Based on previous family studies, an autosomal recessive disease ...

Question

Based on previous family studies, an autosomal recessive disease with alleles A and a is suspected to be linked to an RFLP marker. The RFLP marker has four alleles, R₁, R₂, R₃, and R₄. The accompanying pedigree shows a three-generation family in which the disease is present. The gel shows the RFLP alleles for each family member directly below the pedigree symbol for that person. After determining the genotypes for the RFLP and disease gene for each family member, answer the following questions.

Is there any evidence of recombination in this pedigree? If so, identify the recombinant individuals and illustrate the recombination that has occurred.

Accepted Answer

Hey, everyone. Let's take a look at this question together in a study of a certain plant species. Researchers have identified two genes G and H located on the same chromosome. A di hybrid plant that is heterozygous or both genes is crossed with a homozygous recessive plant. The following offspring were obtained heterozygous for both genes. 50 homozygous recessive for G and heterozygous for H heterozygous for G and homozygous recessive for H and 50 homozygous recessive for both genes using this information, which of the following gives the correct genetic map. And we can see we have four different answer choices containing different map distances. And we have to determine of the following answer choices which one contains the correct distance between the two genes. And to do so, we are going to use the map distance formula, which we know is map distance equals the number of recombinant divided by the total number of offspring multiplied by 100 which means that we need to determine the number of recombinant as well as the number of total offspring. In order to find that map distance value and to determine the number of recombinant. We must first identify our parent genotypes so that we can determine the genotypes of the recombinant. And we know that we have one parent who is heterozygous for both genes with the genotype, that is heterozygous for both genes. So that means that we have 250 individuals with one parent genotype and the other parent genotype is homozygous recessive for both genes. And we have another 250 offspring with the other parent genotype. So that means that we have 50 individuals that are homozygous recessive for G and heterozygous 4h, as well as 50 individuals, heterozygous for G and homozygous recessive 4h making these two genotypes our recombinant which equals a total of 100 offspring. So we have our map distance which equals the number of recombinant, which is 100 divided by that total offspring, which we know is 100 plus the total offspring value for the parent genotypes, which is 500. So our total offspring is 600 multiplied by 100. So we divide 100 by 600 which gives us 0.167, multiplied by 100 which gives us our map distance value of 16.7. And so answer choice C gives us the correct genetic map because we note that the distance between G and H is 16.7, which we calculate by using the map distance formula, which is map distance equals 100 which is our number of recombinant divided by the total number of offspring that were obtained from that cross. And we multiply it by 100 which gives us the final value of 16.7. Making answer choice. C the correct answer. I hope you found this video to be helpful. Thank you and goodbye.

Key Concepts

Autosomal Recessive Inheritance

Restriction Fragment Length Polymorphism (RFLP)

Recombination

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