Question

Many kinds of wild animals have the agouti coloring pattern, in which each hair has a yellow band around it. a. Black mice and other black animals do not have the yellow band; each of their hairs is all black. This absence of wild agouti pattern is called nonagouti. When mice of a true-breeding agouti line are crossed with nonagoutis, the $F_{1}$ is all agouti and the $F_{2}$ has a 3: 1 ratio of agoutis to nonagoutis. Diagram this cross, letting $A$ represent the allele responsible for the agouti phenotype and $a$ nonagouti. Show the phenotypes and genotypes of the parents, their gametes, the $F_{1}$, their gametes, and the $F_{2}$ b. Another inherited color deviation in mice substitutes brown for the black color in the wild-type hair. Such brown-agouti mice are called cinnamons. When wildtype mice are crossed with cinnamons, all of the $\mathrm{F}_{1}$ are wild type and the $\mathrm{F}_{2}$ has a 3: 1 ratio of wild type to cinnamon. Diagram this cross as in part $a$, letting $B$ stand for the wild-type black allele and $b$ stand for the cinnamon brown allele. c. When mice of a true-breeding cinnamon line are crossed with mice of a true-breeding nonagouti (black) line, all of the $F_{1}$ are wild type. Use a genetic diagram to explain this result. d. In the $F_{2}$ of the cross in part $c,$ a fourth color called chocolate appears in addition to the parental cinnamon and nonagouti and the wild type of the $\mathrm{F}_{1}$. Chocolate mice have a solid, rich brown color. What is the genetic constitution of the chocolates? e. Assuming that the $A / a$ and $B / b$ allelic pairs assort independently of each other, what do you expect to be the relative frequencies of the four color types in the $\mathrm{F}_{2}$ described in part $d ?$ Diagram the cross of parts $c$ and $d$ showing phenotypes and genotypes (including gametes). f. What phenotypes would be observed in what proportions in the progeny of a backcross of $\mathrm{F}_{1}$ mice from part $c$ with the cinnamon parental stock? With the nonagouti (black) parental stock? Diagram these backcrosses. g. Diagram a testcross for the $\mathrm{F}_{1}$ of part $c .$ What colors would result and in what proportions? h. Albino (pink-eyed white) mice are homozygous for the recessive member of an allelic pair $C / c,$ which assorts independently of the $A / a$ and $B / b$ pairs. Suppose that you have four different highly inbred (and therefore presumably homozygous) albino lines. You cross each of these lines with a true-breeding wild-type line, and you raise a large $\mathrm{F}_{2}$ progeny from each cross. What genotypes for the albino lines can you deduce from the following $\mathrm{F}_{2}$ phenotypes? $$\begin{array}{cccccc} & {4}{c} {\text {Phenotypes of progeny}} \\ { 2 - 5 } \mathrm{F}_{2} \text { of } \text { line } & \begin{array}{c} \text { Wild } \\ \text { type } \end{array} & \text { Black } & \begin{array}{c} \text { Cinna- } \\ \text { mon } \end{array} & \begin{array}{c} \text { Choco- } \\ \text { late } \end{array} & \text { Albino } \\ \hline 1 & 87 & 0 & 32 & 0 & 39 \\ 2 & 62 & 0 & 0 & 0 & 18 \\ 3 & 96 & 30 & 0 & 0 & 41 \\ 4 & 287 & 86 & 92 & 29 & 164 \\ \hline \end{array}$$ (Adapted from A. M. Srb, R. D. Owen, and R. S. Edgar General Genetics, 2nd ed. W. H. Freeman and Company, 1965.)

Name: Problem 42, Chapter 6: Introduction to Genetic Analysis
Uploaded: 2020-07-28T16:06:09-08:00
Duration: 20 min 3 s
Channel: Jessica Wooten

   Many kinds of wild animals have the agouti coloring pattern, in which each hair has a yellow band around it.
a. Black mice and other black animals do not have the yellow band; each of their hairs is all black. This absence of wild agouti pattern is called nonagouti. When mice of
a true-breeding agouti line are crossed with nonagoutis, the $F_{1}$ is all agouti and the $F_{2}$ has a 3: 1 ratio of agoutis to nonagoutis. Diagram this cross, letting $A$ represent the allele responsible for the agouti phenotype and $a$ nonagouti. Show the phenotypes and genotypes of the parents, their gametes, the $F_{1}$, their gametes, and the $F_{2}$
b. Another inherited color deviation in mice substitutes brown for the black color in the wild-type hair. Such brown-agouti mice are called cinnamons. When wildtype mice are crossed with cinnamons, all of the $\mathrm{F}_{1}$ are wild type and the $\mathrm{F}_{2}$ has a 3: 1 ratio of wild type to cinnamon. Diagram this cross as in part $a$, letting $B$ stand for the wild-type black allele and $b$ stand for the cinnamon brown allele.
c. When mice of a true-breeding cinnamon line are crossed with mice of a true-breeding nonagouti (black) line, all of the $F_{1}$ are wild type. Use a genetic diagram to explain this result.
d. In the $F_{2}$ of the cross in part $c,$ a fourth color called chocolate appears in addition to the parental cinnamon and nonagouti and the wild type of the $\mathrm{F}_{1}$. Chocolate mice have a solid, rich brown color. What is the genetic constitution of the chocolates?
e. Assuming that the $A / a$ and $B / b$ allelic pairs assort independently of each other, what do you expect to be the relative frequencies of the four color types in the $\mathrm{F}_{2}$ described in part $d ?$ Diagram the cross of parts $c$ and $d$ showing phenotypes and genotypes (including gametes).
f. What phenotypes would be observed in what proportions in the progeny of a backcross of $\mathrm{F}_{1}$ mice from part $c$ with the cinnamon parental stock? With the nonagouti (black) parental stock? Diagram these backcrosses. g. Diagram a testcross for the $\mathrm{F}_{1}$ of part $c .$ What colors would result and in what proportions?
h. Albino (pink-eyed white) mice are homozygous for the recessive member of an allelic pair $C / c,$ which assorts independently of the $A / a$ and $B / b$ pairs. Suppose that you have four different highly inbred (and therefore presumably homozygous) albino lines. You cross each of these lines with a true-breeding wild-type line, and you raise a large $\mathrm{F}_{2}$ progeny from each cross. What genotypes for the albino lines can you deduce from the following $\mathrm{F}_{2}$ phenotypes?
$$\begin{array}{cccccc} 
& {4}{c} {\text {Phenotypes of progeny}} \\
 { 2 - 5 }   \mathrm{F}_{2} \text { of }
\text { line } & \begin{array}{c}
\text { Wild } \\
\text { type }
\end{array} & \text { Black } & \begin{array}{c}
\text { Cinna- } \\
\text { mon }
\end{array} & \begin{array}{c}
\text { Choco- } \\
\text { late }
\end{array} & \text { Albino } \\
\hline 1 & 87 & 0 & 32 & 0 & 39 \\
2 & 62 & 0 & 0 & 0 & 18 \\
3 & 96 & 30 & 0 & 0 & 41 \\
4 & 287 & 86 & 92 & 29 & 164 \\
\hline
\end{array}$$
(Adapted from A. M. Srb, R. D. Owen, and R.
S. Edgar General Genetics, 2nd ed. W. H. Freeman and Company,
1965.)

Introduction to Genetic Analysis

Anthony J. F.… 11th Edition

Chapter 6, Problem 42

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Solved by Expert Jessica Wooten

07/28/2020

Step 1

The true-breeding agouti line has the genotype AA, and the nonagouti line has the genotype aa. When these are crossed, all of the F1 generation will have the genotype Aa, which is agouti. When the F1 generation is crossed, the F2 generation will have the genotypes Show more…

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Many kinds of wild animals have the agouti coloring pattern, in which each hair has a yellow band around it. a. Black mice and other black animals do not have the yellow band; each of their hairs is all black. This absence of wild agouti pattern is called nonagouti. When mice of a true-breeding agouti line are crossed with nonagoutis, the $F_{1}$ is all agouti and the $F_{2}$ has a 3: 1 ratio of agoutis to nonagoutis. Diagram this cross, letting $A$ represent the allele responsible for the agouti phenotype and $a$ nonagouti. Show the phenotypes and genotypes of the parents, their gametes, the $F_{1}$, their gametes, and the $F_{2}$ b. Another inherited color deviation in mice substitutes brown for the black color in the wild-type hair. Such brown-agouti mice are called cinnamons. When wildtype mice are crossed with cinnamons, all of the $\mathrm{F}_{1}$ are wild type and the $\mathrm{F}_{2}$ has a 3: 1 ratio of wild type to cinnamon. Diagram this cross as in part $a$, letting $B$ stand for the wild-type black allele and $b$ stand for the cinnamon brown allele. c. When mice of a true-breeding cinnamon line are crossed with mice of a true-breeding nonagouti (black) line, all of the $F_{1}$ are wild type. Use a genetic diagram to explain this result. d. In the $F_{2}$ of the cross in part $c,$ a fourth color called chocolate appears in addition to the parental cinnamon and nonagouti and the wild type of the $\mathrm{F}_{1}$. Chocolate mice have a solid, rich brown color. What is the genetic constitution of the chocolates? e. Assuming that the $A / a$ and $B / b$ allelic pairs assort independently of each other, what do you expect to be the relative frequencies of the four color types in the $\mathrm{F}_{2}$ described in part $d ?$ Diagram the cross of parts $c$ and $d$ showing phenotypes and genotypes (including gametes). f. What phenotypes would be observed in what proportions in the progeny of a backcross of $\mathrm{F}_{1}$ mice from part $c$ with the cinnamon parental stock? With the nonagouti (black) parental stock? Diagram these backcrosses. g. Diagram a testcross for the $\mathrm{F}_{1}$ of part $c .$ What colors would result and in what proportions? h. Albino (pink-eyed white) mice are homozygous for the recessive member of an allelic pair $C / c,$ which assorts independently of the $A / a$ and $B / b$ pairs. Suppose that you have four different highly inbred (and therefore presumably homozygous) albino lines. You cross each of these lines with a true-breeding wild-type line, and you raise a large $\mathrm{F}_{2}$ progeny from each cross. What genotypes for the albino lines can you deduce from the following $\mathrm{F}_{2}$ phenotypes? $$\begin{array}{cccccc} & {4}{c} {\text {Phenotypes of progeny}} \\ { 2 - 5 } \mathrm{F}_{2} \text { of } \text { line } & \begin{array}{c} \text { Wild } \\ \text { type } \end{array} & \text { Black } & \begin{array}{c} \text { Cinna- } \\ \text { mon } \end{array} & \begin{array}{c} \text { Choco- } \\ \text { late } \end{array} & \text { Albino } \\ \hline 1 & 87 & 0 & 32 & 0 & 39 \\ 2 & 62 & 0 & 0 & 0 & 18 \\ 3 & 96 & 30 & 0 & 0 & 41 \\ 4 & 287 & 86 & 92 & 29 & 164 \\ \hline \end{array}$$ (Adapted from A. M. Srb, R. D. Owen, and R. S. Edgar General Genetics, 2nd ed. W. H. Freeman and Company, 1965.)

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Key Concepts

Mendelian Inheritance

This concept refers to the principles originating from Gregor Mendel’s work where traits are determined by single genes that have two alleles, one dominant and the other recessive. Fundamental aspects include segregation of alleles during gamete formation and the appearance of predictable phenotypic ratios (such as 3:1) in the offspring of specific crosses.

Dominance and Recessivity

Dominance describes the phenomenon where one allele masks the expression of a second, recessive allele in heterozygotes. This is key to understanding why crosses between different phenotypes often produce offspring that display only one of the parental traits, while the recessive phenotype only appears when the individual is homozygous for the recessive allele.

Alleles and Genotypes

Alleles are alternative forms of a gene, and the genotype is the combination of these alleles in an organism. Understanding how different alleles combine (e.g., homozygous versus heterozygous states) allows for predicting the phenotypic outcomes of genetic crosses.

Punnett Square Analysis

Punnett squares are a fundamental tool used to visualize and predict the outcome of genetic crosses. They allow one to diagram all possible combinations of alleles from parental gametes, facilitating the calculation of expected genotypic and phenotypic ratios in the progeny.

Independent Assortment

This principle refers to the random distribution of different genes and their alleles during gamete formation. When genes are not linked, they segregate independently, which results in predictable combinations of traits and phenotypic ratios in dihybrid or multigenic crosses.

Epistasis and Gene Interactions

Epistasis occurs when the effect of one gene depends on the presence of one or more 'modifier genes'. This interaction means that the phenotypic expression of one gene can mask or modify the effect of another, leading to phenotypes that are not predictable by simple Mendelian ratios and necessitating more complex analysis.

Genetic Crosses: F1, F2, Backcross, and Testcross

These are specific types of breeding experiments used to determine the mode of inheritance of a trait. An F1 cross involves the first generation of offspring from a parental cross, F2 is the second generation resulting from crossing F1 individuals, a backcross involves mating an F1 individual with one of the parental phenotypes, and a testcross is used to reveal the genotype of an individual by crossing it with a homozygous recessive individual. Each type of cross provides critical insight into the underlying genetic mechanisms governing trait inheritance.

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Transcript

00:01 So if we take 42a, if we breed an agouti mouse with a non -agootie mouse, the f -1 will all be big a, little a, and a goody.

00:28 And then the f -2 will be three -to -one ratio, three agouti, with one individual that has big a, big a, and two individuals that are heterozygous.

00:41 And one individual that's non -a -goody.

00:47 And that one will have a genotype of two little a's.

00:51 For b part, it's similar.

00:55 So we have a wild type mouse, which is black of beauty, crossed with a cinnamon mouse, and then in the f -1, they will all be heterozygous because the gametes are big b and little b, and they'll all have the wild -type color.

01:21 And then in the f -2, you'll have three that are, wild type with one that's big b, big b, two that's heterozygous, and one that is cinnamon.

01:39 And that individual will have the genotype two little bs.

01:45 For c part, if you take a cinnamon mouse, so big a, big a, two little bs, anytime you have two little bs, it's cinnamon, and two little a's and two big b, so that's a non -agutti mouse.

02:09 In the f1, you will have a heterozygot that is a wild type black of duty.

02:20 And then in the f1, i'm sorry, in the f2, when you breathe them, we'll look at what happens in d part.

02:28 So in d part, what we're supposed to do is breed these two individuals together.

02:35 So d part, we're going to do a breeding between two heterozygotes from c.

02:41 And what we get is this ratio.

02:50 We have nine mice that are black agouti, which is wild type.

03:01 We have three mice that are black but non -agooty.

03:13 We have three mice that are cinnamon.

03:17 So remember, cinnamon has these two little bees.

03:22 And we have one individual that's a double recessive and that individual is chocolate.

03:28 Now, that's really important because if you have a chocolate individual that shows up, that tells you a lot of information about your f1.

03:36 So we're going to keep that information sort of handy so that when we move forward, we have it.

03:42 Okay, so that's d part.

03:45 So let's look at e.

03:47 E part, you take, you have parents who are homozygous dominant, homozygous recessive, so that individual is cinnamon, right, because the two little bees.

04:00 Bred with a mouse that is homozygous recessive for one gene and homozygous dominant for the other.

04:06 And that individual is a black mouse, but he's a non -agooty.

04:13 And in the f -1, they should all be black -a -goody because you have a heterozygous.

04:19 Oh, oops, hold on.

04:21 I don't know why i read a little b there.

04:24 You should have a heterozygous, which exhibits the wild type, remember, which is black -a -goody, color.

04:32 And then in the f2, you should have a 9 -3 -3 -1 ratio where nine exhibit the wild type color.

04:41 So a something, b something, wild type.

04:47 And that distribution of genotypes looks like this.

04:49 There should be one big a, big a, big b.

04:53 There should be two heterozygous a, homozygous dominant bs, two homozygous dominant a, heterozygous i'm sorry, four heterozygous for both traits.

05:09 Now, if you're uncertain, just simply do a diehybrid cross between two heterozygous individuals from the f1.

05:18 Three of these individuals will be black non -agooty, and they will look like one, two little a's dominant homozygous b, and two homozygous recessive a's heterogeneous bs.

05:39 Three individuals will be cinnamon.

05:42 So big a, something, two little bs.

05:51 And they will be like this, one homozygous domina a, homozygous recessive b, and two heterozygous a, homozygous recessive b.

06:02 And then the last individual will be chocolate.

06:08 And that genotype will be two, it's a double recessive.

06:12 So two homozygous genes.

06:15 Now, if we continue and we look at f, f says, you know, what happens if you breed a wild type mouse with a cinnamon mouse? and what happens is you get one -fourth homozygous dominant for a, heterozygous for b, one -fourth heterozygous for both.

06:51 And both of those are wild -type.

06:54 So there's 50 % of your mice will be wild -type color.

06:59 And then 50 % will be cinnamon.

07:02 And so they will have big a.

07:04 Big a and two little b's, and then heterozygous for a, but still two little b's because that's the only way you get cinnamon.

07:15 And then the other breeding for f, we take a wild type mouse and we breed it with a black non -a -guity mouse.

07:29 And so here we have big a, little a, big b, cross with two little a's and two big bs.

07:37 And so remember, this is wild type and this is black, non -a -goody.

07:45 And in the offspring, what you get is one -fourth big a, little a, two big b's, and that's wild -type color.

07:53 One -fourth, big a, little a, big b, little b.

07:57 And so that's 50 % wild -type color.

08:03 And then one -fourth, two little a's, big b, big b, and one -fourth, two little a's, heterozycus b.

08:11 And so these are 50 % of your offspring and they're going to be black, non -a -goody.

08:23 Okay.

08:24 And then let's work on g part.

08:31 G is what happens when you take a wild type mouse and you breed it with a chocolate mouse.

08:44 And so what happens is you get one four that are big a little a, big b, little b, and that's going to be wild type.

08:57 One fourth will be big a little a, two little bs, and that's cinnamon.

09:06 One fourth will be two little a's, big b little b.

09:10 And because there's two little a's there, that's black, non -a -goody.

09:17 And then one -fourth will be chocolate.

09:25 Okay.

09:28 And then h part has four parts to it.

09:31 So h cross 1.

09:37 The information we need here is that to be albino, they must have two little cs.

09:46 Now, i underline my recessive cs because you can't tell the difference between my upper and lowercase c's.

09:53 So an underlined c is recessive.

09:56 So in cross 1, the parents are albino, somewhat underline those recessive traits.

10:02 We don't know what they are for a or b.

10:07 So we're going to leave them blank.

10:09 And they're crossed with a parent that has two, that's homozygous dominant for all three genes.

10:16 In the f1, we know that they're not albino, but they carry the albino gene.

10:23 They have at least one big a and one big b.

10:27 In the f2, you get 87 mice that are wild type in color.

10:35 32 that are cinnamon and 39 that are albino.

10:44 And so there's a couple things we can make a point of, and i'm going to just write them over here.

10:49 Or maybe i'll just put them in, i'll put them in different color where we're next to it...

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