A form of dwarfism known as Ellis–van Creveld syndrome was first discovered in the late 1930s, when Richard Ellis and Simon van Creveld shared a train compartment on the way to a pediatrics meeting. In the course of conversation, they discovered that they each had a patient with this syndrome. They published a description of the syndrome in 1940. Affected individuals have a short-limbed form of dwarfism and often have defects of the lips and teeth, and polydactyly (extra fingers). The largest pedigree for the condition was reported in an Old Order Amish population in eastern Pennsylvania by Victor McKusick and his colleagues (1964). In that community, about 5 per 1000 births are affected, and in the population of 8000, the observed frequency is 2 per 1000. All affected individuals have unaffected parents, and all affected cases can trace their ancestry to Samuel King and his wife, who arrived in the area in 1774. It is known that neither King nor his wife was affected with the disorder. There are no cases of the disorder in other Amish communities, such as those in Ohio or Indiana.

From the information provided, derive the most likely mode of inheritance of this disorder. Using the Hardy–Weinberg law, calculate the frequency of the mutant allele in the population and the frequency of heterozygotes, assuming Hardy–Weinberg conditions.

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Accepted Answer

Hi, everyone. Welcome back. Let's look at our next problem. It says, and population genetics consider a population of Monash is deployed with two alleles of a single locus. Noted big A and little A. The frequency of big a is given a 0.25 and the frequency of little a is 0.75. Find the percentage of the offspring that are homos, I guess for the dominant allele homos, I guess the recessive allele and hetero sickness respectively using the Hardy Weinberg method. Um One quick note here, um We said that in this population, it's manassas, meaning in plants that means capable of self fertilization. So there's male and female flowers on the same plant and the plant can fertilize itself. This can affect our ability to use the Hardy Weinberg method because the Hardy Weinberg method to be used, you have to have random mating and a population of Manischewitz plants will not necessarily have random mating because they'll be slightly more likely to fertilize themselves. Um And this could throw off the frequency of homesickness versus hetero sickos offspring since you'll be slightly more likely to have homogeneous offspring. However, since our problem doesn't give us any factor by which we need to adjust this, we can assume that in this population, um it's close enough to random mating that we can apply the hardy Weinberg equilibrium. So with that in mind, let's go on and calculate this. We have our frequency of big A which will call P Is equal to 0.25 and our frequency of little a which is equal to Q is 0.75. So to calculate the percent of the different Jenna types here, we know the frequency of our homes, I guess dominant big a big a bowl equal P squared. So 0.25 squared, which is going to give us zero point 0625. But when we look over our answer choices, we see that listed as percents. So we'll go ahead and convert that to 6. than the frequency of the recessive allele frequency of little a little A well equal Q squared. So 0.75 squared, which is going to equal 0.56 Or 56.25%. And then finally, the frequency of our hedgerows, I get big a little A Must equal to PQ. So two times 0.25 times 0.75, which is going to equal zero point 375, which is going to equal 37.5%. So our three values in order would be 6.25%, and 37.5%. When we look at our answer choices, we see that these numbers in this order are found in answer choice. A see you in the next video.

Verified Solution

Key Concepts

Ellis–van Creveld Syndrome

Mode of Inheritance

Hardy–Weinberg Principle