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CU-Boulder PSYC 2841 - Practice Test 4 No Answers

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MCDB 2150 – Practice Test 4 1. Below is shown a complementation table from Neurospora mutants that cannot make the amino acid tryptophan 1 2 3 4 5 6 1 - + - - + + 2 - + + + + 3 - - + + 4 - + + 5 - - 6 - Given this information, how many genes are involved in the synthesis of tryptophan? A) 1 B) 2 C) 3 D) 4 E) 5 3. You have the seeds from inbred tomato plants and seeds from genetically diverse tomatoes. You grow both types of seeds in the same greenhouse, measure the stem lengths and plot distribution curves as shown to the right. The total variance in this distribution curve is due to genetic variance (VG) and environmental variance (VE). Looking at the 2 distribution curves for the genetically identical and genetically diverse plants, which of the following statements is true? A) A = VG and B = VG B) A = VE and B = VG C) A = VG and B = VE D) A = VE and B = VE 3. You find several additional mutant fly lines with lethal patterning defects, which you maintain as heterozygotes. You suspect that at least one of the mutant lines is the result of a mutation in a maternal effect gene, while the other is the result of a mutation in a Hox gene. Which of the following experiments and outcomes would allow you to determine which is the maternal effect mutant line? A) cross two heterozygotes together; if the offspring are all dead, it is the maternal effect mutation B) cross two heterozygotes together; if the offspring are all alive, it is the maternal effect mutant C) cross two heterozygotes together; if ¼ of the offspring are dead, it is the maternal effect mutantMCDB 2150 – Practice Test 4 4. Hox genes, as described in class, give cells important positional information in all organisms. One Hox gene, Antp, defines the T2 segment, which normally produces wings in Drosophila. Suppose you have 2 mutant flies. One mutant has a loss of function Antp mutation, and the other has a gain of function Antp mutation in which Antp is now expressed in the T3 segment in addition to the T2 segment. What would you expect to see? A) no wings in both loss of function and gain of function mutants B) wings on both the T2 and T3 segments in both loss of function and gain of function mutants C) no wings in the loss of function, wings on both the T2 and T3 segments in the gain of function D) no wings in the gain of function, wings on both the T2 and T3 segments in the loss of function 5. Frogs from line A have a mutation in the A gene. When examined more closely 'aa' frogs have white spots on their bellies. Frogs from the B line have a mutation in the B gene. When examined more closely 'bb' frogs have white stripes on their bellies. Wild type frogs do not have any stripes or spots. The A and B genes follow a recessive epistasis inheritance pattern and A is epistatic to B. If two frogs heterozygous for both A and B are mated, what is that probability that an offspring have white spots? A) 9/16 B) 4/16 C) 3/16 D) 1/16 E) 0 6. Magnus puer disorder is caused by a recessive mutation in a maternally imprinted gene. The maternal copy of the gene that causes this disorder is methylated and is silenced. What is the probability that the male in the third generation will have Magnus puer disorder? A) 1/4 B) 1/2 C) 2/3 D) 1 E) 0 7. Muscular dystrophy is an X linked recessive disease. Men are usually more severely affected than women. A woman who is a carrier for muscular dystrophy but has no symptoms and a man who does not have muscular dystrophy have a daughter who has clear symptoms of muscular dystrophy. What is the most likely explanation? A) The X chromosome she got from the father was inactivated in all the cells in her body. B) The X chromosome she got from her mother was inactivated in all the cells in her body. C) The X chromosome she got from her father was inactivated in the cells that gave rise to all the muscle cells of her body. D) The X chromosome she got from her mother was inactivated in the cells that gave rise to all the muscle cells of her body. E) This is impossible; a woman with one copy of the disease allele and one copy of the normal allele will not have symptoms.MCDB 2150 – Practice Test 4 8. Quarter’s disease is an autosomal dominant disorder. In a population of 600 people in an isolated community, 150 have Quarter’s disease, and 450 do not have Quarter’s disease. What is the frequency of the dominant Quarter’s allele? A) 0.13 B) 0.27 C) 0.5 D) 0.67 E) 0.87 9. The peppered moth (white with small black speckles) is well camouflaged when resting on lichen-covered tree trunks. In the nineteenth century, domestic coal fires (part of the industrial revolution) caused sooty air pollution that killed off the lichens and blackened urban trees. As a result, a solid black peppered moth came to outnumber the speckled form within the cities. More recently, with improved regulation of air pollution, the lichen is returning to the trees and the white moth is returning to predominance. Which of the following statements is true? A) The air pollution preferentially caused mutations in the genes that specify the moth’s coloring. B) The mutation leading to the black moth is an example of a neutral mutation. C) The ability of the white peppered moths to camouflage themselves in the lichen leads to a selective advantage over the solid black form of the moth. D) This is an example of stabilizing selection. 10. Which of the following is correct concerning events such as natural disasters and habitat destruction? A) They have a greater impact on small populations B) They are examples of genetic drift C) They eliminate members of a population without regards to genotype D) The resulting population has a different genetic makeup from the starting population E) All of the answers are correct 11. What type of selection occurs as a population of bacteria in a TB patient’s lungs becomes resistant to antibiotics? A) Directional selection B) Stabilizing selection C) Disruptive selection D) Heterozygote advantageMCDB 2150 – Practice Test 4 Short Answer 1. The gene H19 in humans is involved in cell division. No functional copies of H19 result in a syndrome called BWS syndrome that produces a form of gigantism. Mark is normal phenotypically but carries a recessive mutation in H19. Mary is also phenotypically normal but carries a deletion of H19. They carry out genetic testing on their unborn child and learn that the fetus has inherited


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