Brachydactyly type D is a human autosomal dominant condi...

Question

Brachydactyly type D is a human autosomal dominant condition in which the thumbs are abnormally short and broad. In most cases, both thumbs are affected, but occasionally just one thumb is involved. The accompanying pedigree shows a family in which brachydactyly type D is segregating. Filled circles and squares represent females and males who have involvement of both thumbs. Half-filled symbols represent family members with just one thumb affected.

Is there any evidence of variable expressivity in this family? Explain.

Is there any evidence of variable expressivity in this family? Explain.

Accepted Answer

Hi, everybody. Welcome back. Here's our next problem. Familial polydactyly is an autosomal dominant trait that is characterized by extra fingers or toes. One of the types of polydactyly is postaxial polydactyly type B, which has a penetrance of approximately 40%. What is the probability of a child being born with polydactyly? Two parents who are carriers for the trait. Choice? A one half, choice B 3/4 choice C 1/10 or choice D 3/10. Well, when we're talking about the probability of a child being born with polydactyly. So expressing the phenotype of this trait. In this case, we have two factors going on. We have to look at the probability of the child inheriting the alley. So inheriting the genotype for this trait. In this case, we have a dominant trait. So the child only needs to inherit one affected allele, but we have a variable penetrance. And as we can recall from our content video, that means that not everyone that inherits the genotype for the trait will express it in their phenotype. So we also have to look at the probability ability of the child having the condition and that's where we talk about our penetrance. And now we're talking about the probability of two different things happening. So to calculate the probability of both of those things happening, we're going to multiply the two probabilities and get the product. So let's start with probability one, the probability of the child inheriting the ally for polydactyly. So we're just talking about a mono hybrid trait. So we're looking at a single four square pune square. So I've put up my four square grid and where I'm gonna call this gene P, we know that uh capital P, the dominant gene is the gene for polydactyly. And the recessive P gene is the unaffected gene. Now, for my parental genotypes, my problem says the parents are carriers for the treat. Well, this is a little bit confusing because we don't normally talk about carriers when we have dominant traits. Because if you have one allele, you express the trait. So there's kind of no such thing as carrier. But in this case, because we have this variable penetrance, you could technically have people that have an alley but don't express the trait. So in this case, we'll assume that by carriers, our problem means what you usually have in the context of a recessive trait, a heterozygous genotype. So we'll assume the parents are both heterozygous for the P trait. So big P little P across the top, big P little P on the left side. And now let's fill in the possible genotypes for the offspring. So we'd have homozygous dominant, top, left heterozygous bottom, left heterozygous top, right, and homozygous recessive bottom. Right. And again, we're looking at what is the probability of the child inheriting the dominant polydactyly gene. In this case, there are three possibilities. We know that our trait is autosomal dominant. You only need one copy of the all. So that would mean that a genotype of homozygous dominant or heterozygous will have inherited the genotype to make polydactyly a possibility only a child whose homozygous recessive will not have the possibility of expressing that trait. So our first probability is 3/4. So 34 is the chance the child has the genotype necessary. Now, we want to look at given that our child has the necessary genotype, what is the probability they will actually express the trait? But we're told that the penetrance is 40%. So that would mean that there is a 4/10 chance a child with the alley, we'll have polydactyly. So overall our probability of the child having polydactyly is going to equal the product of our two probabilities. So 3/4 multiplied by 4/ that will equal 12 40th, which could we can reduce to 3/10. So when we look for our answer, choices, choice D is indeed 3/10. And that is going to be our answer. Just a quick side note here, when you have a case of variable penetrance and you have a dominant trait like this, you're gonna see a homozygous person probably expressing, even though we express the penetrance as a single value of 40% you'll probably see that being a bit higher in practice than someone who's homozygous for the gene. Just because with two affected alleles, there's a little more chance of actually being affected. Uh in actuality when you're looking at people with that genotype versus a heterozygote who just has one alley. But for the purposes of solving this problem, we'll just treat them all the same. So again, that answer choice is D 3/10. See you in the next video.

My Courses

Chemistry

Biology

Math

Physics

Business

Social Sciences

Programming

Product & Marketing