In lecture you were shown hybrid crosses involving 2 and 3 loci. If you were to

Question

In lecture you were shown hybrid crosses involving 2 and 3 loci. If you were to conduct a hybrid cross with 4 loci, what would the dimensions of your Punnett square be? If the alleles were purely codominant and additive to yield a quantitative phenotypic trait, how many different phenotypic classes would you expect? Explain your reasoning. Hint: You do NOT need to actually construct a Punnett square to figure out the answer and in fact it is not recommended to construct a Punnett square. Look at the similar example of a polygenic trait with 3 loci from your online activities.

Explanation / Answer

Part I

First we should calculate number of possible gametes.

The formula for calculating number gametes is 2k.

Here k is number of genes involved in the cross.

For, monohybrid cross, there will be 21 =2 gametes.

For, dihybrid cross, there will be 22 =4 gametes.

For, trihybrid cross, there will be 23 =8 gametes.

For a cross involving 4 genes, there will be 24 =16 gametes.

So, in a cross involving 4 genes, each parent produces 16 different gametes. Therefore, dimensions of Punnett square will be 16 x 16.

Part II

A codominant pair of alleles gives 3 possible genotypes.

Formula to be used here is 3k

Here k is number of codominant loci.

For, monohybrid cross, there will be 31 =3 phenotypes.

For, dihybrid cross, there will be 32 =9 phenotypes.

For, trihybrid cross, there will be 33 =27 phenotypes.

For a cross involving 4 genes, there will be 34 =81 phenotypes.

Therefore, If there are 4 genes each one having two codominant alleles, there will be 3 x 3 x 3 x 3 = 81 different phenotypes.

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.