Hello, my name is Alvino. I have a question in biology about evolution. The question asks this:
In a population in Hardy-Weinberg equilibrium, the frequency of the recessive allele is 0.6. What is the frequency of individuals expressing the dominant phenotype?
It also says to assume Mendelian genetics but I'm unsure what this means exactly. I used the p squared + q squared equation and subtracted 1 - 0.6 and got 0.4 for the dominant so p = 0.4. I squared that and got 0.16 for dominant but its saying my answer is wrong and should be 0.64. There is something I am missing in the steps and I think it is because it says assume Mendelian genetics. Thank you for your help
Alvino! Thanks for the question. You can think of Mendelian genetics as the context in which genes are propagated, or passed from one generation to another. Mendelian genetics follow the laws of segregation and independent assortment. The law of segregation states that the alleles of a single gene will separate randomly during gamete formation; independent assortment states that the allele pairs of multiple genes on different chromosomes separate randomly during gamete formation.
You were very close to getting the answer! There was only one more step to go. We know that p + q = 1, and we're given q = 0.6. Therefore, p = 0.4, which you calculated. The p + q equation tells us the frequency of dominant and recessive alleles in a population. Right now, we know that 40% of the alleles in this population are dominant and 60% are recessive. These alleles could be combined in different ways, so the frequencies don't necessarily give us the phenotypes which is what the question is asking for. Phenotypes are the physical traits (brown hair, blue eyes, height, etc.) that result from the pairing of alleles; alleles combine to form gene pairs, or genotypes, which are conceptualized as: AA, Aa, aA, or aa.
We now have 0.4 + 0.6 = 1. It looks like you squared 0.4 which is how you got the answer of 0.16. Remember, we need to use the equation p^2 + 2pq + q^2 to determine the frequencies of phenotypes. You completed the first part (the p^2), but we now need to determine the value of 2pq and find that 2 * 0.4 * 0.6 = 0.48. The sum of p^2 + 2pq equals the frequency of the dominant phenotype, which is 0.16 + 0.48 = 0.64. If we had been asked to find the frequency of the recessive phenotype, we would need to calculate the value of q^2, which is 0.6 * 0.6 = 0.36. Notice that 0.64 + 0.36 = 1.
Hope this helps. You were very close to getting the answer!
THat makes sense thank you