College Human population genetics problem (biology)

Question

Consider a population with a frequency of the M allele of p = 0.3 and the frequency of the N allele of q = 0.7. Ten parents are randomly selected to produce the next generation of 10 offspring. a) What is the probability of exactly zero M alleles being present in the offspring population? b) What is the probability of exactly 5 M alleles being present in the offspring population. c) How many M alleles in the offspring population do we expect on average? a) _____________ b)_____________ c) ______________

I never ever feel confident doing these problems. Any help is super appreciated!

For part A, would I just need to use the formula p(x=k)=(n/k) (p)^k (q)^n-k where k=0 and the factorial of 10 over 0 is equal to 1?

For part B, should I do the same thing as part A where k=5 and the factorial 10 over 5 is equal to 252? I get really confused with factorials... would it be written as 10! divided by (5!)*(5!) ?

And then for part C... should I just do E(x)=n*p ? where n=10 and p= 0.3 ??

Answer 1

Hi Annie,

I think that this is a binomial situation, which means you would use:

P(X=x) = C(n,x) * p^x * q^(n-x)

From a pure statistics standpoint, this makes sense since we have two outcomes and independent trials, but, as a disclaimer, I don't know the biological aspects.

a. P(X=0) = C(10,0) * 0.3^0 * 0.7^10

Both 10!/(10!*0!) and 0.3^0 reduce to 1, so:

P(X=0) = 0.7^10 = 0.028

b. Same procedure, as you suspected. Change x to 5.

P(X=5) = C(10,5) * 0.3^5 * 0.7^5

Yes, your factorial expression is correct:

C(n,x) = n!/x!(n-x)! = 10!/(5!*5!) = 252

P(X=5) = 252*0.3^5*0.7^5 = 0.103

c. You're exactly right: E(x) = np = 10*0.3 = 3

Assuming I'm not missing some biological aspect, your methods were completely correct. Practice more problems like this and I suspect it will come to you more naturally, including the factorial part. I hope this helps.