Question 4 (6 points): Microsatellite marker Z has been found to be linked to Hu

Question

Question 4 (6 points): Microsatellite marker Z has been found to be linked to Huntington's disease, an autosomal dominant disease in humans. Therefore, microsatellite marker Z has the potential to be used to follow the inheritance of the disease. The reliability of microsatellite marker Z as a predictor of the disease depends on the map distance between the marker and the disease causing gene called huntingtin You have a male patient affected with Huntington's disease (genotype Aa) who has heterozygous alleles Z1 and Z2 of microsatellite marker Z; Z1 is on the same chromosome as genotype A of huntingtin gene and Z2 is on the same chromosome as genotype a. His wife has homozygous alleles of 23 with aa genotype (note: Z3 is another allele for microsatellite marker Z). She is pregnant and underwent a genetic testing by amniocentesis to predict whether her fetus will have the disease. The results showed that the fetus has Z1 and Z3 alleles of the microsatellite marker Z. If the distance between huntingtin gene and fetus? Briefly explain your answer. Question 5 (5 points) microsatellite marker Z is 0.00001 map unit, what prediction can be made for her The human haploid cell contains about 3.1 x 109 bp of DNA. Assuming that 200 bp of DNA is wrapped around each nucleosome, how many molecules of histone H2A are present in one normal somatic human cell?

Explanation / Answer

Since the distance between the Z marker and huntingtin gen is very small, they are likely to be linked and are inherited together and thus the fetus will be have the disseased allele

5. Each nucleosome is an octamer and has two copies of 4 different histones including histone 2A.

Numberof nucleosome = 3.1 X 109 / 200 = 15 X 106 nucleosomes.

Number of H2A = 2( 15 X 106 ) = 30, 000,000

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