Question

Part A: What is the probability of producing and F1 offspring that phenotypically resemble either parent from the cross?

AaBbCC x aabbCc

Note: Assume that capital letters represent completely dominant alleles.

Part B: Same cross. But in this case the alleles are incompletely dominant. What is the probability of producing and F1 offspring that phenotypically resemble either parent from the cross?

Answer 1

Answer Part A:

The possible genotypes for two allele A and a are— AA, Aa and aa. It is said that capital letters represent completely dominant alleles so A allele is completely dominant over a allele. So the possible phenotype will be phenotype-1 (for AA and Aa) and phenotype-2 (for aa). [When one allele is completely dominant over another then both homozygous dominant and heterozygous genotypes produce same phenotype.]

	A	a
a	Aa	aa
a	Aa	aa

Genotypic ratio in F1 generation— Aa:aa = 1:1; probability of Aa = 50% or 0.5 and probability of aa = 50% or 0.5

The possible genotypes for two allele B and b are— BB, Bb and bb. It is said that capital letters represent completely dominant alleles so B allele is completely dominant over b allele. So the possible phenotype will be phenotype-3 (for BB and Bb) and phenotype-4 (for bb). [When one allele is completely dominant over another then both homozygous dominant and heterozygous genotypes produce same phenotype.]

	B	b
b	Bb	bb
b	Bb	bb

Genotypic ratio in F1 generation— Bb:bb = 1:1; probability of Bb = 50% or 0.5 and probability of bb = 50% or 0.5

The possible genotypes for two allele C and c are— CC, Cc and cc. It is said that capital letters represent completely dominant alleles so C allele is completely dominant over c allele. So the possible phenotype will be phenotype-5 (for CC and Cc) and phenotype-6 (for cc). [When one allele is completely dominant over another then both homozygous dominant and heterozygous genotypes produce same phenotype.]

	C	C
C	CC	CC
c	Cc	Cc

Genotypic ratio in F1 generation— CC:Cc = 1:1; probability of CC = 50% or 0.5 and probability of Cc = 50% or 0.5

Genotype of the parents = AaBbCC and aabbCc. Phenotype for the genotype AaBbCC is phenotype-1, phenotype-3, phenotype-5 and Phenotype for the genotype aabbCc is phenotype-2, phenotype-4, phenotype-5.

So, if cross is done between AaBbCC and aabbCc the phenotypes in F1 generation which are resemble with the phenotype of either parent are— AaBbCC, AaBbCc (phenotype-1, phenotype-3, phenotype-5) and aabbCc, aabbCC (phenotype-2, phenotype-4, phenotype-5).

Now, if we imply the multiplicative rule of probability for genetic analysis the probability of having these genotypes can be calculated.

Probability of AaBbCC = 0.5 0.5 ​​​​​​​0.5 = 0.125

Probability of AaBbCc = 0.5 ​​​​​​​0.5 ​​​​​​​0.5 = 0.125

Probability of aabbCc = 0.5 ​​​​​​​0.5 ​​​​​​​0.5 = 0.125

Probability of aabbCC = 0.5 ​​​​​​​0.5 ​​​​​​​0.5 = 0.125

Now, if we imply the additive rule of probability for genetic analysis the probability of having these phenotypes can be calculated.

Probability of (phenotype-1, phenotype-3, phenotype-5) = AaBbCC + AaBbCc = 0.125 + 0.125 = 0.25

Probability of (phenotype-2, phenotype-4, phenotype-5) = aabbCc + aabbCC = 0.125 + 0.125 = 0.25

So, if cross is done between AaBbCC and aabbCc the probability the phenotypes in F1 generation which are resemble with the phenotype of either parent = 0.25 + 0.25 = 0.5

So, the probability = (0.5 100)% = 50%

Thus, the probability of producing and F1 offspring that phenotypically resemble either parent from the cross AaBbCC x aabbCc is 50% or 0.5

Answer Part B:

The possible genotypes for two allele A and a are— AA, Aa and aa. It is said that the alleles are incompletely dominant, i.e., neither allele is dominant over another. So A and a alleles are incompletely dominant. So the possible phenotype will be phenotype-1 (for AA) phenotype-2 (for Aa) and phenotype-3 (for aa). [When two alleles are incompletely dominant then homozygous dominant and heterozygous genotypes produce two different phenotypes.]

	A	a
a	Aa	aa
a	Aa	aa

Genotypic ratio in F1 generation— Aa:aa = 1:1; probability of Aa = 50% or 0.5 and probability of aa = 50% or 0.5

The possible genotypes for two allele B and b are— BB, Bb and bb. It is said that the alleles are incompletely dominant, i.e., neither allele is dominant over another. So B and b alleles are incompletely dominant. So the possible phenotype will be phenotype-4 (for BB) phenotype-5 (for Bb) and phenotype-6 (for bb). [When two alleles are incompletely dominant then homozygous dominant and heterozygous genotypes produce two different phenotypes.]

	B	b
b	Bb	bb
b	Bb	bb

Genotypic ratio in F1 generation— Bb:bb = 1:1; probability of Bb = 50% or 0.5 and probability of bb = 50% or 0.5

The possible genotypes for two allele C and c are— CC, Cc and cc. It is said that the alleles are incompletely dominant, i.e., neither allele is dominant over another. So C and c alleles are incompletely dominant. So the possible phenotype will be phenotype-7 (for CC) phenotype-8 (for Cc) and phenotype-9 (for cc). [When two alleles are incompletely dominant then homozygous dominant and heterozygous genotypes produce two different phenotypes.]

	C	C
C	CC	CC
c	Cc	Cc

Genotypic ratio in F1 generation— CC:Cc = 1:1; probability of CC = 50% or 0.5 and probability of Cc = 50% or 0.5

Genotype of the parents = AaBbCC and aabbCc. Phenotype for the genotype AaBbCC is phenotype-2, phenotype-5, phenotype-7 and Phenotype for the genotype aabbCc is phenotype-3, phenotype-6, phenotype-8.

So, if cross is done between AaBbCC and aabbCc the phenotypes in F1 generation which are resemble with the phenotype of either parent are— AaBbCC (phenotype-2, phenotype-5, phenotype-7) and aabbCc (phenotype-3, phenotype-6, phenotype-8).

Now, if we imply the multiplicative rule of probability for genetic analysis the probability of having these genotypes can be calculated.

Probability of AaBbCC = 0.5 ​​​​​​​0.5 ​​​​​​​0.5 = 0.125

Probability of aabbCc = 0.5 ​​​​​​​0.5 ​​​​​​​0.5 = 0.125

So, the probability of (phenotype-1, phenotype-3, phenotype-5) = AaBbCC = 0.125 and probability of (phenotype-2, phenotype-4, phenotype-5) = aabbCc = 0.125

So, if cross is done between AaBbCC and aabbCc the probability the phenotypes in F1 generation which are resemble with the phenotype of either parent = 0.125 + 0.125 = 0.25

So, the probability = (0.25 100)% = 25%

Thus, the probability of producing and F1 offspring that phenotypically resemble either parent from the cross AaBbCC x aabbCc is 25% or 0.25