Genetics question on possible gametes

[virtualmaster999]

Hey everyone!

I got stuck on this question (which is easy!!)

Which of the following is true for the gametes of an individual with the genotype AaBb? Assume independent assortment but no crossing over.

I thought it would be just two possibilities. Since there isn't any crossing over, wouldnt there be less gametes? I usually do 2^n, but when I saw the wording of this question, I guess I got tripped up. Could someone explain this for me?

Do you always do 2^n?

Thanks!!

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[FeralisExtremum]

From the chromosome with Aa you can potentially produce two different gametes: A or a
From the separate chromosome with Bb you can potentially produce two different gametes: B or b

(2 possibilities from Aa) * (2 possibilities from Bb) = 4 gamete possibilities total: AB, Ab, aB, ab.

It's not always as simple as 2^n. For example, try to solve this question:

"How many gametes are possible from an individual with the genotype AABbCCddEe? Assume independent assortment but no crossing over."

 

[virtualmaster999]

From the chromosome with Aa you can potentially produce two different gametes: A or a
From the separate chromosome with Bb you can potentially produce two different gametes: B or b

(2 possibilities from Aa) * (2 possibilities from Bb) = 4 gamete possibilities total: AB, Ab, aB, ab.

It's not always as simple as 2^n. For example, try to solve this question:

"How many gametes are possible from an individual with the genotype AABbCCddEe? Assume independent assortment but no crossing over."

So would it be 4 for that example you gave?

Wouldnt it always have to be 2^n though, where n represents number of heterozygous pair?

 

[FeralisExtremum]

Yep, 4 is the right answer. 2^n will work as long as you restrict n to the # of heterozygous pairs.

 

[virtualmaster999]

Yep, 4 is the right answer. 2^n will work as long as you restrict n to the # of heterozygous pairs.

Awesome, thanks I appreciate it!