Question

If two parents are the following genoype AaBoCcDd x AaBboODD mate, predict the fraction of offspring which would be the following Assume all the genes assort independently and that the uppercase letters represent dominant alleles a. How many offspring are expected to express the phenotype aBCD? b. How many offspring are expected to have the genotype AABCDD?

Answer 1

Given: The cross AaBbCcDd x AaBbccDD.

Genotype of parent 1: AaBbCcDd

Genotype of parent 2: AaBbccDD

During gametogenesis, each parent produces haploid gametes.

It is given that all genes assort independently. What this means that the presence of any one gene in a given haploid gamete is independent of the presence of any of the other 3 genes.

Parent 1:

Number of ways in which alleles for gene A can be inherited = 2 (the gamete can either have A or a).

So, probability of inheriting allele A = probability of inheriting allele a = 1/2.

Number of ways in which alleles for gene B can be inherited = 2 (the gamete can either have B or b).

So, probability of inheriting allele B = probability of inheriting allele b = 1/2.

Number of ways in which alleles for gene C can be inherited = 2 (the gamete can either have C or c).

So, probability of inheriting allele C = probability of inheriting allele c = 1/2.

Number of ways in which alleles for gene D can be inherited = 2 (the gamete can either have D or d).

So, probability of inheriting allele D = probability of inheriting allele d = 1/2.

Therefore, total number of possible combinations of alleles in the gametes from parent 1 = 2 x 2 x 2 x 2 = 16.

Parent 2:

Number of ways in which alleles for gene A can be inherited = 2 (the gamete can either have A or a).

So, probability of inheriting allele A = probability of inheriting allele a = 1/2.

Number of ways in which alleles for gene B can be inherited = 2 (the gamete can either have B or b).

So, probability of inheriting allele B = probability of inheriting allele b = 1/2.

Number of ways in which alleles for gene C can be inherited = 1 (the gamete can only have allele c).

So, probability of inheriting allele C = 0.

Probability of inheriting allele c = 1.

Number of ways in which alleles for gene D can be inherited = 1 (the gamete can only have allele D).

So, probability of inheriting allele D = 1.

Probability of inheriting allele = 0.

Therefore, total number of possible combinations of alleles in gametes from parent 2 = 2 x 2 x 1 x 1 = 4.

So, in the cross between the 2 parents, the total number of combinations of the alleles in the offspring = 16 x 4 = 64.

a) To express the phenotype aBCD, the offspring must inherit the allele a from both parents (since a is recessive and can only be expressed in the homozygous condition) and must inherit the alleles B, C and D from atleast one parent (since B, C and D are dominant, they can be expressed in either the homozygous or the heterozygous condition).

Probability of inheriting allele a from both parents = Probability of inheriting allele a from parent 1 x Probability of inheriting allele a from parent 2 = 1/2 x 1/2 = 1/4.

Probability of inheriting allele B from atleast 1 parent = Probability of inheriting allele B from parent 1 and allele b from parent 2 + Probability of inheriting allele b from parent 1 and allele B from parent 2 + Probability of inheriting allele B from both parents = 1/2 x 1/2 + 1/2 x 1/2 + 1/2 x 1/2 = 3/4.

Probability of inheriting allele C from atleast 1 parent = Probability of inheriting allele C from parent 1 and allele c from parent 2 + Probability of inheriting allele c from parent 1 and allele C from parent 2 + Probability of inheriting allele C from both parents = 1/2 x 1 + 1/2 x 0 + 1/2 x 0 = 1/2.

Probability of inheriting allele D from atleast 1 parent = Probability of inheriting allele D from parent 1 and allele d from parent 2 + Probability of inheriting allele d from parent 1 and allele D from parent 2 + Probability of inheriting allele D from both parents = 1/2 x 0 + 1/2 x 1 + 1/2 x 1 = 1.

Therefore probability of expressing the phenotype aBCD = Probability of inheriting allele a x Probability of inheriting allele B x Probability of inheriting allele C x Probability of inheriting allele D = 1/4 x 3/4 x 1/2 x 1 = 3/32.

Number of offspring expressing the phenotype aBCD = Total number of offspring x probability of expressing the phenotype aBCD = 64 x 3/32 = 6.

b) Since the 4 genes are assorted independently, the probability of inheriting genotype AABbCcDD can be broken down into the probabilities of inheriting genotype AA, Bb, Cc and DD.

Probability of inheriting genotype AA = Probability of inheriting allele A from both parents = 1/2 x 1/2 = 1/4.

Probability of inheriting genotype Bb = Probability of inheriting allele B from parent 1 x Probability of inheriting allele b from parent 2 + Probability of inheriting allele b from parent 1 x Probability of inheriting allele B from parent 2 = 1/2 x 1/2 + 1/2 x 1/2 = 1/2.

Probability of inheriting genotype Cc = Probability of inheriting allele C from parent 1 x Probability of inheriting allele c from parent 2 + Probability of inheriting allele c from parent 1 x Probability of inheriting allele C from parent 2 = 1/2 x 1 + 1/2 x 0 = 1/2.

Probability of inheriting genotype DD = Probability of inheriting allele D from parent 1 x Probability of inheriting allele D from parent 2 = 1/2 x 1 = 1/2.

Probability of inheriting genotype AABbCcDD = Probability of inheriting genotype AA x Probability of inheriting genotype Bb x Probability of inheriting genotype Cc x Probability of inheriting genotype DD = 1/4 x 1/2 x 1/2 x 1/2 = 1/32.

Number of offspring with the genotype AABbCcDD = Total number of offspring x Probability of inheriting genotype AABbCcDD = 64 x 1/32 = 2.