Solutions For An Introduction to Genetic Analysis Chapter 6 Problem 42P

So, the cross of Agouti and non-agouti results in production of one pure agouti, 2 intermediate agoutis and 1 non-agouti.

b. The genotype and phenotypes of the cross between wild type and cinnamon are:

Parent generation B/B (wild type) X b/b (cinnamon)

Gametes B and b

↓

F₁ generation B/b (wild type) X B/b (wild type)

Gametes B and b

↓

F₂ generation 1 B/B (wild type): 2 B/b (wild type): 1 b/b (cinnamon)

So, the cross of wild type and cinnamon results in the production of one wild type, 2 intermediate wild types, and 1 cinnamon.

c. The genotype and phenotypes of the cross between cinnamon or brown agouti and Black non-agouti are:

Parent generation A/A; b/b X a/a; B/B

(Cinnamon or brown agouti) (Black non-agouti)

↓

Gametes A; b and a; B

↓

F₁generation A/a; B/b (All are wild type or black agouti)

So, the cross of cinnamon or brown agouti and black non-agouti results in the production of all wild types and black agouti in F1 generation.

d. The self-cross of Wild type or black agouti results in producing variety of offsprings. They are as below:

F₁ generation AaBb × AaBb

↓

F₂ generation: Genotypes and phenotypes are as follows:

9 A/-; B/- black agouti

3 a/a; B/- black non-agouti

3 A/-; b/b cinnamon

1 a/a; b/b chocolate

So, the self-cross of Wild type or black agouti results in production of 9 black agouti, 3 black non-agouti, 3 cinnamon, and 1 chocolate.

e. The genotype and phenotypes of the cross between cinnamon and black non-agouti are:

Parent generation A/A; b/b (cinnamon) X a/a; B/B (black non-agouti)

Gametes A; b and a; B

F₁ generation: A/a; B/b (all are wild type)

Gametes ABAb × aBab

F₂ generation

The genotype and phenotypes of the above cross is:

Dihybrid ratio	Alleles	Genotypes	Phenotypes
9	A/-; B/-		wild type
		1 A/A; B/B
		2 A/a; B/B
		2 A/A; B/b
		4 A/a; B/b
3	a/a; B/-		black non-agouti
		1 a/a; B/B
		2 a/a; B/b
3	A/-		cinnamon
		1 A/A; b/b
		2 A/a; b/b
1	a/a; b/b		chocolate

f. The genotype and phenotypes of the cross between wild type and cinnamon are:

Parent generation AaBb × AAbb

(Wild type) (Cinnamon)

F₁

¼ A/A; B/b wild type

¼ A/a; B/b wild type

¼ A/A; b/b cinnamon

¼ A/a; b/b cinnamon

The resultant genotype and phenotype of the above cross is:

The genotype and phenotypes of the cross between wild type and black non-agouti are:

Parent generation A/a; B/b X a/a; B/B

(Wild type) (Black non-agouti)

F₁

¼ A/a; B/B wild type

¼ A/a; B/b wild type

¼ a/a; B/B black non-agouti

¼ a/a; B/b black non-agouti

g. The genotype and phenotypes of the cross between wild type and black chocolate are:

Parent generation A/a; B/b X a/a; b/b

(Wild type) (Chocolate)

F₁

¼ A/a; B/b wild type

¼ A/a; b/b cinnamon

¼ a/a; B/b black non-agouti

¼ a/a; b/b chocolate

h. The mice must be c/c to be albino but only the F₂ progeny only would aid us in determining the genotype with regards to A and B genes

Cross 1

Parent generation c/c; ?/?; ?/? X C/C; A/A; B/B

F₁ generation C/c; A/-; B/-

F₂ generation C/c; A/-; B/- X C/c; A/-; B/-

The genotypes and phenotypes are as follows:

No. of Phenotypes	Phenotypes	Genotypes
87	wild type	C/-; A/-; B/-
32	cinnamon	C/-; A/-; b/b
39	albino	c/c; ?/?; ?/?

In order to get cinnamon in F₂ the parents in F₁ must be B/b (as the wild type is B/B while the albino is b/b). Thus the F₁ parent could be written as C/c; A/-; B/b gives off 1/4^th of the progeny as albino (c/c) and the remaining 3/4^th of the remaining progeny would be as the black either agouti or non-agouti while the rest of the 1/4^th would be cinnamon (if agouti) or chocolate (if non-agouti). Since chocolate is not formed the F₁ parents should be devoid of the allele for non-agouti. This proves that F₁ is A/A and the original albino much be c/c; A/A; B/B.

Cross 2

Parent generation c/c; ?/?; ?/? X C/C; A/A; B/B

F₁ generation C/c; A/-; B/-

F₂ generation C/c; A/-; B/- X C/c; A/-; B/-

The genotypes and phenotypes are as follows:

No. of Phenotypes	Phenotypes	Genotypes
62	wild type	C/-; A/-; B/-
18	albino	c/c; ?/?; ?/?

The ratio of wild type to albino is in the ratio of 3:1 showing that the gene is heterozygous in F₁. Hence, the albino must be c/c; A/A; B/B.

Cross 3

Parent generation c/c; ?/?; ?/? X C/C; A/A; B/B

F₁ generation C/c; A/-; B/-

F₂ generation C/c; A/-; B/- X C/c; A/-; B/-

The genotypes and phenotypes are as follows:

No. of Phenotypes	Phenotypes	Genotypes
96	wild type	C/-; A/-; B/-
30	black	C/-; a/a; B/-
41	albino	c/c; ?/?; ?/?

The presence of black non-agouti in F₂ shows that the F₁ should be heterozygous for the gene A. Hence, the gene could be written as C/c; A/a; B/- and the albino parent must be c/c; a/a; ?/?. The colored F₂ generation showing a ratio of 3:1 shows only those two genes is heterozygous of F₁. Hence, F₁ should be C/c; A/a; B/B, while the albino should be c/c; a/a; B/B.

Cross 4

Parent generation C/c; ?/?; ?/? X C/C; A/A; B/B

F₁ generation C/c; A/-; B/-

F2 generation C/c; A/-; B/- X C/c; A/-; B/-

The genotypes and phenotypes are as follows:

No. of Phenotypes	Phenotypes	Genotypes
287	wild type	C/-; A/-; B/-
86	black	C/-; a/a; B/-
92	cinnamon	C/-; A/-; b/b
29	chocolate	C/-; a/a; b/b
164	albino	c/c; ?/?; ?/?

For getting chocolate in F₂ the F₁ should be heterozygous for all genes while the albino parent must be c/c; a/a; b/b.