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Suppose that gene A is on the X chromosome, and genes B, C, and D are on three different autosomes. A- signifies the dominant phenotype. An equivalent situation holds for B-. C-, and D-. The cross AABBCCDD females by aYbbccdd males is made.

A. What is the frequency of 'A-' individulas in the F1? Explain

B. What is the frequency of 'a' males in the F1? Explain

C. What is the frequency of A- B- C- D- among the femals in the F1? Explain

D. How many F2 genotypes will there be? Explain

E. What proportion of F2s will be heterozygous for the four genes? Explain

F. How often will you obtain each of the following genotypes in the F2: (1) A-bb CC dd (female); (2) aY BB Cc Dd; (3) AY bb cc dd; (4) aa bb Cc Dd. Show rationale.

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Tod Thiel
Tod ThielLv2
29 Sep 2019

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Related questions

Suppose gene A is on the X chromosome, and genes B, C and D are on three different autosomes. Thus, A- signifies the dominant phenotype in the male or female. An equal situation holds for B-, C- and D-. The cross AA BB CC DD (female) x aY bb cc dd (male) is made.

A) probability of obtaining A- individual in F1

B) probability of obtaining an a male in the F1 progeny

C) Probability of A- B- C- D- female in F1

D) How many different F1 genotypes are there?

E) Probability of F2 individuals will be heterozygous for the four gens?

F) Determine a probability of each of the following types in the F2 individuals

(1) A- bb CC dd (female);

(2) aY BB Cc Dd (male);

(3) AY bb CC dd (male);

(4) aa bb Cc Dd (female)

Suppose gene A is on the X chromosome, and genes B, C, and D are onthree different autosomes. Thus, A_ signifies the dominantphenotype in the male or female. An equivalent situation hold forB_, C_, and D_. The cross AA BB CC DD female x aY bb cc dd male ismade.

a. what is the probability of obtaining an A_ individual in the F1progeny?
b. What is the probability of obtaining an a male in the F1progeny?
c. What is the probability of obtaining an A_B_C_D_ female in theF1 progeny?
d. How many different F1 genotypes will there be?
e. what proportion of F2 individuals will be heterozygous for thefour genes?
f. determine the probabilities of obtaining each of the followingtypes in the F2 individuals (1) A_bbCCdd (female) (2) aYBBCcDd(male) (3) AYbbCCdd (male) (4) aabbCcDd (female)

Homozygous recessive loss of function mutations in any one of 4 different genes in C. elegans worms (genes A, B, C, or D) results in female worms that don't form proper egg-laying structures, called vulvas, as shown in Lines 2-5 of the data table below. In vulvaless females, the fertilized eggs hatch and baby worms develop inside the mother, killing her in the process.

genotype phenotype
1 AA BB CC DD (wild type) normal vulva
2 aa BB CC DD No vulva
3 AA bb CC DD no vulva
4 AA BB cc DD no vulva
5 AA BB CC dd no vulva

You would like to figure out the order in which these genes act during the creation of the vulva (you cannot assume it will be A--->B--->C--->D as the gene names are arbitrary). Your friend tells you to perform epistasis analysis by making double mutants between the different homozygous recessive mutants and analyzing the phenotype of the double mutant. For example, she asks you to examine the phenotype of aabbCCDD (double mutant of genes A and B) and compare this to the single mutants to figure out whether gene A acts earlier than gene B, or vice versa. You know this won't work. Why not? Pick the ONE BEST choice:

a. The phenotype of the single mutant is already pretty severe. The double mutant will most likely be dead, therefore epistasis analysis won't be possible.

b. Each of the single mutants (aaBBCCDD) and (AAbbCCDD) has the same mutant phenotype i.e., no vulva. Epistasis analysis between any 2 genes is only possible when the mutant phenotypes for each gene is different.

c. Epistasis analysis involves making triple mutants (such as aabbccDD) in order to learn something about how the genes are ordered.

d. Epistasis analysis can never be carried out with null loss of function mutations. The mutations being analyzed all have to be dominant gain of function alleles.

To address your concern, you first generate overactive alleles of either gene A (denoted as A*) or gene B (denoted as B*). As shown in lines 6-7 of the table below, C. elegans that have one of these overactive alleles produce multiple vulvas.

Genotype Phenotype
6 A*A BB CC DD multiple vulvas
7 AA B*B CC DD multiple vulvas

To find out the order in which these genes act, you combine the overactive alleles with different loss of function alleles and observe the phenotype in double mutants (see Lines 8-11).

Genotype Phenotype
8 A*A bb CC DD multiple vulvas
9 A*A BB cc DD multiple vulvas
10 A*A BB CC dd no vulva
11 AA B*B cc DD multiple vulvas

Based on the data in the table, what is the order in which these 4 genes normally act in wild-type C. elegans in order to produce a wild-type/normal vulva?

a. A----> B----->C----->D ---> Vulva

b. D----> B----->C----->A----> Vulva

c. B----> C----->A----->D----> Vulva

d. C----> B----->A----->D----> Vulva

e. C----> B----->D----->A----> Vulva

f. don't have enough data to make any conclusions

Based on the vulva formation phenotype of the double mutants, which of the following statement(s) accurately describes the genetic interactions between the A* allele and alleles of other genes affecting vulva formation? Pick ALL that apply:

a. The A* allele is epistatic to homozygous recessive loss of function mutations in gene B

b. A homozygous recessive loss of function mutation in gene B is epistatic to the A* allele

c. Homozygous recessive loss of function mutation in gene D is epistatic to the A* allele

d. The A* allele is epistatic to homozygous recessive loss of function mutation in gene D

e. The A* allele enhances the homozygous recessive loss of function mutation in gene D

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