Nicole asked in Science & MathematicsBiology · 8 years ago

# Biology PTC question?

In Fisher, Ford, and Huxley's experiment with chimpanzees that tested to see if they were tasters or nontasters of PTC, they found that out of the sample size of 27, 20 were tasters and 7 were nontasters.

Based on this information, they found that the allele frequencies were 49 and 51% (taster and nontaster, respectively). But, I don't understand how they got these values. Can someone explain to me how the experimenters got 49 and 51% out of this sample?

Update:

I didn't think Hardy-Weinberg could be used for this, because when they did the trait they thought it was either that you could taste it or you could not, and they had no way to determine if one was heterozygous or not. Thus, the assumption would be that one that could taste it would be dominant, while those that could not taste it were recessive. Of course, the nontasters have to homozygous, but how do you know what is heterozygous and what is homozygous for the tasters?

Just to clarify, the total population tested was 27, there were not 27 heterozygous tasters. I'm still confused how they could have gotten to 49% and 51% for the allele frequencies from these values.

Relevance
• 8 years ago

First understand that an organism can has a pair of alleles. Those alleles can have different genes. A gene can be dominant meaning if the organism has one allele with that gene the gene is "active". Recessive is just the opposite. Homozygous means the organism has 2 alleles with that gene. Heterozygous means the 2 alleles have different genes (the dominant gene will be "active").

Now... What I am figuring out is that those 20 are homozygous dominant (for whatever allele makes them tasters), the other 27 are are heterozygous (meaning they have both alleles), and the remaining 7 were homozygous recessive (since they show the phenotype or physical trait of being non-tasters, and are a minority). Using that information, I figured that 41 (27+7*2) are non-taster. Then, 40 (20*2) are taster. So 40:41 (taster to non-taster) is about 49% to 51%.

• ?
Lv 7
8 years ago

You should use Hardy-Weinberg to solve this. p^2 + 2pq +q^2 = 1.

The frequency of the recessive genotype is q^2 = 7/27 = 0.26, so q = 0.51, which is the frequency of the recessive allele.

As p+q = 1, then p = 0.49.

You could then go on to work out that f(homozygous dominant) p^2 = 0.24

and f(heterozygotes) = 2pq = 0.5

Source(s): I'm a geneticist