94ex1a - Exam #1 Student's name _ AGR 3303 - Genetics 26...

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Exam #1 Student's name ________________________________________ AGR 3303 - Genetics 26 September 1994 Multiple choice . (60 pts.) Please read these carefully . One and only one response (a, b, c, d, or e) completely and correctly answers the question, or completes the statement. Circle the appropriate response and turn in this exam. Make sure your circle is unambiguous. Take time to relax. (Suggestion: put the problem into gene symbols, if that helps you to visualize the problem and its solution.) 1. The alleles determining human ABO blood types are an example of: a. dominance/recessiveness b. codominance c. multiple alleles d. all of the above e. none of the above 2. What single information would most help in predicting genotypic frequencies of the progeny resulting from a cross? a. the allelic relationship, i.e. dominance, incomplete dominance, codominance b. the frequency of the various possible parental phenotypes c. the frequency of the various possible gametes d. the number of chromosomes e. whether sex linkage was involved 3. Duchenne muscular dystrophy in humans is a rare, sex-linked, recessive trait. A man has two biological sons, #1 and #2. The sons have different mothers. Son #1 has Duchenne muscular dystrophy. Which value most closely estimates the chance that son #2 has the same condition? a. 100% b. 50% c. 25% d. 0% e. cannot be determined unless the father's phenotype is known University of Florida - Fort Lauderdale 4. In the following metabolic pathway for anthocyanin (purple coloration) the enzymes C and D are coded for by the dominant alleles at the gene loci C and D,
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respectively. The C and D loci are not linked, but assort independently. What would be the frequency of green (purple deficient) progeny from the cross ccDd X CCdd ? enzyme C enzyme D anthocyanin precursor (green) anthocyanin intermediate (green) anthocyanin (purple) a. 100% b. 50% c. 25% d. 12.5% e. 0% 5. A chi-square value was calculated from a genetics experiment, based on a comparison of observed phenotypes with their expected distribution based on Mendelian genetics. The tabular probability level for chi-square was P = 0.90, thus we: a. reject the null hypothesis b. accept the null hypothesis c. cannot conclude anything because we do not know the null hypothesis d. cannot conclude anything because we do not know the chi-square value e. cannot conclude anything because we do not know alternative hypotheses(is) 6. Polyploidy a. is common among animal species, such as humans. b.
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This note was uploaded on 07/30/2011 for the course AGR 3303 taught by Professor Gallio during the Fall '08 term at University of Florida.

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94ex1a - Exam #1 Student's name _ AGR 3303 - Genetics 26...

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