Question

Four independent assorting genes, A, B, C and D, interact to affect the development of the retina in humans. Individuals who are homozygous recessive for either one of the genes will be able to see, individual who are homozygous recessive for either two or three genes will have diminishing sight. If an individual who has four homozygous recessive genes will be blind. In a cross between two individuals with these genotypes: AaBBccDd x AabbccDd

a. What is the probability that the child will be blind? Show calculations

b. What is the probability that the child AaBbccDD? Show calculations

Answer 1

as the given genotype of individuals are AaBBccDd and AabbccDd

let see the different probability of individual gene allele combination

Both individuals are heterozygous for gene A

so total number of combination= 4 (1 AA, 2Aa, 1aa) ,

so, probability of AA=1/4, probability of Aa= 1/2 and probability of aa= 1/4

similarly in case of B gene, one is homozygous dominant while other is homozygous recessive

so total number of combination= 4 ( 4 Bb)

probability of BB=0, probability of Bb=1 and probability of bb=0

in case of gene C, both individuals are homozygous recessive

so total number of combination= 4 ( 4 cc)

probability of CC=0, probability of Cc=0 and probability of cc=1

in case of gene CD both individuals are heterozygous

so total number of combination= 4 (1 DD, 2Dd, 1dd) ,

probability of DD=1/4, probability of Dd= 1/2 and probability of dd= 1/4

now

a) child must have to possess all four homozygous recessive genes, in order to show a bind trait

so probability of having gene aabbccdd = probability of aa x probability of bb x probability of cc x probability of dd =1/4 x 0 x 1 x 1/4 = 0

b) the probability that the child having gene combination AaBbccDD= probability of Aa x probability of Bb x probability of cc x probability of DD = 1/2 x 1 x 1 x 1/4 = 1/8