4) Suppose that Gen e A corresponds to a human disease gene. Many patients had t

Question

4) Suppose that Gen e A corresponds to a human disease gene. Many patients had their genomic ki. The following pattern was found for the different patients endonucleaslitied for the Gene A region. The PCR products were digested with the EcoRI. The following pattern normal Pt 1 Pt 2 Pt 3 Pt 4 PtS Pt6 The patients do NOT necessarily have the disease. From the pattern of digestion, answer the following: A. Which patient(s) could be normal for the disease? B. Which patient(s) are likely to have the disease? C. Which patient(s) could be haploinsufficient and show the disease? D. Assuming no haploinsufficency among the alleles, suppose that patient 1 and 6 had a child. What is the probability that a child will be affected? E. Assuming no haploinsufficency among the alleles, suppose that patient 3 and 4 had a child. What is the probability that a child will be affected? F. Draw a map of gene A and indicate possible sites for the mutations and EcoRI site(s). 5) Suppose that there was Marshak sighting of a novel rodent species, called the Marshak scientificus. A white, sharp nosed M. scientificus was mated with a black, flat nosed M scientificus. The F1 progeny were as follows: 500 pepper colored, sharp nosed 500 pepper colored, flat nosed. A. How many independent traits are represented with this cross? What are the trait(s)? B. What is/are the dominance pattern(s) of the traits? C. Give ONE possible set of genotypes of the parents. 6) Recessive mutations in the p-globin gene can give rise to p-thallassemia or sickle cell anemia; the p-globin gene is on chromosome 11. Recessive mutations of the phenylalanine hydroxylase (PAH) gene on chromosome 12 gives rise to phenylketonuria (PKU). A couple is thinking of

Explanation / Answer

Please find the answers below:

Answre A: Pateints 1, 2, 3 and 6 seem to be normal for the disease. This is because their banding pattern is very similar to that of the normal individual and hence, restriction digestion yielded almost similar pattern of bands in the gel electrophoresis. Presnece of larger band fragments does not pose a problem here since this suggests that a particular patient lacks the restriction site and hence does not carry diseased allele.

Answer B: Patients 4 adn 5 are very likely to have this disease since their banding pattern changes as compared to the nromal version. This suggests that the allelic isoform of these patients is different from normal individuals and hence, diseased in nature.

Answer C: Haploinsufficient patients are those which contain a single allele for the disease and are very prone to expression of the diseased allele. Since patients 4 adn 5 show presence of a single normal band as compared to the normal individual, it shows that these patients might be haploinsufficient in nature.

Answer D: Zero percent. Both patients 1 adn 6 have the normal copies of the restriction pattern as compared to the normal individual. Thus, their child too will have these two normal copies of the genes. Under no haploinsufficiency, these alleles would be fully expressed and hence, there are no chances that their child would be affected from the disease.

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