Question

A maize plant with genotype Aa Bb produces the following gametes 100 Ab 45 AB 106 aB 50 ab This tells us that the two genes are likely on different chromosomes. True False

Answer 1

Answer:

True

Explanation:

The linkage relation of two genes can be determined on the basis of calculating recombination frequency.

If the recombination frequency value is greater than 50% then the genes are not linked and if the value is less than 50% then the genes are linked.

In the given question, the  crossover gametes formed will be Ab = 100

and aB= 106.

Therefore,

$\dfrac{recombinant\ progenies}{ Total\ progeny}\times 100\\\\\dfrac{100 +106}{100+106+45+50}\times 100\\\\=\dfrac{206}{301}\times 100\\\\= 68.43\%$

Since the recombinant frequency value is larger than 50% therefore the genes do not show linkage.

Thus, True is the correct answer.