Question

40. Pink-eye and albino are two recessive traits found in the deer mouse Peromyscus maniculatus. In pink-eyed mice, the eye is devoid of color and appears pink because of the blood vessels within it. Albino mice are completely lacking color, both in their fur and in their eves. F. H. Clark crossed pink-eyed mice with albino mice; the resulting Fi had normal coloration in their fur and eyes. He then crossed these Fį mice with mice that were pink eyed and albino and obtained the following progeny. It is hard to distinguish between mice that are albino and mice that are both pink eyed and albino, so he combined these two phenotypes (F. H. Clark. 1936. Journal of Heredity 27:259–260). 12 Number of Phenotype progeny wild-type fur, wild-type eye color wild-type fur, pink-eye 62 albino 78 albino, pink-eyes Total 152 a. Give the expected numbers of progeny with each phenotype if the genes for pink-eye and albino assort independently. b. Use a chi-square test to determine if the observed numbers of progeny fit the number expected with independent assortment.

Answer 1

a .-

In this case, albinism and pink eye color are recessive traits found within the deer mouse Peromyscus maniculatus. So, the wild type fur color is expressed by gene A where a corresponds to the recessive allele.

P corresponds to dominant wild type eye color allele and p corresponds to the recessive allele.

At first, F. H. Clarke crossed albino mice with pink-eyed mice and the F1 (first filial) generation showed normal coloration in fur and eyes.

Parental – ppaa X PPAA

F1 generation – PpAa (wild type fur and wild type eye color)

Again, the F1 offspring were crossed to pink-eyed, albino mice, so the cross is between PpAa and ppaa.

The four genotypic combinations produced in F2 generation will be PpAa (wild type eye color and wild type fur), Ppaa (wild type eye color and albino), ppAa (pink eyed wild type fur) and ppaa (pink eyed albino).

Following independent assortment of Mendel, phenotypic ratio for a dihybrid cross is 1:2:1.

Hence, expected frequencies for homozygous dominant individuals among the overall population = ¼

Expected frequencies for heterozygous individuals among total population = ½

Expected frequencies for homozygous recessive individuals among total population = ¼

Here, total mice in population is given as 152.

Therefore, wild type fur and wild type eye colored mice = ¼ X 152 = 38.

Expected frequency of wild type, pink eyes mice = ¼ X 152 = 38.

Expected frequency of albino, wild type or pink eyes = ½ X 152 = 76

Phenotype	Observed	Expected
wild type, wild type	12	38
wild type, pink eyes	62	38
albino, wild type or pink eyes	78	76
Total	152	152

b.

	Observed (O)	Expected (E)	(O-E)2	(O-E)2/E
Wild type, wild type	12	38	676	17.78
Wild type, pink eyes	62	38	576	15.15
albino, wild type, or pink eyes	78	76	4	0.052
Total	152	152		32.982

So, the chi-square value is 32.982 that is almost equal to 33.

There are four alleles involved in this case and hence, the degrees of freedom will be (4-1) = 3.