The accompanying pedigree below shows a family in which an autosomal recessive disorder is present. Family members I-2 and II-2 are affected by the disorder and have the genotype dd. A pregnancy involving II-4 has just undergone genetic testing for a VNTR that is linked to the disease gene. The VNTR has a recombination frequency of r = 20 with the disease gene. The VNTR has two alleles, V₁ and V₂. The gel electrophoresis patterns for each family member are shown, including the VNTR genotype for II-4. Based on the information given, answer the following questions about the family.
What are the syntenic disease gene and VNTR alleles in I-1 and I-2?

Question

The accompanying pedigree below shows a family in which an autosomal recessive disorder is present. Family members I-2 and II-2 are affected by the disorder and have the genotype dd. A pregnancy involving II-4 has just undergone genetic testing for a VNTR that is linked to the disease gene. The VNTR has a recombination frequency of r = 20 with the disease gene. The VNTR has two alleles, V₁ and V₂. The gel electrophoresis patterns for each family member are shown, including the VNTR genotype for II-4. Based on the information given, answer the following questions about the family.

What are the syntenic disease gene and VNTR alleles in I-1 and I-2?

Pedigree chart illustrating inheritance patterns in a different family.

What are the syntenic disease gene and VNTR alleles in I-1 and I-2?

Accepted Answer

Hi, everyone. Let's take a look at this practice problem together, an autosomal recessive disorder with a Lil's capital A and lowercase A has been linked to A V N T R. The recombination frequency of the disease gene and the V N T R is 10.7%. The company pedigree shows a two generation family in which the disease is present. The V N T R has two alis V one and V two and the gel electrophoresis pattern of each family member has been shown based on this information. Identify the syn disease gene and V N T R A lies in individuals 23 and five of the second generation. The answer options are a individual two is homozygous. V two capital A individual three is heterozygous for both V N T R and disease gene and individual five is V one lower case A. Option B individuals 2 and 5 are heterozygous for both B N T R and the disease gene. Individual three is homozygous B 2 capital a option C individual two is homozygous V one capital A individual three is heterozygous for B N T R and disease gene. Individual five is homozygous V 2 lowercase a option D individual two, 3 and 5 are all heterozygous for both B N T R and disease gene. Now, let's take a look at the pedigree and the gel electrophoresis. The pedigree is a two generation family. See, first generation has two parents that are unaffected by the autosomal recessive disorder and they have five Children. The 1st and 5th individuals are affected by disord the disorder and the individuals 2, 3 and 4 are unaffected. The gel electrophoresis results are the parents in the first generation have bands for both V one and V two individual. One of the second generation has only a V two band individual. Two of the second generation has only a V one band. Individuals three and four have both V one and V two bands and the fifth individual of the second generation has only the V two band. Let's talk about the gel electrophoresis results a little bit more. So test samples are loaded into wells and they are taken from each individual in the pedigree. Then it shows whether the bands are present for the V and T R lil. So V one and V two, the lil run or migrate through the gel at different speeds. V two moves faster than V one. So individuals with the band on top have the V one a little and individuals with the bottom band have the V two A lil. OK. Let's remind ourselves what we're actually solving for, we need to identify the synthetic disease gene and V N T R all lies. Now, recall that synthetic allys are lil that are found on the same chromosome. This occurred between de the disease gene and B N T R through recombination. Remember they had a recombination frequency of 10.7%. So essentially, we are determining what disease gene a lil and what B N T R A lil are found on the same chromosome. Let's begin by solving for the first generation parents. We know they are carriers of the disease or the disorder, they pass on the autosomal recessive disorder to some offspring, but they are not affected themselves. Therefore, their disease gene genotype is heterozygous. So capital a lowercase A, the gel results for both parents of the first generation are that they have both the V one and V two bands. So the V and T R genotype is V one V two. For both parents. With only this information, we cannot tell what the syndic ails are yet. But we can, if we look at individual one, uh excuse me, individual, one of the second generation, this individual is affected by the disorder as seen on the pedigree. Therefore, they have a homozygous recessive disease gene or lowercase a lowercase A, the gel results show that this individual only has a V two band. So the V N T R genotype is V 2, v 2. This tells us that the synthetic alleles are V two and lower case A. So let's write this out. We have the Centene ails the recessive a little of the disease gene. Lower case A and the V two a little for the V and T R O must be synthetic allys the dominant disease gene A lil capital A and the V one a little are also synthetic oils. Therefore, if we rewrite this individual one is homozygous recessive B 2 lowercase a. Now we know enough information to solve the question. When the V one band is present on the gel, we know that individual has a disease gene dominant A lil. And when the V two band is on the gel, we know that individual has a disease gene recessive a little. This means if we rewrite the parental genotypes up here, they both are V 1 capital a V 2 lowercase a. So let's solve for the second individual in the second generation, the gel results for them show only a V one band. Therefore, they are V 1 capital a V one capital a individual three of the second generation has bands for both V one and V two. So using the knowledge about the syn allys, this individual is V 1 capital a V 2 lowercase a. Now lastly individual five of the second generation, the gel results are A V two band present only. Therefore, they are B two lowercase A B two lowercase A this means if we identify the correct answer, the correct answer is option C individual two is homozygous V one capital A individual three is heterozygous for both B N T R and disease gene. And individual five of the second generation is homozygous V two lowercase A. All right, everyone. Thanks for hanging in with me. I know this one is a bit of a longer one. I hope you found this helpful and I'll see you soon for the next practice problem.

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Key Concepts

Autosomal Recessive Inheritance

Variable Number Tandem Repeats (VNTR)

Gel Electrophoresis