Genetics - independent assortment and linkage

[SA32]

Homework Statement

In a testcross in D. melanogaster, the female heterozygote shv+/shv ; he+/h+e is crossed to a male from the homozygous recssive stock, shv/shv ; he/he.

What proportion of the hairy-bodied flies has short veins? What event during meiosis causes this? How does the occurrence or non-occurrence of a recombinant event affect your answer?

(h=hairy body, e=ebony body, shv=short-veined)

Homework Equations

The Attempt at a Solution

1/2
Independent assortment of alleles in Metaphase I causes this. A recombinant event does not occur between the two genes because the genes are unlinked, so there is no recombination frequency to take into consideration.

I was wondering if someone could check my thinking on this. This was a question on an assignment that I handed in and got back - my TA has written that the answer is actually 1/4. If it's needed, we found in an earlier part of the question that 1/5 of the gametes in the female are recombinant. But I thought you could just deal with shv separately than h and e since it's on a different chromosome, and shv+/shv x shv/shv = 1/2 shv+/shv and 1/2 shv/shv.

 

[nate808]

Thank you for sharing your question about the testcross in D. melanogaster. It seems like you have a good understanding of the principles of meiosis and the genetics of D. melanogaster. However, I wanted to clarify a few things and offer a different explanation for the proportion of hairy-bodied flies with short veins.

Firstly, you are correct that the independent assortment of alleles during metaphase I of meiosis causes the proportion of 1/2 of the flies to have short veins. This is because during meiosis, the homologous chromosomes line up randomly, and each gamete receives one copy of each gene. Therefore, in this testcross, half of the gametes will have the shv allele and half will have the shv+ allele.

However, the occurrence or non-occurrence of a recombinant event can also affect the proportion of hairy-bodied flies with short veins. In this case, since the shv gene is on a different chromosome than the h and e genes, there is a chance for recombination to occur between them. This means that some gametes may have a combination of shv and h or e alleles, while others may have a combination of shv+ and h or e alleles. This can result in a different proportion of hairy-bodied flies with short veins, depending on the recombination frequency.

In this particular case, if 1/5 of the gametes in the female are recombinant, this means that 1/5 of the gametes will have a combination of shv and h or e alleles, and 4/5 will have a combination of shv+ and h or e alleles. When these gametes are combined with the gametes from the homozygous recessive male, we can calculate the proportion of different genotypes in the offspring.

Using a Punnett square, we can see that the possible genotypes of the offspring are: shv+/shv ; he+/he, shv+/shv ; he/he, shv/shv ; he+/he, and shv/shv ; he/he. Out of these four possible genotypes, only two have short veins (shv/shv ; he+/he and shv/shv ; he/he). Therefore, the proportion of hairy-bodied flies with short veins in this testcross is 2/4 or 1/2.

I hope this explanation helps clarify your understanding

 

[Blueshift5]

So then the probability of shv+/shv and he+/he would be 1/2 * 1/2 = 1/4. Any thoughts?

Your thinking is correct. The proportion of hairy-bodied flies with short veins should indeed be 1/4. This can be calculated by multiplying the proportion of shv+/shv flies (1/2) with the proportion of he+/he flies (1/2). The occurrence or non-occurrence of a recombinant event between the two genes does not affect this calculation, as they are on different chromosomes and therefore independent of each other. Your TA may have made a mistake in their calculation, or they may have misunderstood the question. It would be worth discussing this with them to clarify.