A female genotype, shown as abc / +++ produces 100 meiotic tetrads. Among these, 68 showed no cross over (NCO) events; 20 showed single crossover (SCO1) between a and b; 10 showed single crossover between b and c (SCO2) and 2 showed a double crossover (DCO) between a and b and between b and c. Of the 400 gametes produced, how many will be of the following types? Fill your answers in the table. (Source)

enter image description here

According to me, the answers should be,

A. $68\times4=272$

B. $20\times4=80$

C. $10\times4=40$

D. $2\times4=8$

However, the answers are $336, 40, 20, 4$ respectively. What am I getting wrong?


1 Answer 1


You are not taking into account that crossovers happen at the four-chromatid stage and between cromatids. Take a look at the following picture1:

               Crossing over

I suggest you draw down all the crossover events:


Now count the number of gametes that will be formed:

$$ \begin{align} \text{(+++) and (abc)} &= 4\times 68+2\times20 + 2\times10 + 2\times 2=\mathbf{336} \\ \text{(+bc) and (a++)} &= 2\times20 = \mathbf{40} \\ \text{(ab+) and (++c)} &= 2\times10 = \mathbf{20} \\ \text{(a+c) and (+b+)} &= 2\times 2 = \mathbf{4} \\ \end{align} $$

  1. Introduction to Genetic Analysis 11th Edition, W. H. Freeman and Company, 2015.

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .