question---seems too easy...?

Question

Okay

It says that some bears have orange noses. 64 out of 100 births have orange noses. The orange nose gene is recessive. There are 7357 bears in Aust., compute the gene frequency of the normal gene and the orange gene. How may koala bears posses the orange gene?

I just thought it would be a .64 frequency for orange, .36 for normal....

then i just multiplied .64 by 7357 to get 4708.48 orange possessors. is this right?

Answer 1

I think you made it too easy...You need to use the Hardy-Weinberg equations to deal with this question. I THINK that when it says "how many bears possess the orange gene" it is asking for both homozygous recessive individuals AND heterozygotes.

So, from the information given, the frequency of homozygous recessive individuals is 0.64. This corresponds to q^2 in the H-W equation. From that, you can calculate the frequency of the recessive allele in the population (q) by taking the square root of 0.64=0.8. Since q=0.8, p=1-q=0.2.

Now, you have correctly calculated the number of bears in the population which are homozygous recessive. To get to the number of heterozygous bears, you need to calculate the frequency of heterozygotes, which is 2pq=2(0.2)(0.8) = 0.32. Multiplying that by the total number of bears gives you 2354.2 additional bears which are heterozygotes, and thus do also have a single copy of the orange gene.

Adding your number to this one gives 7063 bears which have at least one copy of the orange gene.

Answer 2

I'm not sure about the question but if the gene is recessive than many Koalas could have the gene without having the orange nose. Only those where both parents had the gene would the child have an orange nose. Those that had only one parent would not have the orange nose but would still have the gene.

Sorry I couldn't be more help.

Answer 3

q^2 = 0.64, so q = 0.8

p + q = 1, so p = 0.2

Number of OO individuals = (0.2 ^2)*(7357) = 294.28
Number of Oo individuals = 2*0.8*0.2*7357 = 2354.24
Number of oo individuals = (0.8^2)*7357 = 4708.48

So there are 7062.72 bears with at least one orange nose allele.

Just to double check, (294.28+2354.24)/7357 = 0.36

Answer 4

not sure..