Biology 201 (Genetics) Exam #1 120 points 22 September 2006
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1 Name KEY Section Biology 201 (Genetics) Exam #1 120 points 22 September 2006 Read the question carefully before answering. Think before you write. You will have up to 50 minutes to take this exam. After that, you MUST stop no matter where you are in the exam. Use clear gene symbols (i.e. Aa is easier to distinguish that Oo). If I can not read your handwriting, I will count the question wrong. Sign the honor pledge if applicable. Good luck! I pledge that I have neither given nor received unauthorized assistance during the completion of this work. Signature: 1
2 12 pts. 1. Genetic variation, provided through sexual reproduction, offers more opportunity for offspring to survive in changing environments. List two aspects of meiosis provide sources of such variation. Recombination/crossover between non-sister chromatids Independent assortment of different homologous chromosome sets into gametes (random assortment of homologous chromosomes / lining up of homologous chromosomes on metaphase plate and subsequest assortment into gametes did not get full credit unless you clearly indicated that set 1 assorts into gametes independently of set 2). 12 pts. 2. Consider three independently assorting gene pairs, which control three different characters. A is dominant to a, B is dominant to b, and D is dominant to d. For a cross between parents that are AaBbDD and AABbDd, write the answer to the questions below in the blank to the left of the question. 1/16 a. what is the probability of obtaining an offspring that is AaBBDd? 1/16 b. what is the probability of obtaining an offspring that is homozygous for all three dominant traits? 3/4 c. what is the probability of obtaining an offspring that will express all three dominant traits? 2
3 12 pts. 3. In rabbits, the alpha gene, which is on chromosome 9, controls leg formation: AA is normal, Aa results in short legs, and aa results in no legs. The beta gene, which is on chromosome 10, controls fur color: BB results in black fur, Bb results in brown fur, and bb results in white fur. a) If a cross is made between brown rabbits with short legs and white rabbits with short legs, what percentage of the offspring would be expected to have white fur and short legs? Parents BbAa x bbaa Progeny we are looking for bbaa Probability of bb in cross = ½ Probability of Aa in cross = ½ Probability of bb AND Aa = ½ * ½ = ¼ = 25% b) what type of genetic inheritance is described above for leg length? incomplete dominance 12 pts. 4. You are a spider geneticist. You cross a pure breeding white spider and a pure breeding green spider and you get F1 spiders that are green. Like a good geneticist, you continue your experiment and you mate the F1 spiders with each other and get the following F2 progeny: 8 green 4 yellow 3 blue 1 white What is the mode of inheritance for spider color? epistasis or 2 genes controlling 1 character 3
4 12 pts. 5. In the plant species Flowerus richmondus, known as the spider plant, individuals produce fruits that are red or blue. There are two alleles of one gene involved in fruit color. Researchers collected an array of individuals from the field and performed a series of crosses. The results are given in the table below. Parental Phenotype(s) Number of Offspring with Red Fruits Number of Offspring with Blue Fruits Parental Genotype(s) blue (self-fertilized) 0 80 aa **red (self-fertilized) Aa X Aa **red blue 46 0 AA x aa red red 44 0 A- X AA a) Which trait is dominant? red b) Put a star by the cross(es) that helped you determine which trait was dominant. c) Construct a legend (using A and a) for spider plant color. A = red a = blue d) Using your legend from part c, complete the last column of the table above by writing down the genotype of the parent or parents involved in each cross. 6. For ABO blood type inheritance, the A allele is codominant to the B allele; the O allele is recessive to both the A and B alleles. In a particular family, the father Joe s blood type in AB and the mother Ann s blood type is O. Ann has a child that is blood type O. a) Is Joe the father? NO** b) Explain your answer briefly. The child s genotype must be oo since the o allele is recessive to both the A and B alleles. The father only carries A and B alleles since he is AB. So none of the child s alleles for blood type could have come from Joe and so he can not be the father of the child. ** You could have answer possibly yes IF you presented a scenario where epistasis was in play such that a second gene controlled whether or not the A and B sugars even got attached to the cell surface. 4
5 18 pts. 7. You are a mouse geneticist and you believe that you have isolated the gene that controls ear size. You isolate mutant mice that have exceptionally large ears, and you have named the gene that encodes ear size the dumbo gene. You have also established a pure breeding strain of these dumbo mice that have exceptionally large ears. How would you use the scientific method to set up a genetic experiment to determine whether the ear size gene in mice is located on the X-chromosome? Make sure in your answer that you clearly indicate how you are using the scientific method. Several possible answers are shown here. You could also have taken one cross thought the F2 generation provided you gave predictions for both crosses. Step 1: make a hypothesis: Ear size gene is located on X-chromosome and large ears are dominant to normal-sized ears. Step 2: Design an experiment: The best way to test for X-linkage is to do reciprocal crosses. So, let D = large ears and d = normal ears. Cross pure breeding DY large-eared males with pure breeding dd normal-eared females (cross 1) and vice versa (cross 2). Examine the F1 progeny of these crosses for sex and ear size. Step 3: Make a prediction: If ear size gene is located on X-chromosome and large ears are dominant to regular ears, then I would expect for cross 1 that the males will have normal ears and the females will have large ears. If ear size gene is located on X-chromosome and large ears are dominant to smaller ears, then I would expect for cross 2 that the males and the females will have large ears. Step 4: Do experiment and collect data Step 5. Accept or reject hypothesis based on the data you obtain. OR Step 1: make a hypothesis: Ear size gene is located on X-chromosome and normal sized ears are dominant to large ears. Step 2: Design an experiment: The best way to test for X-linkage is to do reciprocal crosses. So, let d = large ears and D = normal ears. Cross pure breeding dy large-eared males with pure breeding DD normal-eared females (cross 1) and vice versa (cross 2). Examine the F1 progeny of these crosses for sex and ear size. Step 3: Make a prediction: If ear size gene is located on X-chromosome and regular ears are dominant to large ears, then I would expect for cross 1 that the males will have large ears and the females will have regular ears. If ear size gene is located on X-chromosome and regular ears are dominant to large ears, then I would expect for cross 2 that the males and the females will have regular ears. Step 4: Do experiment. Step 5. Accept or reject hypothesis. 5
6 B A C C Multiple choice section: (30 points total 6 points per question) Write your answer in the blank provided to the left. If you want to explain your answer, you can do so next to the question 1. The somatic cells of a particular male organism contain only one pair of homologous chromosomes (call these chromosomes A 1 A 2,) and one additional chromosome without a homolog (call it chromosome Q). What chromosomal combination would be likely in the four meiotic products of a single cell? a. A 2 Q, A 2 Q, A 2 Q, A 2 Q b. A 1, A 1, A 2 Q, A 2 Q c. A 2 Q, A 1 Q, A 1, A 2 d. A 1 Q, A 1 Q, AQ, A 2 Q e. A 1 Q, A 1 Q, A 1 Q, A 1 Q f. None of the above. 2. Which of the following best reflects a phylogenetic relationship between chimpanzees and baboons? a. They are grouped together because they have a common ancestor in recent history. b. They are grouped together because they both have an opposable thumb. c. They are grouped together because they both feed their young breast milk. d. They are grouped together because they live in similar habitats. e. All but A 3. In the scientific method, a hypothesis is a. another name for a theory. b. a theory that has been proven to be true. c. an idea that is being tested. d. an idea that has no basis in fact or observation. e. only useful if it is shown to be true. 4. You cross a pure breeding curly-haired, long-tailed mice with pure breeding straight-hair, short-tail mice. The F1 generation all have straight hair and long tails. You take a female F1 mouse (straight hair and long tail) and cross it with a pure breeding curly-hair, short-tailed mouse and get the following F2 ratios: 495 curly-haired, long-tailed 495 straight-hair, short-tailed 5 straight-hair, long-tailed 5 curly-haired, short-tailed Which of the following inheritance patterns is consistent with the data given here? a. Non-disjunction b. Independent assortment of unlinked genes c. Genetic linkage d. X-linkage e. Equal segregation E 5. The most fundamental feature that two members of a homologous chromosome pair share is that they have a. the same length. b. the same banding pattern. c. the same centromere position. d. alleles of the same genes at the same loci, where the chromosomes are genetically identical. e. alleles of the same genes at the same loci, where the chromosomes are genetically similar but not necessarily genetically identical. f. a, b, and c 6
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