A farmer purchases f1 hybrid seeds. These seeds contain four heterozygous loci that give a significant increase in crop yield when compared to either homozygote. Two of these loci are on the same chromosome and show partial linkage. If the farmer allows the f1 plants to self-fertilize, what is the probability that a seed from the resulting f2 generation has all four heterozygous loci?

a) at least 1/2

b) between 1/2 and 1/4

c) between 1/4 and 1/8

d) between 1/8 and 1/16

e) less than 1/16

I think it's either d or e. For the first two genes (unlinked), there is 1/2 chance to get heterozygote. So, 1/2 * 1/2 = 1/4. For the next two linked genes, if they were unlinked, it would be 1/2*1/2 = 1/4 for a total of 1/16. But since they are linked, I'm thinking that the chance for heterozygote might be reduced, so maybe the answer is e) less than 1/16?? Not sure where to end, though...

  • 2
    $\begingroup$ Try to draw a Punnet square! If you have tried already, what difficulties have you encountered? What brought you to think the answer is either 'd' or 'e'? $\endgroup$
    – Remi.b
    Jan 6, 2018 at 0:18

1 Answer 1


I agree with your reasoning! If the two genes are perfectly linked, then the answer is $\frac{1}{8}$. If they are perfectly unlinked, then the answer is $\frac{1}{16}$. The answer therefore must be somewhere between $\frac{1}{8}$ and $\frac{1}{16}$. Good job!

Of course the answer e) is pretty silly as any other answer ('a','b','c' and 'd') would yield to 'e' being necessarily true. So, if you have to choose only one, I would just go with 'd' and highlight that the answer 'e' is silly.

I still don’t understand how if the two are completely linked, the answer would be 1/8.

Let's consider that at both loci, we have two alleles A and B. Here is a representation of the first F1 individual

-------A----A----- -------B----B-----

Here is a representation of the second F1 individual

-------A----A----- -------B----B-----

Of course, they are the same as the parental lines are pure breed. Now, if the two loci are perfectly linked, then the offspring can have the following possible genotypes

-------A----A----- -------A----A-----,

-------B----B----- -------A----A-----,

-------A----A----- -------B----B-----


-------B----B----- -------B----B-----

, but the offspring cannot be

-------B----A----- -------B----B-----

or an other combination requiring recombination between the two loci. Therefore the probability to be heterozygote for the F2 at two perfectly linked loci is the same as the probability to be heterozygote at one of these two loci (if we assume the mutation rate is negligible).

Why can't parents be: parent1: -------B----A----- -------A----B----- parent2: -------A----A----- -------B----B-----[?]

Because parental lines are pure lines. When we talk about an F1 generation, we talk about the hybrid between two pure line. By pure line, we mean that we got rid of all genetic variance. So, one parental line is fixed for the A alleles and the other line is fixed for the B alleles. In other words here are the genotypes of the pure parental lines.

-------A----A----- -------A----A-----


-------B----B----- -------B----B-----

One parent can therefore only give


while the other one can only give


so, the F1 generation is therefore necessarily heterozygote at all loci.

  • $\begingroup$ I still don’t understand how if the two are completely linked, the answer would be 1/8. It would mean that in order for these two linked genes to produce a heterozygote, they would have a 1/2 chance to do so. Is this necessarily true? $\endgroup$ Jan 6, 2018 at 6:59
  • $\begingroup$ Please see edit. $\endgroup$
    – Remi.b
    Jan 6, 2018 at 15:08
  • $\begingroup$ Why can't parents be: parent1: -------B----A----- -------A----B----- parent2: -------A----A----- -------B----B----- $\endgroup$ Jan 6, 2018 at 17:52
  • $\begingroup$ @LittleDragon Pleas see edit $\endgroup$
    – Remi.b
    Jan 6, 2018 at 17:54
  • $\begingroup$ Wow, speedy. So you're saying that all f1 parents must have the same alleles in the same order on chromosomes? If so, then I'll mark your response as the answer. $\endgroup$ Jan 6, 2018 at 18:05

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