Question

5. Consider the following tetrahybrid self-cross:

Aa Bb Cc Dd    X    Aa Bb Cc Dd

Calculate the probability for each of the following offspring. Show your work.

1. Genotype Aa BB CC Dd

2. Heterozygous for ALL FOUR genes

3. Heterozygous for ANY gene

4. Dominant phenotype for ALL FOUR traits

5. GIVEN that an individual offspring has the dominant phenotype for all four traits, what is the probability that individual is heterozygous for all four genes?

6. GIVEN that an individual offspring has the recessive phenotype for all four traits, what is the probability that individual is homozygous for all four genes?

Answer 1

Given tetrahybrid cross is :

AaBbCcDd × AaBbCcDd

To determine the probability of a particular genotype we need to determine the probability of passing allele at each locus. For gene A, the croos is Aa × Aa.

	A	a
A	AA	Aa
a	Aa	aa

We can observe that, probabilities of each gamete is:

p(AA) = 1/4

p(Aa) =2/4

p(aa) = 1/4

Also, probability of passing dominant phenotype would be the sum of AA and Aa, i.e. p(A_).

p(A_) =3/4

The same would apply to other genes and gametes formed.

Part A:

Genotype Aa BB CC Dd

= p(Aa) × p(BB) × p(CC) × p(Dd)

= (2/4) × (1/4) × (1/4) × (2/4)

= (1/64)

Part B:

Heterozygous for ALL FOUR genes: i.e. AaBbCcDd

= p(Aa) × p(Bb) × p(Cc) × p(Dd)

= (2/4) ×(2/4) ×(2/4) ×(2/4)

= (16/256)

=(1/16)

Part C:

Heterozygous for any gene:

The required probability can be determined by subtracting the sum of probability of homozygous conditions from 1.

= 1 - { p(AABBCCDD)+p(aabbccdd) }

= 1- { (1/4) 4 + (1/4) 4}

= 1 - {(1/256) + (1/256) }

= 1-(2/256)

= (254/256)

Part D:

Dominant phenotype for ALL FOUR traits

For dominant traits, the required genotype would be A_B_C_D_

=p(A_) × p (B_) × p (C_) × p(D_)

=(3/4) ×(3/4) ×(3/4) ×(3/4)

= (81/256)

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