MCQ: Genetics- Recombination in Drosophila

[TytoAlba95]

Homework Statement

The given answer to this question is (2)

Relevant Equations

NA

Here (A) is a product of recombination between a-&-b, (B) is a product of recombination between b-&-c and (C) a parent

1. If 1000 is the total number of progenies, then 10% of it is 100 (A), 5% is 50 (B) and the rest should be <850 because the other parent has not been included. So I reject 1.
2. Here the total is 500, maybe they have considered the other P to be 500.
3. Not possible because for the given total (B) cannot be more than 50.
4. Similarly, it cannot be 425 either.

So I have to go back to 2. if we take to total population as 500, it is correct, but the total pop is not 500?!

I'm very confused about the concept, please guide me. Also, tell me if there's a simpler or shorter way to arrive at the correct answer.

SS141

 

[Ygggdrasil]

If you have a crossover between a&b, you will produce two different chromatids, a+/b/c+ and a/b+/c, and each chromatid will have equal probability of being inherited. In the test cross, 50 would inherit a+/b/c+ and 50 would inherit a/b+/c.

 

[TytoAlba95]

... two different chromatids, a+/b/c+ and a/b+/c, and each chromatid will have equal probability of being inherited. In the test cross, 50 would inherit a+/b/c+ and 50 would inherit a/b+/c.

But doesn't the ratio 1:1 hold true for independently assorting gene combinations/genotypes? For linked genes, recombinant genotypes are rare and the parental ones are frequent.

 

[Ygggdrasil]

Try drawing out the possible gametes that can be produced by the F1 individual and assigning probabilities to each of the six possible gamete genotypes.

 

[TytoAlba95]

I'm not sure if I'm right. Please help.

 

[Ygggdrasil]

You've correctly listed out the possible gametes. Now, for each individual gamete's genotype (e.g. a+/b+/c), what is the probability?

 

[TytoAlba95]

# SC_ab= Single crossover between a & b

 

[Ygggdrasil]

The math here disagrees with the math above. In your previous post, you correctly said that the two parental genotypes sum to 85%, but here your two parental genotypes sum to 70%.

 

[TytoAlba95]

I agree, my recent calculation contradicts the previous one. I think 70% makes more sense. Because in the question though they have provided only one genotype from each crossover type, the chances of occurrence of both the genotypes should be equal, P(a/b+/c)=P(a+/b/c+)=10%.

 

[Ygggdrasil]

Here's how I approach the problem:

 

[sysprog]

Here's how I approach the problem:
View attachment 263835

What tool(s) did you use to make the nice tree drawing?

 

[Ygggdrasil]

I just put it together in PowerPoint.

 

[TytoAlba95]

Here's how I approach the problem:
View attachment 263835

Okay, I got this.
So the gametes arising out of non-recombination and recombination have the same chances of occurrence as its counterpart. The sum of the frequencies of a pair of recombinant gametes is the distance between them in map units.

Thank you. Thank you so much.