The frequency of an autosomal recessive condition is 0.001 (1 in ...

Question

The frequency of an autosomal recessive condition is 0.001 (1 in 1000) in a population.

Assuming individuals mate at random, what is the chance that two heterozygous individuals will mate?

Accepted Answer

Welcome back everyone. Let's look at our next problem. It says what is the frequency of the heterozygous genotype in a population where an autosomal recessive condition occurs at a frequency of 0.1 or one in 1000. Choice A 0.1 choice B 0.61 choice C 0.980 or choice D 0.999. Well, this one's a bit of a trick question. We do have a mathematical formula for calculating this frequency. But in this case, we actually don't need to go through all the math. We can kind of logic our way to the correct answer. So we'll talk through that. But then I will also do the calculations to confirm it. But again, if you were in a rush, you could use logic to go a little bit quicker and avoid the math here. So let's think about what this means. If the autosomal recessive condition has this frequency of one in we know that the autosomal recessive condition will be homozygous recessive, we'll call our gene A. So the genotype of people with the condition would be homozygous recessive. So the frequency of the homozygous recessive condition equals 0.1. Well, we have equations to relate the frequency of the dominant recessive all and the frequency of the three different genotypes. And that would be that if the frequency of our dominant Allal equals P and the frequency of our recessive allele equals Q, then we know the frequency of homozygous recessive, which is what we're given, which equals 0.1 must equal Q squared. And the frequency of our heterozygous genotype must equal two P Q. So when we look about, look at that relation and we think about what does it mean that our homozygous recessive genotype is only occurring in one in of the people in this population. So it's a very, very low frequency. Well, that would mean we wouldn't expect a large number of heterozygote in this population. We don't have a lot of the recessive values circulating in this population. So when we look at our answer choices and we see that choice C is 0. and choice D is 0.999, that would be extremely unlikely that those numbers are nearly one and that would mean nearly the entire population is heterozygous. And you definitely wouldn't expect to see that in a population where the frequency of this recessive all is going to be quite low. So choice C and choice D, I'm going to eliminate just based on logic here and then choice A is 0.1. Well, I'm not going to expect that my frequency of heterozygous is going to be the same as my frequency of homozygous recessive people again, with such a discrepancy in the frequency of the recessive allele and the dominant allele. So choice A can be eliminated for that reason. So that just gives me by process of elimination. That choice B 0.61 is the correct answer. And that makes sort of uh gut level sense or intuitive sense because that would give me a frequency of 6% heterozygous in this population, which would feel a little more correct given the tiny number of people that are homozygous recessive. But let's go ahead and do the math to confirm that. So I need P to be able to calculate my frequency of my heterozygote, which would be two P Q and I have Q squared. That's my given information. So I need to calculate P. Well, I have a relationship between P and Q because P plus Q must equal one. Since there are only two alleys in the population, everybody in the population has to have these alleys. So then P must equal one minus Q, I'm given Q squared since that's the frequency of the homozygous recessive genotype. So that must equal one minus the square root of Q squared or the square root of 0.1. So that will equal one minus 0.316. And that is going to equal 0.9684. And this makes sense to us because again, this is virtually all this is 97% of the population which makes sense with that very low number of people that are autism that are homozygous recessive, excuse me. So now I have P I have Q I can calculate the frequency of heterozygous in this population will equal two P Q which equals two multiplied by 0.9684, multiplied by 0.316. And when I do that multiplication, I get the answer of 0. as I would expect corresponding to choice B see you in the next video.

The frequency of an autosomal recessive condition is 0.001 (1 in 1000) in a population. Assuming individuals mate at random, what is the chance that two heterozygous individuals will mate?

Key Concepts

Hardy-Weinberg Principle

Genotype Frequencies

Heterozygous Mating Probability

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