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Biology 321 QUIZ#1 Spring 2013 25 pts. NAME______________________________ 1. (9 pts) In tomatoes, stem color is controlled by a gene that is located on chromosome 5 and plant height is controlled by a gene located on chromosome 1. Examine the data below from Cross 1 and Cross 2. The green dwarf parent in Cross2 is from a true-breeding line, but the ancestry of all other plants is unknown. Trait Alternative phenotypes Stem color purple or green Plant Height Tall vs dwarf Number of progeny in each class Parental phenotypes 1 2 purple tall X sP sG ht hd green tall sG sG ht hd purple dwarf X sP sG hd hd green dwarf sG sG hd hd purple tall purple dwarf green tall green dwarf 300 sP sG ht h- 100 s sG hd hd 300 sG sG ht h- 100 sG sG hd hd P 400 sP sG hd hd 400 sG sG hd hd a. Which stem color is dominant? purple Dominance for this trait could be determined from (circle correct answer) Cross 2 only (since green dwarf is true-breeding, green can’t be dominant) b. Which plant height is dominant ? tall Dominance for this trait could be determined from (circle correct answer) Cross 1 only c. For each cross indicate the genotypes of the parents and progeny in the space underneath the phenotype. If there is more than one possible genotype for a parent or class of progeny, be sure to indicate this. Use these allele designations OR define your own alleles symbols below. sp= purple stem sG = green stem ht = tall hd = dwarf OPTIONAL: OR use your own alleles symbols. Define them here: 1 2. (3 pts.) True or False. The first statement is true and you are to decide if the second statement is True or False. Consider a statement false if any part of it is false. It is known that many genes are required for pigment formation in the skin or fur of a mammal including those that specify proteins important in the formation of pigment cells, the biosynthesis of pigment (melanin) and the deposition of melanin granules. This means that albinism (complete lack of pigment) cannot be inherited as a standard Mendelian single-gene trait. Circle: False One-two sentence explanation/defense of your answer (no credit if no explanation): A trait is said to be monogenic if the phenotypic variation between two individuals or two strains of organisms can be explained by differences in a single gene. In this example, a single malfunctioning gene can result in an unpigmented animal; for example, if no pigment cells are made (because a gene required for specification this type of cell is faulty), then it doesn’t matter if the pigment biosynthesis & deposition pathway genes are intact. 3. (3 pts.) Examine the diagram to the right . Each line represents a single molecule of doublestranded DNA, which will segregate as indicated. The dotted line indicates where the cytoplasm will be divided during cytokinesis. Fill in the 2 blanks for parts a & b. a. This diagram is consistent with mitosis in a 1 n = 3 organism b. This diagram is consistent with Meiosis II in a 2 n = 6 organism c. If the genotype the organism in (ii) is AaBbDD, what fraction of the all of the Meiosis II divisions occurring in the gonad of the organism would resemble the diagram? [Ignore complications from crossing over events & no explanation needed.] Circle the correct answer: (i) 1.0 (ii) ½ (iii) ¼ (iv) 1/8 (v) not possible to determine from info given 4. (2 pts.) The diagram shown below is consistent with: Choose all possible correct interpretations. a) a 2n=4 cell undergoing mitosis b) a 4n=8 cell undergoing mitosis c) Meiosis I in a 2n=4 organism d) Meiosis I in a 4n=4 orgnism e) Meiosis II in a 2n=4 organism f) Meiosis II in a 4n=8 organism g) none of the above Ó All B chromosomes consist of two dsDNA molecules Ô 2 5. (3 pts.) Circle True or False for each statement. Answer false any part of the statement is false. If there are two statements, the first statement is true and you are to decide if the second statement is T or F. 1 pt if no explanation required. 2 pts if an explanation is required; no credit if no explanation. . T F The genome of the domestic cat is 2n= 38. Each product of meiosis II in the cat gonad would contain 19 chromosomes and 38 double-stranded DNA polymers. T F Commercially valuable strains of strawberries can be 2n=14, 4n=28, 6n=42 or 8n=56. This information implies that these strains differ with respect to the number of different chromosomes (double-stranded DNA polymers) per single genome copy. One sentence defense/explanation: Although n (the number of different chromosomes per single genome copy) is the same for each of these strains (7), they differ with respect to the number of copies of each chromosome (number of genome copies in a somatic cell). 6. (2 pts.) Your mom is an avid sportsperson and often browses the WDFW site. She wants to know why the triploid rainbow trout are sterile. Explain in two- three (max) sentences what the problem is here. Use proper terminology. No need to fill up this space…… Stocking a lake with triploid fish is an efficient way to prevent expansion of the introduced species. Because of the uneven number of homolog copies, the regular mechanism for ensuring that complete genome copies are distributed to the products of meiosis fails. The products of meiosis in these animals have unbalanced genome copies (chromosome numbers). So triploid animals (or plants) cannot produce viable offspring because either the eggs or sperm are nonfunctional, or, even if the gametes are functional, the offspring are inviable. quiz continues on the next page 3 NOTE: For 7, just set up the answer for each question. Also, be sure to indicate the logic behind your answer -- with a very, very brief explanation. You will not get any credit for an answer that does not have a brief explanation. 7. (3 pts.) You self an F1 plant of genotype AaBbCc and select one F2 plant at random. Assume all three genes are on different chromosomes. Uppercase allele = dominant; lowercase allele = recessive. a. (1 pt.) What is the probability that your selected F2 plant will be truebreeding for either the dominant phenotype or the recessive phenotype conferred by gene A? Aa X Aa = ¼ AA ½ Aa ¼ aa Probability that F2 plant is either AA or aa = ¼ + ¼ = 1/2 b. (1 pt.) What is the probability that the F2 plant has the same genotype as the parent? ½ Aa X ½ Bb X ½ Cc = 1/8 c. (1 pt.). What is the probability that the F2 plant has a different genotype from the parent. Use your answer from part b. [ 1 – 1/8] = 7/8 = all other possible genotypes other than AaBbCc Problem (pts) 1 (9 pts) Score 2, 3 & 4 (8 pts.) 5 & 6 (5 pts.) 7 (3 pts.) Total out of 25 4