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The albinism gene is shown below. The disease is caused by a point mutation that blocks the first restriction site. Fill in the gel electrophoresis with the results of the RFLP analysis. Use this information to answer the following questions and fill in the gel and the Southern. Keep in mind how thick the bands should be!What size would you expect from a homozygous with no disease? (lane 1)What size would you expect from a heterozygote carrier? (lane 2)What size would you expect from a homozygous recessive with the disease? (lane 3)2kb0kb------------------------------------------------------------10kbSouthern Molecular Probe4kb123Southern Blot

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Related questions

Southern blot analysis of the three genotypes found at the β-globin locus revealed a distinctive pattern of RFLP variation among the genotypes, while northern blotting reveals no difference. What results would you expect from native electrophoresis of the protein from the three β-globin genotypes?

(a) A western blot analysis performed on protein from individuals with the three β-globin genotypes shows that βAβA individuals have a single protein band, whereas βSβS individuals have no protein. βAβS heterozygotes have half as much protein as a homozygous individual.

(b) A western blot analysis performed on protein from individuals with the three β-globin genotypes detects a single protein band, so all genotypes look identical by western blotting.

(c) When a western blot analysis is performed on protein from individuals with the three β-globin genotypes, βAβA and βSβS individuals each have one distinct band, whereas βAβS individuals have both bands.

(d) A western blot analysis performed on protein from individuals with the three β-globin genotypes looks very similar to the Southern blot results. βAβA individuals have two bands, βSβS individuals have one band, and βAβS individuals have all three bands.

Answer is C because western blotting uses proteins.

Please explain the northern, southern, and western blots and why a, b, and d were incorrect. Use drawings when necessary.

Southern blot analysis of the three genotypes found at the β-globin locus revealed a distinctive pattern of RFLP variation among the genotypes, while northern blotting reveals no difference. What results would you expect from native electrophoresis of the protein from the three β-globin genotypes? (a) A western blot analysis performed on protein from individuals with the three β-globin genotypes shows that βAβA individuals have a single protein band, whereas βSβS individuals have no protein. βAβS heterozygotes have half as much protein as a homozygous individual. (b) A western blot analysis performed on protein from individuals with the three β-globin genotypes detects a single protein band, so all genotypes look identical by western blotting. (c) When a western blot analysis is performed on protein from individuals with the three β-globin genotypes, βAβA and βSβS individuals each have one distinct band, whereas βAβS individuals have both bands. (d) A western blot analysis performed on protein from individuals with the three β-globin genotypes looks very similar to the Southern blot results. βAβA individuals have two bands, βSβS individuals have one band, and βAβS individuals have all three bands. Answer is C because western blotting uses proteins. Please explain the northern, southern, and western blots and why a, b, and d were incorrect. Use drawings when necessary.

INSTRUCTIONS

Go to the OMIM database on the NCI portal. Search for PAH, select the first record on the list The following page will appear: Read the initial information Phenylketonuria (PKU) is an autosomal recessive inborn error of metabolism resulting from a deficiency of the enzyme PAH (Phenylalanine Hydroxylase). Lack of PAH enzyme activity results in elevated level of phenylalanine. On the other hand, mutations in the PAH gene can also result in a milder disease called hyperphenylalaninemia (HPA) characterized by an elevated level of phenylalanine without the cognitive impairment associated with PKU. QUESTIONS

Question 1: Consider two actual DNA mutations found as 2 PAH mutant alleles. a. Mutation # 0017 changes codon 414 from a tyrosine to a cysteine (TYR414CYS). A homozygous genotype leads to HPA. Indicate the type of mutation for 0017 at the DNA level. Identify the class of mutation at the protein level. Based on the disease phenotype, would you expect the mutated protein to be partially functional or non-functional? (I think it is functional..? http://www.omim.org/entry/612349) Explain your reasoning briefly b. . Mutation #0018 is a mutation that changes the splice site of intron 4 (IVS4). Explain the consequence of mutation 0018 at the level of transcription of the gene, splicing and protein synthesis. Homozygous 0018 leads to PKU phenotype Question 3: Assume an individual heterozygote for 0017/0018. ?Which special term is used to describe a heterozygote with two different mutant alleles - Compound heterozygote Where along the dominance-recessivity scale would the phenotype of this individual falls in relation to the phenotype of the two mutant homozygotes? Explain briefly.

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