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midterm12002answers - Molecular Genetics 640 - Midterm I -...

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Molecular Genetics 640 - Midterm I - 2002 NAME _____________________________ (5) 1. You are studying the Amish population of Ohio for the Secretor locus, which is related to the occurrence of the antigens of the ABO blood group in the saliva of individuals. The system show dominant/recessive inheritance patterns. Two alleles exist, Se and se , with the allele Se (secretion of antigens in saliva) being dominant to se (no secretion). You have developed a new molecular tool to identify the alleles. In a very small sample of the Ohio Amish population you use your molecular test to determine that the allele frequencies for the two alleles are 0.65 for the se and 0.35 for the Se allele. a. If these allele frequencies are accurate, predict the genotype and phenotype frequencies for the secretor system in the entire Amish population of Ohio. (3POINTS) genotype frequencies = p 2 + 2pq + q 2 SS = (.35) 2 = 0.1225 Ss = 2(.35)(.65) = 0.455 ss = (.65) 2 = 0.4225 (2 POINTS) phenotype secretor = 0.1225 + 0.455 = 0.5775 non-secretor = 0.4225 Three significant digits is acceptable -2 points for reversing allele frequencies
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Molecular Genetics 640 - Midterm I - 2002 NAME _____________________________ 2. The following phenotypes represent all types (and the numbers of each kind) that were observed on DNA gels analyzing samples taken in a study of 1000 individuals. size # observed: 252 110 10 304 36 6 188 4 90 (3) a. From the information provided, what type of genetic marker is represented by the gel? Microsatellite or short tandem repeat or STR (3) b. How many nucleotides are present in a single copy of the sequence which constitutes the repeat unit? four (tetranucleotide) (3) c. How many distinct allele types appear on the gel? Four (4) d. What are the allele frequencies in the population sample? f(154) = (252 x 2) + 10 + 304 + 188 = 1006/2000 = 0.503 f(150) = 10 + 6 + 4 = 20/2000 = 0.01 f(146) = 110 + 304 + 6 + (90 x2) = 600/2000 = 0.30 f(142) = 110 + (36x2) + 188 + 4 = 374/2000 = 0.187 (2) e. Are any genotypic classes missing from among the genotypes which were observed in this study, and if there are one or more genotypes missing, explain why they were probably not observed. Yes, 150/150 homozygous class missing. Expect only (0.01) 2 x 2000 = one-tenth of an individual. Not surprising that we do not see any.
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Molecular Genetics 640 - Midterm I - 2002 NAME _____________________________ (10) 3. You are collaborating with the College of Optometry in a study of the population of central Mexico. You are studying colorblindness. For the purpose of this question, colorblindness is caused by a recessive allele at one sex-linked locus. In a population of individuals in an isolated Mexican region, 15% of grandfathers and 1.8% of grandmothers are found to be colorblind upon testing. (Grandparents have completed their reproduction). You find that 2% of the reproductive age mothers in the region are colorblind. (These women are the daughters of the group that we classify as grandparents in the population). The reproductive age
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midterm12002answers - Molecular Genetics 640 - Midterm I -...

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