Donkey

[donkey]

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[Russ]

Welcome to VN

Since the genes are linked and close together, you would expect approximately equal amounts of SW and sw gametes from the heterozygous parent (since this is apparently the arrangement on the chromosome). Given that it's a test cross, the gametes from the homozygous recessive don't contribute to the phenotype of the offspring. Therefore, if the SW and sw gametes are present in equal amounts, the SW and sw phenotypes will be present in equal amount.

SW = long, straight
sw = short, wavy

Option D.

[donkey]

Welcome to VN

Since the genes are linked and close together, you would expect approximately equal amounts of SW and sw gametes from the heterozygous parent (since this is apparently the arrangement on the chromosome). Given that it's a test cross, the gametes from the homozygous recessive don't contribute to the phenotype of the offspring. Therefore, if the SW and sw gametes are present in equal amounts, the SW and sw phenotypes will be present in equal amount.

SW = long, straight
sw = short, wavy

Option D.

Thank you Russ. What a nice welcome!

I'm also wondering why is the alternative 'A' wrong in the question attached. The correct answer is C. While this is also consistent with the diagram according to my analysis, I assume the 'most likely to be part' raised it above alternative 'A'. Could you please explain why this is?

[shinny]

Likely because of the phrase 'Individuals marrying into the family have no history of the disease', hence meaning you can't assume that they could be heterozygous (i.e. must assume they are homozygous negative for the disease) along with II-2 and II-5 to produce homozygous children. Not 100% sure of this though.

[Russ]

Yeah that's it. If they have no history of the disease, then the only way for them to be a carrier is if the same (recessive) allele has been passed down in every generation. Whilst it's not impossible, it's certainly more likely that the allele was never present. Additionally, if the question states that the disease being considered is "rare" you should generally assume that there are no carriers - ie. nobody is heterozygous.

In this case, the fact that only males are affected also lends support to C, since that's a hallmark of X linked recessive disease.