
Solution
When excess water is present, the roots will be deprived of oxygen because the water fills the
pores in the soil. This mostly occurs in clay soils. Also, excess water drain out the nutrients and
promote root rot disease, which gradually kills the plant..
In biology, the roles of transition metal ions can be broadly groupe.pdf
In biology, the roles of transition metal ions can be broadly grouped into two classifications:
redox chemistry (catalysis, electron transfer, etc.) and acid-base chemistry (catalysis, structural,
etc.).
A) List three metal ions that commonly participate in biological redox chemistry, and give
examples of how they participate.
B) Compare this list to metal ions commonly participating in acid-base chemistry. Discuss the
overlap of these two groups of metal ions. (e.g. Why are some used for one of these functions
and not for the other?)
C) List some common biological ligands of these metal ions. Can you see patterns that would
allow you to group these ligands into sets with common properties, or that bind common metals?
D) Why are trace metals used for these biological functions, and not other more-abundant
elements like C,N,O, etc?
Solution
(a) iron, copper, cobalt, manganese etc are used in biology redox reactions like molecular di
oxygen, superoxide, hydrogen peroxide thus inhibiting the effect or formation of these
compound by use of the certain other enzymes..
(c) the ions and the Ligands for them are Ike gluthaione peroxide which can bind to chromium,
arsenic etc.. Iron binds to ferrodoxin which is associated with the photosynthesis..
(d) transition metals are used because they have empty d-orbitals in with them due to which they
can easily transfer or accept the electrons or exchange the electrons among the metals and
enzymes.. Which is not possible in other trace metals like carbon, nitrogen etc...
More Related Content
Similar to 1.) The normal eye color of Drosophila is red, but strains in which .pdf
Chromosomal mutations
Variations in chromosome structure or number can arise spontaneously or be induced. There are four main types of chromosomal mutations: rearrangements, aneuploidy, polyploidy, and changes in chromosome number. Chromosomal mutations can cause genetic disorders and contribute to human miscarriages and stillbirths. Specific examples of chromosomal mutations include deletions, duplications, inversions, translocations, and changes in chromosome number like trisomy, monosomy, and polyploidy. These mutations begin with breaks in DNA that are repaired, sometimes incorrectly.
AP Biology Ch 12 gene linkage groups and chromosome maps
1. The document discusses linkage groups and chromosome mapping in genetics. It provides examples of crosses involving fruit flies that demonstrate genes located on the same chromosome tend to be inherited together, while genes on different chromosomes assort independently.
2. The document explains how observing recombinant vs parental offspring types allows geneticists to determine if genes are linked and calculate the distance between genes on a chromosome.
3. It provides practice problems demonstrating crosses to determine linkage and asks the reader questions about linkage and chromosome mapping concepts.
Genetics Notes
1. The document explains genetic crosses and Mendelian inheritance through examples of monohybrid and dihybrid crosses in pea plants.
2. It discusses key genetics concepts like genotype, phenotype, probability, Punnett squares, and inheritance patterns like complete dominance, incomplete dominance, and codominance.
3. As an example, a monohybrid cross between a homozygous dominant purple pea plant and a homozygous recessive white pea plant would result in all heterozygous purple offspring.
Geneticschapter5part1 140222103829-phpapp02
Morgan established the chromosome theory of inheritance through his work with genetic linkage and mapping in fruit flies. His student Sturtevant created the first genetic linkage map by ordering genes based on their recombination frequencies on fruit fly chromosomes. Genetic linkage mapping is based on the principle that genes located closer together on a chromosome assort together more often than genes farther apart due to less recombination between linked genes. This technique allows genes to be ordered and distances between genes to be measured in centiMorgans based on recombination frequencies observed in test crosses.
AP BIOLOGY MENDELIAN GENETICS AND 2 Student Packet.pdf
Gregor Mendel conducted experiments on pea plants that demonstrated basic principles of heredity. He found that traits are transmitted through discrete factors (now called genes) and that these factors assort and segregate independently during the formation of gametes. Mendel's work established the foundations of classical or Mendelian genetics, including the laws of segregation and independent assortment. Later, scientists developed additional genetic concepts and tools, such as genetic linkage maps and the chi-square test, which allow for more sophisticated analysis of inheritance patterns.
Biology Chromosomes fertilization
Fertilization occurs when male and female gametes combine. In animals, the male gamete is sperm and the female is an ovum. In plants, the male gamete is a pollen cell and the female is an egg cell. When gametes combine, the resulting cell is a zygote. Meiosis reduces the number of chromosomes in gametes so that the zygote has the normal number after fertilization. This, along with genetic recombination during meiosis, introduces variation among offspring.
The genetics eye color drosohila 15 pollock
- This document outlines a genetics experiment using Drosophila melanogaster to study eye color inheritance.
- Students will set up crosses between fly strains with different eye colors and analyze the results over multiple generations.
- In week 3, chromatography will be used to separate pterin eye pigments from fly strains, which will then be identified under UV light based on their fluorescent colors.
Genetics
Genetic crosses can be used to predict the likelihood of offspring inheriting different traits from their parents. A monohybrid cross looks at inheritance of one trait, while a dihybrid cross examines two traits. Punnett squares are used to display all possible combinations of alleles from each parent. For monohybrid crosses involving homozygous parents, the offspring will all be heterozygous. Dihybrid crosses have multiple genotype and phenotype ratios depending on if the parents are homozygous or heterozygous.
Ch 12 gene linkage groups and practice problems
1. The document discusses genetics concepts like linkage groups, chromosome mapping, parental and recombinant types. It provides examples of genetic crosses in pea plants and fruit flies to demonstrate these concepts.
2. Recombinant types result from crossovers during meiosis, where alleles switch positions on homologous chromosomes. The frequency of recombinants allows calculation of distance between genes.
3. A sample genetics problem is presented, asking the reader to analyze a dihybrid cross in pea plants and determine the chromosome arrangement and distance between genes.
Genetics Review.pdf
1. The document provides an overview of genetics concepts including: Mendelian inheritance patterns; DNA structure and replication; gene expression and regulation; and modern applications like biotechnology.
2. Key topics covered include alleles and inheritance ratios; meiosis; linkage and crossing over; mutations and their effects; the central dogma of DNA to protein; gene regulation in prokaryotes and eukaryotes; and recombinant DNA techniques.
3. Sample multiple choice and free response questions assess understanding of inheritance patterns, molecular genetics concepts, and experimental techniques.
Genetics Review.pdf
1. This document provides a review of genetics concepts including: Mendelian inheritance patterns like monohybrid and dihybrid crosses; non-Mendelian inheritance like incomplete dominance; DNA and RNA structure; the central dogma of molecular biology; gene regulation; and modern techniques like genetic engineering and gel electrophoresis.
2. It also includes sample multiple choice questions testing these concepts, such as questions about inheritance patterns demonstrated in genetic crosses, DNA replication, transcription and translation processes, and recombinant DNA techniques.
3. The review concludes with sample free response questions asking students to explain meiosis, chromosomal abnormalities resulting from defects in meiosis, and how the human genome demonstrates both continuity and variation.
Recombination and LinkageA Three point test cross in Drosophil.docx
A three-point test cross was conducted in Drosophila to determine the order and distance between three linked genes (y, cv, and f) on the X chromosome. F2 progeny were scored for phenotypes to determine the number of each genotype. Recombination frequencies and a genetic map were constructed based on the data. Chi-square tests compared the observed and expected values, and some deviations from expectations were observed. The results provided insight into genetic mapping and factors that influence mapping accuracy.
Rflp
The document discusses restriction fragment length polymorphism (RFLP) analysis, which refers to variations in restriction sites between individuals in a population. These variations are useful for genetic mapping as they allow researchers to distinguish between chromosomes from different individuals. The document provides examples of how RFLP analysis can be used to map genes and determine genetic distances between loci by examining patterns produced by restriction enzyme digestion and southern blotting.
Genetics practice q key
- One or more of the three children will have disease 1.
- All three children will have the disease.
- At least one child out of three will be phenotypically normal.
Biology Variation 2
There are two main ways heritable variations can arise: independent assortment of genes during meiosis and fertilization, which introduces new combinations of parental genes, and spontaneous gene and chromosome mutations. Independent assortment occurs when parental chromosomes separate during meiosis and recombine in new configurations during fertilization. Gene mutations happen due to errors in DNA replication, while chromosome mutations result from changes like gains or losses of chromosomes. Both types of mutations can lead to new traits.
Genetics
This document contains genetics concept questions and their answers related to topics like Mendelian inheritance, sex determination, extensions of Mendelian inheritance, and blood types. Some example questions include the probability of offspring genotypes from given parental crosses, determining heterozygotes vs homozygotes, inheritance patterns like sex-influenced traits, and solving genetics problems using rules of probability. The responses provide detailed explanations and use techniques like Punnett squares, product rule, and binomial/multinomial expansions to calculate the requested probabilities.
BIO Virtual fly lab (final draft)
The document summarizes Gregor Mendel's experiments with pea plants and how they laid the foundation for genetics. It then describes how the author will simulate Mendel's experiments virtually using fruit flies. The author performs a series of crosses between fruit flies with different traits and uses a chi-square analysis to determine if the results match expected Mendelian ratios. The crosses are designed to show traits behaving as dominant, recessive, recessive sex-linked, and dihybrid traits.
CODOMINANCE.pptx
Codominance occurs when two alleles are simultaneously expressed in a heterozygote. For example, some chickens have black and white feathers that are speckled due to codominance - the alleles for black (B) and white (W) feathers are both expressed at the same time in chickens with genotype BW. If two speckled chickens were crossed, there would be a 50% probability that their offspring would be speckled and a 25% probability of being either black or white.
Pre-AP_Biology_Unit_04_-_Genetics_Practice_Set (1).pdf
This document provides genetics practice problems and examples related to monohybrid and dihybrid crosses, inheritance of sex-linked traits, blood typing, and other genetics concepts. It begins with practice problems identifying genotypes and predicting phenotypes based on genetic crosses. The document then provides step-by-step examples and explanations of monohybrid and dihybrid crosses, inheritance patterns such as incomplete dominance and codominance, sex-linked traits, and multiple alleles as seen in human blood typing. Students are to complete related practice problems and predict outcomes of genetic crosses based on provided genotypes.
Pre-AP_Biology_Unit_04_-_Genetics_Practice_Set.pdf
This document provides genetics practice problems and instructions for solving monohybrid and dihybrid cross problems. It includes examples of setting up Punnett squares and predicting offspring genotypes and phenotypes for crosses involving one or two gene traits. Students are asked to solve sample genetics problems involving traits such as pea pod shape, eye color in mice, and fur color/texture in guinea pigs. The document emphasizes following a five step process: determining parent genotypes, setting up a Punnett square, predicting offspring genotypes and phenotypes, and working backwards from given offspring data.
Similar to 1.) The normal eye color of Drosophila is red, but strains in which .pdf (20)
AP Biology Ch 12 gene linkage groups and chromosome maps
AP Biology Ch 12 gene linkage groups and chromosome maps
AP BIOLOGY MENDELIAN GENETICS AND 2 Student Packet.pdf
AP BIOLOGY MENDELIAN GENETICS AND 2 Student Packet.pdf
Recombination and LinkageA Three point test cross in Drosophil.docx
Recombination and LinkageA Three point test cross in Drosophil.docx
Pre-AP_Biology_Unit_04_-_Genetics_Practice_Set (1).pdf
Pre-AP_Biology_Unit_04_-_Genetics_Practice_Set (1).pdf
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It was noted that excess water can kill plants. If water is not toxi.pdf
It was noted that excess water can kill plants. If water is not toxic, and if all plants require water,
then how can excess water kill plants? What is actually killing the plants?
Solution
When excess water is present, the roots will be deprived of oxygen because the water fills the
pores in the soil. This mostly occurs in clay soils. Also, excess water drain out the nutrients and
promote root rot disease, which gradually kills the plant..
In biology, the roles of transition metal ions can be broadly groupe.pdf
In biology, the roles of transition metal ions can be broadly grouped into two classifications:
redox chemistry (catalysis, electron transfer, etc.) and acid-base chemistry (catalysis, structural,
etc.).
A) List three metal ions that commonly participate in biological redox chemistry, and give
examples of how they participate.
B) Compare this list to metal ions commonly participating in acid-base chemistry. Discuss the
overlap of these two groups of metal ions. (e.g. Why are some used for one of these functions
and not for the other?)
C) List some common biological ligands of these metal ions. Can you see patterns that would
allow you to group these ligands into sets with common properties, or that bind common metals?
D) Why are trace metals used for these biological functions, and not other more-abundant
elements like C,N,O, etc?
Solution
(a) iron, copper, cobalt, manganese etc are used in biology redox reactions like molecular di
oxygen, superoxide, hydrogen peroxide thus inhibiting the effect or formation of these
compound by use of the certain other enzymes..
(c) the ions and the Ligands for them are Ike gluthaione peroxide which can bind to chromium,
arsenic etc.. Iron binds to ferrodoxin which is associated with the photosynthesis..
(d) transition metals are used because they have empty d-orbitals in with them due to which they
can easily transfer or accept the electrons or exchange the electrons among the metals and
enzymes.. Which is not possible in other trace metals like carbon, nitrogen etc...
Image that a researcher examined how the people’s heights are associ.pdf
Image that a researcher examined how the people’s heights are associated with their salaries. She
found that the correlation between heights (measured in inches) and salary range (measured in
dollars) was .36. Had she measured the height in cm instead, what would the correlation have
been?
Solution
Correlation would not change because it is unaffected by the units..
I need help with this one Question 18 of 18 Sapling Learning macmill.pdf
I need help with this one Question 18 of 18 Sapling Learning macmillan earning Integral
membrane proteins have different characteristics and functions than peripheral membrane
proteins. Sort the statements into those that are properties of integral membrane proteins and
those that are properties of peripheral membrane proteins. Integral membrane proteins Peripheral
membrane proteins can move laterally (sideways) in the hydrophobic part of the membrane
found at the surface of the cell membrane reach through all or part of the membrane pump
substances across the cell membrane
Solution
Integral membrane proteins ( intrinsic proteins)
Peripheral membrane proteins, or extrinsic proteins
Mainly found either fully or partially submerged in the phospholipids bilayer of the plasma
membrane. They have hydrophobic side chains that interact with fatty acyl groups of the
membrane phospholipids, thus anchoring the protein to the membrane.
Integral proteins represent around 70% of plasma membrane proteins.
Present at the surface ie on the innermost and outermost of phospholipids bilayer. These proteins
are loosely bound to the plasma membrane either directly by interactions with polar heads of
phospholipids bilayer or indirectly by interactions with integral proteins. These proteins
constitute about 20-30 % of total membrane proteins.
Reach through all or part of the membrane (Most of the integral proteins span the entire
phospholipid bilayer -from the inner surface to the outer surface. These transmembrane proteins
contain one or more membrane-spanning domains as well as domains, from four to several
hundred residues long, extending into the aqueous medium on each side of the bilayer
Pump substances across the cell membrane (Peripheral proteins at the cytosolic face of the
plasma membrane include the cytoskeletal proteins spectrin and actin in erythrocytes and the
enzyme protein kinase C. This enzyme shuttles between the cytosol and the cytosolic face of the
plasma membrane and plays a role in signal transduction)
Can move laterally (sideways) in the hydrophobic part of the membrane. Integral proteins are
tightly bound and are directly interacting with the hydrophobic core of the plasma membrane
Do not interact with the hydrophobic core in between two layers of phospholipids.
Integral membrane proteins ( intrinsic proteins)
Peripheral membrane proteins, or extrinsic proteins
Mainly found either fully or partially submerged in the phospholipids bilayer of the plasma
membrane. They have hydrophobic side chains that interact with fatty acyl groups of the
membrane phospholipids, thus anchoring the protein to the membrane.
Integral proteins represent around 70% of plasma membrane proteins.
Present at the surface ie on the innermost and outermost of phospholipids bilayer. These proteins
are loosely bound to the plasma membrane either directly by interactions with polar heads of
phospholipids bilayer or indirectly by interactions with integral proteins. These pro.
Explain how recombination increases the amount of genetic variation i.pdf
Explain how recombination increases the amount of genetic variation in offspring: Explain why
it is not possible to have a recombination frequency of greater than 50% (half recombinant
progeny): A second pair of Drosophila are mated. The female is Cucu YY (straight wing, gray
body), while the male is Cucu yy (straight wing, yellow body). Assuming recombination,
perform the cross and list the offspring genotypes and phenotypes.
Solution
Que-3:
The recombination frequency is the measure of genetic linkage at different loci. The value of
recombination frequency is 50% if the chromosomal genes are located at different regions on the
different chromosome because of independent assortment. If the genes are located closely
together on the same chromosome, then those genes considered as genetically linked; because
they have not assorted independently due to \"crossing over\", then the recombination frequency
value is less than 50%. Linked genes are due to linkage often do not obey Mendel\'s laws of
inheritance such as independent assortment because of \"crossing over with reciprocal exchange
of hereditary genetic material meticulously in between non-sister chromatids\" at the time of
\"synapsis formation\" (propahse -I) in reduction division or meiosis 1. These events are leading
to formation of gamete cells with complete different genotypically from diploid parent
reproductive cell via meiosis finally induce \"genetic variation\" in phenotype due to
\"fertilization\" events between sperm cells & ovum.
The recombination of chromosomes enables meiosis process of cell division with the generation
of novel recombinant nucleotide -sequences by crossing over during the cell division in both
unicellular and multicellular species. Sometimes, double strand breaks in DNA during
homologous chromosomal recombination may be produced due to exposure any harmful
radiation finally may cause higher genetic variation via duplication followed by fertilization to
form a phenotype with higher genetic variation
Que-4:
Crossing over is somewhat randomly distributed over the length of the chromosome because if
the two genes are close together then it is very change to get exchange of chromosomal breaks
therefore, if the two genes are located farther apart then there will be higher room for the
exchange of genetic material with chromosomal breaks. Therefore, when two genetic are far
apart on the same chromosome are more likely to have a crossover between them than two loci
that are close together. This recombination frequency is considered as a measure of genetic
linkage during the crossover of the homologous chromosomes. Linkage disequilibrium defined
as the existence of alleles at different loci with the absence of genetic linkage between them even
though there is no equilibrium with allelic frequencies independently. The recombination
frequency is the measure of genetic linkage at different loci. The value of recombination
frequency is 50% \"if the chromosomal genes are .
exclange I ren peretr) ol and eleaced by the maxtit wadr Ho t Col pox.pdf
exclange I ren peretr) ol and eleaced by the maxtit wadr Ho t Col poxio pon 2 the stol sle shiage
type head eschane
Solution
since there are two different fluids takes place that is hot body and cold body initially
condensation takes place in the condensation process the hot body temperature will maintain low
temperature than that of surroundings and afterwards evaporation takes palces to maintain the
temperature inside the system so that the pressure variation does not takes place and gets
maintained to the differntlevel of fluids..
DNA is a macromolecule that is critical for life. Much of the functi.pdf
DNA is a macromolecule that is critical for life. Much of the functionality of DNA lies in
itscomplex structure. What are the 2 main forces responsible for holding together the structureof
DNA and what does each force hold together?
Solution
The double helical structure of DNA (deoxyribonucleic acid) is formed by the coiling of two
polynucleotide chains, similar to a spiral staircase.
Of the two polynucleotide chains, the sugar-phosphate backbones are exposed to the outside of
the spiral staircase. While the base pairs (side chains) extend towards inside of the helix. The two
strands are connected together by hydrogen bonds and hydrostatic forces.
Of the two polynucleotide chains, the sugar-phosphate backbones are exposed to the outside of
the spiral staircase. While the base pairs (side chains) extend towards inside of the helix. The two
strands are connected together by hydrogen bonds..
Chapter 7 , book Health in the Later Years, 5th EditionGive a his.pdf
Chapter 7 , book: Health in the Later Years, 5th Edition
Give a historical perspective on advocacy efforts for the mentally ill in the U.S.
Solution
Answer:
Though it is not clear what was the student/client really asked for. As per the studies concern
making the fact easily understandable is the goal of our company. So, here is the brief of the fact
asked for-
Mental health services throughout the world are in the midst of unprecedented change. New
thinking is emerging about what constitutes \"best practice\" in mental health services.
Community mental health treatment and support services are now the primary locus of care for
people with serious mental illness in many areas. Fundamental components of mental health
service systems such as methods of designing and delivering services, mechanisms for
establishing and funding service priorities, methods of ensuring system accountability, and
definitions of staff role, needed competencies, and training approaches, and so forth, are
undergoing scrutiny and restructuring. Historic distinctions between clinical treatment services
and disability support or rehabilitation services are no longer simple and examples of integrated
service systems are becoming more common. In this context, traditional psychiatric hospital
services are undergoing a parallel process which is re-shaping their role, services, and position as
partners with community treatment and rehabilitation services, disability services, and self-
help/peer support services in an integrated system. Mental health consumers and their family
members are exercising their \"voice\" and are assuming increasingly vital roles in governance,
policy formulation, and service design and delivery which is helping to shape how these changes
will occur in our thinking and practice.\'.
Discuss the benefits of event-driven programming.Contrast event-dr.pdf
Discuss the benefits of event-driven programming.
Contrast event-driven programming with procedural programming.
Discuss how messages from other programs can be viewed as events.
Give examples. Give examples of tasks that require the program to guide the user.
Should programs be purely event driven, procedural, or a mixture of both? Why?
Solution
The benefit of event-driven programming is that it\'s very intuitive and naturally well-suited to
applications whose control flow are based, not on its structure, but rather on internal or external
events. This is why event-driven programming is so popular with GUI programming. Likweise,
SOA and client-server architectures are naturally event-driven as well.
We if we find contrast between Procedural and event-based then applications written in
procedural languages execute by proceeding logically through the program code, one line at a
time. Logic flow can be temporarily transferred to other parts of the program through the GoTo,
GoSub, and Call statements, directing the program from beginning to end.
In contrast, program statements in an event-driven application execute only when a specific
event calls a section of code assigned to that event. Events can be triggered by keyboard input,
mouse actions, the operating system, or code in the application. For example, consider what
happens when the user clicks a command button named Command1 on a form. The mouse click
is an event. When the Click event occurs, Visual Basic executes the code in the
Sub procedure named Command1_Click. When the code has finished running, Visual Basic
waits for the next event.
Messages from other programs/threads can be seen as events, as they trigger some instance of
code to execute..
Before 1900, despite its weaknesses in effective management of worke.pdf
Before 1900, despite its weaknesses in effective management of workers, manufacturing
leadership was well provided by top management. They were technological entrepreneurs,
archictects of productive systems, veritable lions of industry. But when they delegated their
production responsibilities to a second-level department, the factory institution never recovered
its vitality. The lion was tamed. It\'s management systems became protective and generally were
neither enterpreneual nor strategic. Production managers since then have typically had little to do
with initiating substantially new process technology-in contrast to their predecessors before 1900
(skinner 1985).
D) how is Japan (or Germany) different from (or the same as) America with regards to this trend
in manufacturing leadership?
E) taking the structural charestaristics of manufacturing enterprises (e.g., scale, complexity, pace
of technological change) as given, what can be done to revitalize manufacturing leadership?
Solution
Strategic Windows: their nature.
The nature and purpose of strategy and how it is formulated. The nature of marketing strategy
and how this should take account of the interests of various stakeholders when involving such
things as, product/service development and delivery, promotional mix, support services,
manufacturing and production processes, R&D, and material purchasing affect the stakeholders.
Other factors in the business environment that influence marketing strategy: political, economic,
socio-cultural and technological (PEST).
Marketing and competitors: how a firm must be able to position itself competitively in the minds
of its customers so that its products and services stand out very favourably in important respects
in relationship to competitors.
Matching the firm’s products / services with opportunities and threats in the market place. The
limited periods during which the fit between the key requirements of a market and the particular
competencies of a firm competing in that market are at an optimum. Investment in a product line
or a market area should be timed to coincide with periods during which a strategic window is
open. Correspondingly, withdrawal should be considered where something which was a good fit,
is no longer a good fit. Ways in which a market can evolve and how firms might develop a
competitive strategy to take advantage of Strategic Windows.
Portfolio Analysis
How organisations create their own environments rather than simply adapt to existing ones. How
they select the strategic windows of opportunities and threats through which they want to look
out into the world and develop and market product and services to meet the needs of what they
observe to be required in the face of environmental turbulence.
How well the fit between an organization’s products/services meet the needs presented by the
windows of opportunities and threats is a fitting start for exploring the subject of strategic
marketing. It introduces the many factors t.
Alpha particle radiation sensor. A silicon diode radiation sensor is .pdf
Alpha particle radiation sensor. A silicon diode radiation sensor is used to check the radioactive
source in a smoke detector. According to the label, the source is given as 1 music. The source is
Ain-241 with an energy of 486 times 10^6 e V and an activity of 195 times 10^10 B q. The
energy needed to generate an electron-hole pair is 61 e V. The diode is connected to an inverse
potential of 12 V, has a thickness of 0.5 mm, and the mobility\'s are 450 cm^2/V s for holes and
1350 cm^2/V s for electrons. The radiation from Ain-241 is mostly a particles that penetrate very
little into the silicon and hence may be assumed to be absorbed at the surface of the diode For
this reason the junction is exposed directly so radiation does not pass through the electrodes.
What is the current through the diode if all particles emitted are trapped by the diode?
Solution
Alpha particles, like helium nuclei, have a net spin of zero. Due to the mechanism of their
production in standard alpharadioactive decay, alpha particles generally have a kinetic energy of
about 5 MeV, and a velocity in the vicinity of 5% the speed of light. (See discussion below for
the limits of these figures in alpha decay.) They are a highly ionizing form of particle radiation,
and (when resulting from radioactive alpha decay) have low penetration depth. They are able to
be stopped by a few centimeters of air, or by the skin. However, so-called long range alpha
particles from ternary fission are three times as energetic, and penetrate three times as far. As
noted, the helium nuclei that form 10–12% of cosmic rays are also usually of much higher
energy than those produced by nuclear decay processes, and are thus capable of being highly
penetrating and able to traverse the human body and also many meters of dense solid shielding,
depending on their energy. To a lesser extent, this is also true of very high-energy helium nuclei
produced by particle accelerators..
A ship was caught in a storm and driven to a small island in the mid.pdf
A ship was caught in a storm and driven to a small island in the middle of the South Pacific. For
generations, no one encountered this small colony, who despite the odds made a thriving
settlement on their island. After many years, what characteristics would be unique about this
population?
A: The people on the island would no longer be considered humans after evolving independently
for many generations.
B: The people on the island would all look very different from each other, and they may exhibit
abnormally high occurrences of traits that are rare in the general human population.
C: The people on the island would all look very different from each other, and they may have
evolved some odd characteristics that aren\'t seen in other humans.
D: The people on the island would all look very similar to each other, and they may exhibit
abnormally high occurrences of traits that are rare in the general human population.
Solution
There is a small group of population who lived seperately from the human for many many years .
In this case, the small population would have bred amongst themselves are continued their
colony.
This small group say after a thousand years look more like each other than any other human on
the earth as they had been isolated from the rest of the world.
So the answer should be D)The people on the island would all look very similar to each other,
and they may exhibit abnormally high occurrences of traits that are rare in the general human
population.
8. List 4 GENERAL ways that a CELL, can DECREASE the rate of its rea.pdf
8. List 4 GENERAL ways that a CELL, can DECREASE the rate of its reactions?
1. 2. 3.
4.
10. What is the BASIC difference between substrate-level phosphorylation and oxidative
phosphorylation? (do not tell me which pathways are involved).
11. Exactly….and I mean exactly…..which chemical reaction in the aerobic metabolic pathways,
actually uses oxygen, and what does it do to the oxygen that it uses.
Solution
8).
Enzymes are the proteins, which decrease the activation energy of reaction and catalyse the
reaction process. All enzymes are catalysts because,
Cells decrease the rate of reaction by
1). Altering pH
2). Accumulation of products
3). Altered temperature
4). Inactivation of catalyst.
As leaf temperature increases from 20°C to 35°C, the quantum yield o.pdf
As leaf temperature increases from 20°C to 35°C, the quantum yield of C3 photosynthesis is
directly affected by
The decreasing ratio CO2 to O2 in the gas phase
The increased reaction rates of photorespiration
The decreasing ratio CO2 to O2 in the aqueous phase
Degradation of the thylakoid membrane and proteins
RubisCO’s improved affinity for O2.
Solution
Answer is :
The decreasing ratio CO2 to O2 in the aqueous phase.
Reason: When the leaves are hot, CO2 becomes less soluble in the aqueous phase faster than
O2, and the CO2/O2 ratio drops.
Among 10 people traveling in a group, 2 have outdated passports. It .pdf
Among 10 people traveling in a group, 2 have outdated passports. It is known that inspectors will
check the passports of 20% of the people in any group passing their desks. The group can go as
a
whole (all 10) to one desk or can split into two groups of 5 and use two different desks. How
should
members of the group arrange themselves to maximize the probability of getting by the
inspectors
without having the outdated passports detected? I do not know how to figure out each
probability... The section is on counting rules - i am guessing it is without replacement and order
does not matter. Thank you for your help
Solution
Inspectors would check 20% of people in the group
Hence if 10 people, 2 would be checked or if 5 one person would be checked
If 10 people go to one desk, p for having outdated passport = 0.2
X - no of people having correct passports in 10 is binomial with p = 0.8
They would be caught only if Inspector selects any one person from the two.
Y - no of people having outdated passports and p for Y =0.2
P(Y>=1) = 1-P(Y=0)
= 1-(0.8)10
= 0.8926
---------------------------------------------
If goes into two groups then either each group can have one outdated or one group no outdated
and other groups 2 outdated.
For each group one outdated prob of being caught
= P(y=1) in I group + P(y=1 in second group)-P( P(y=1) in I group *P(y=1 in second group)
as these two events are independent
= {1-(0.8)5}+{1-(0.8)5}-{1-(0.8)5}{1-(0.8)5}
=2(0,67232)-0.672322
= 0.8926 (same as the first case)
----------------------------------------------
If one group has both outdated and other has none then prob of being not caught
= 1(Prob of atleast one in the group having outdated 2)
Here p for good passport =0.6
= P(Y>=1)
= 1-P(y=0)
= 1-(0.6)5
=0.92224
Hence this is the best method to go.
Divide into two groups with 2 outdated in one group.
29. Code an application program that keeps track of student informat.pdf
29. Code an application program that keeps track of student information at your college. Include
their names, identification numbers, and grade point averages in a fully encapsulated,
homogeneous sorted singly linked list structure. When launched, the user will be asked to input
the initial number of students and the initial data set. Once this is complete, the user will be
presented with the following menu: Enter: 1 to insert a new student\'s information, 2 to fetch and
output a student\'s information, 3 to delete a student\'s information, 4 to update a student\'s
information, 5 to output all the student information in sorted order, and 6 to exit the program.
The program should perform an unlimited number of operations until the user enters a 6 to exit
the program. If the user requests an operation on a node not in the structure, the program output
should be “node not in structure.” Otherwise, the message “operation complete” should be
output. in java
Solution
Dear Asker,
Following 3 classes consist the colution to this problem. This will give you an idea to approach
this problem:
//Student Class
package studentinfo;
public class Student implements Comparable {
private String name;
private String id;
private double gpa;
public Student(String id, String name, double gpa) {
this.name = name;
this.id = id;
this.gpa = gpa;
}
/**
* @return the name
*/
public String getName() {
return name;
}
/**
* @param name the name to set
*/
public void setName(String name) {
this.name = name;
}
/**
* @return the id
*/
public String getId() {
return id;
}
/**
* @param id the id to set
*/
public void setId(String id) {
this.id = id;
}
/**
* @return the gpa
*/
public double getGpa() {
return gpa;
}
/**
* @param gpa the gpa to set
*/
public void setGpa(double gpa) {
this.gpa = gpa;
}
@Override
public int compareTo(Student o) {
if (this.id.compareTo(o.id) < 0) {
return -1;
} else if (this.id.compareTo(o.id) > 0) {
return 1;
} else if (this.gpa < o.gpa) {
return -1;
} else if (this.gpa > o.gpa) {
return 1;
}
return 0;
}
public boolean equals(Student o) {
if (this.name == o.name && this.id == o.id && this.gpa == o.gpa) {
return true;
}
return false;
}
}
//Student container class
package studentinfo;
import java.util.Collections;
import java.util.LinkedList;
public class StudentContainer {
LinkedList studentList;
public StudentContainer() {
this.studentList = new LinkedList();
}
public boolean hasStudent(String id, String name, double gpa) {
return this.studentList.contains(new Student(id, name, gpa));
}
public boolean insertStudent(String id, String name, double gpa) {
if (!this.isValidString(id) || !this.isValidString(name) || !this.isValidGpa(gpa)) {
return false;
}
this.studentList.add(new Student(id, name, gpa));
return true;
}
public boolean deleteStudentById(String id) {
Student toRemove = null;
for (Student e : studentList) {
if (e.getId() == id) {
toRemove = e;
break;
}
}
if (toRemove != null) {
this.studentList.remove(toRemove);
return true;
}
return false;.
Write the code for a small function called myStack, which creates a .pdf
Write the code for a small function called myStack, which creates a Stack object of String
elements, named cities. Add \"Sydney\", then \"London\", then \"New York\" and then
\"Beijing\" to cities. Then print each element on the screen, one per line. Print each element in a
way that causes cities to be empty (i.e. have 0 elements) afterwards. (6 mks)
Solution
PROGRAM CODE:
package stack;
import java.util.Stack;
public class MyStackClass {
//stack function
public static void myStack()
{
//creating a stack object called cities
Stack cities = new Stack();
//Adding cities into the stack
cities.push(\"Sydney\");
cities.push(\"London\");
cities.push(\"New York\");
cities.push(\"Beijing\");
//using while loop to loop through the stack till its empty
while(!cities.isEmpty())
{
// pop function is used to remove elements from the stack
// pop() returns the elemets which is printed on the csreen
System.out.println(cities.pop());
}
}
// Main function for the class
public static void main(String args[])
{
myStack(); // function call to the stack method
}
}
OUTPUT:
Beijing
New York
London
Sydney.
Why does the neutralization of an acid by a base often produce water.pdf
Neutralization often produces water because acids release H+ ions and bases release OH- ions. When an acid and base react, the H+ and OH- ions combine to form water. For example, when HCl and NaOH react, they produce NaCl and water as H+ from HCl combines with OH- from NaOH.
Which statement about phloem transport is falseIt takes place in .pdf
Which statement about phloem transport is false?
It takes place in sieve tubes
It requires metabolic energy
A high pressure potential is maintained in the sieve tubes
At sinks, solutes are actively transported into sieve tube elements
The mycorrhizal relationship allows a plant’s roots to ??
increase the amount of soil than can be accessed for nutrients.
grow in nutrient-poor soils.
avoid direct interaction with the soil.
None of the above
Solution
1. At sinks, solutes are actively transported into sieve tube elements. (False)
Phloem is a vascular tissue in plants that actively transports synthesized food from the source
tissue ( photosynthetic) to sink tissue (non-photosynthetic). The sieve tube elements (STEs) are
the phloem tissue compartments which help in the transportation of food from source to sink.
Phloem transport requires metabolic energy and a high-pressure potential is maintained at the
sieve tubes.
2. Increase the amount of soil that can be accessed for nutrients
Mycorrhiza is the symbiotic relationship between fungi and the plant roots on which they
extensively grow. The network of fungal growth or mycorrhiza helps in increasing the surface
area of the roots so that they can access more amount of soil and take up nutrients essential for
plant growth..
Which of the following is not a primitive data type Which of th.pdf
Which of the following is not a primitive data type
Which of the following is not a primitive data type
Solution
Hi, PLease find primitive data type list in C:
The primitive data types in c language are the inbuilt data types provided by the c language
itself. Thus, all c compilers provide support for these data types.
The following primitive data types in c are available:
Integer Data Type
int , short, long
Float data Type
float
Double Data Type
double
Character Data Type
char
Void Data Type
void.
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1.) The normal eye color of Drosophila is red, but strains in which .pdf
- 1. 1.) The normal eye color of Drosophila is red, but strains in which all flies have brown eyes are available. Similarly, wings are normally long, but there are strains with short wings. A female from a pure line with brown eyes and short wings is crossed with a male from a normal pure line. The F1 consists of normal females and short-winged males. An F2 is then produced by intercrossing the F1. Both sexes of F2 flies show phenotypes as follows: 3/8 red eyes, long wings 3/8 red eyes, short wings 1/8 brown eyes, long wings 1/8 brown eyes, short wings Deduce the inheritance of these phenotypes; use clearly defined genetic symbols of your own invention. State the genotypes of all three generations and the genotypic proportions of the F1 and F2. 2.) When a cell of genotype A/a ; B/b ; C/c having all the genes on separate chromosome pairs divides mitotically, what are the genotypes of the daughter cells? 3.) A presumed dihybrid in Drosophila, B/b ; F/f, is test-crossed with b/b ; f/f. (B = black body ; b = brown body; F = forked bristles; f = unforked bristles.) The results are black, forked 230 brown, forked 240 black, unforked 210 brown, unforked 250 Use the 2 test to determine if these results fit the results expected from testcrossing the hypothesized dihybrid. 4.) In mice, dwarfism is caused by an X-linked recessive allele, and pink coat is caused by an autosomal dominant allele (coats are normally brownish). If a dwarf female from a pure line is crossed with a pink male from a pure line, what will be the phenotypic ratios in the F1 and F2 in each sex? (Invent and define your own gene symbols.) Solution Answer 2. If a cell divides mitotically (equational division), the genetic constitution of the daughter cells remains the same as the parents. So a parent with the genotype A/a; B/b; C/c will give rise to daughter cells with the genotype A/a; B/b; C/c. However, if the cell was to divide meiotically (reductional division), the genetic constitution would be different as only one half of the parental chromosomes goes to the daughter cell. So the genotypes would be as follows:
- 2. ABC or AbC or ABc or Abc or aBC or abC or aBc or abc Answer 3. For the given cross BbFf X bbff For such a test cross the resultant phenotypic ration obtained is 1 (BF):1 (Bf):1 (bF):1 (bf). For dihybrid cross, the degree of freedom will be 4 – 1 = 3 Chi square test is as follows BF Bf bF bf Totals Observed phenotype 230 210 240 250 930 Expected phenotype 930x ¼= 232.5 930x ¼= 232.5 930x ¼= 232.5 930x ¼= 232.5 930 Observed – expected -2.5 -22.5 7.5 17.5 (Obs – Exp)2 Expected 6.25/232.5= 0.0268 506.25/232.5= 2.1774 56.25/232.5= 0.2419 306.25/232.5= 1.3172 3.7633
- 3. Looking at the probability value table for degree of freedom = 3, df .90 .70 .50 .30 .20 .10 .05 .01 3 0.58 1.42 2.37 3.67 4.64 6.25 7.82 11.35 Hence we can deduce that there is more than 20% probability for the occurrence of this event. Answer 4. Let XD = Normal phenotype (tall) Xd = Dwarf phenotype P = Pink coat p = Brown coat According to the question, the following cross is done Xd Xdpp x XDYPP F1 XDP YP Xdp XDXdPp (Tall pink female) XdYPp (Dwarf pink male) On intercrossing, XDXdPp x XdYPp
- 4. F2 XDP XDp XdP Xdp XdP XDXdPP Tall pink female XDXdPp Tall pink female XdXdPP Dwarf pink female XdXdPp Dwarf pink female Xdp XDXdPp Tall pink female XDXdpp Tall brown female XdXdPp Dwarf pink female XdXdpp Dwarf brown female YP XDYPP Tall pink male XDYPp Tall pink male XdYPP Dwarf pink male XdYPp Dwarf pink male Yp XDYPp Tall pink male XDYpp
- 5. Tall brown male XdYPp Dwarf pink male XdYpp Dwarf brown male So the phenotypic ratios are as follows: F1- Females: All tall and pink; Males: All dwarf and pink F2- Females: 3 Tall pink: 3 dwarf pink: 1 tall brown : 1 dwarf brown Males: 3 Tall pink: 3 dwarf pink: 1 tall brown : 1 dwarf brown BF Bf bF bf Totals Observed phenotype 230 210 240 250 930 Expected phenotype 930x ¼= 232.5 930x ¼= 232.5 930x ¼= 232.5 930x ¼= 232.5 930 Observed – expected -2.5 -22.5 7.5 17.5 (Obs – Exp)2 Expected 6.25/232.5= 0.0268 506.25/232.5= 2.1774 56.25/232.5= 0.2419