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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)

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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?

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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:

               Crossovers

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.
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