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Biol 3301 3 rd Midterm -Pink December 3 rd , 2009 There are a total of 8 pages in this exam. This exam consists of 27 questions worth a total of 100 points. All questions are multiple choice and there is one best answer for each question. Record your answers on the scantron sheet. Answer only once for each question or you will be automatically marked wrong. Use a #2 pencil. Name_________________________ Peoplesoft ID#__________________________

Genetics 2009 3rd Exam

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Page 1: Genetics 2009 3rd Exam

Biol 3301 3rd Midterm -PinkDecember 3rd, 2009

There are a total of 8 pages in this exam. This exam consists of 27 questions worth a total of 100 points. All questions are multiple choice and there is one best answer for each question. Record your answers on the scantron sheet. Answer only once for each question or you will be automatically marked wrong. Use a #2 pencil.

Name_________________________

Peoplesoft ID#__________________________

Page 2: Genetics 2009 3rd Exam

1. Hardy-Weinberg Equilibrium refers toa. the stability of the allele structure within a large randomly mating population from one generation to the next b. the maintenance of discontinuous variation within a populationc. heterozygous advantaged. the absence of selectione. none of the above

2. Which of the following does NOT affect Allele frequencies in a Hardy-Weinberg Equilibrium:

a. Incomplete dominanceb. Sampling errors (e.g., Founder’s Effect)c. Mating Patternsd. Migratione. All effect allele frequencies

3. A wild-type chromosome can be represented as ABC * DEFGH and from this a chromosomal aberration arises that can be represented as ABC * DEFGGH ( * = centromere). This is known as a:

 

a. deletion.b. translocation.c. duplication.d. pericentric inversion.e. paracentric inversion

4. In a large randomly mating population of Mexican Red knee Tarantulas, there are two alleles at a single gene that control the number of red spots found in the adult – a dominant 16 spot allele and a recessive 8 spot allele. In this population, 0.49 % of the adult Millipedes have 8 spots (fa/a =0.0049). What is the frequency of the heterozygotes (fA/a)?

a. 0.065b. 0.07c. 0.13d. 0.500e. 0.93

Page 3: Genetics 2009 3rd Exam

5. Beta Thalassemia is an autosomal recessive human disorder in which there is a reduced production of hemoglobin. This is a life threatening disorder for infants as it results in severe anemia and requires lifelong transfusions. The frequency of this disease, also known as Cooley’s Anemia, is very high in people from the Western Mediterranean area. In Greece, this disease is occurs in about 1 in every 100 births. Assuming Hardy-Weinberg equilibrium for this trait, the frequency of carriers is:

a. 0.01b. 0.09c. 0.10d. 0.18e. 0.81

6. It is thought that the Beta Thalassemia mutation, similar to the sickle cell anemia mutation, is an example of balancing polymorphisms. Thus, the heterozygotes may display a greater fitness than either of the two homozygotes in the presence of mosquitoes carrying a species of Plasmodium. Let us imagine that the relative finesses are:

Waa = 0.01WAa = 1WAA = 0.7

What would be the frequency of the Beta Thalassemia mutation at equilibrium (Hint the mutation frequency is = 1- the wild type allele frequency)?

a. 0.23b. 0.77c. 0.53d. 0.014e. None of the above

Q 7-9: In a population of pumpkins, the mean seed weight is 100 mg. Since a major supplier of pumpkin seeds will pay double for seeds that weigh 125 mg, you begin a selective breeding program. You select seed that average 140 mg as parents for a selective breeding program. s2

e = 95, s2g =605, s2

d =205

7. What is the additive genetic variance of this trait?a. 400b. 700c. 1000d. 500e. Not enough data

Page 4: Genetics 2009 3rd Exam

8. What is the Broad Sense heritability (H2) for this trait?a. 0.7833b. 0.29c. 0.86d. 0.57e. 0.01

9. What is the mean weight of the next generation of pumpkin seeds?a. 100b. 111.7c. 122.9d. 125e. 140

10. Dissociation Transposable elements:a. are enzymesb. require reverse transcriptasec. are non-autonomousd. found throughout the human genomee. can induce transposition of Activator elements

11. A retrotransposon such as the copia element of Drosophila:a. Transposes through an RNA intermediateb. Encodes for Capsid proteinsc. Encodes an reverse transcriptased. Does not generate mutable locie. All of the above.

12. Salvador Luria and Max Delbruck designed the fluctuation experiment to show that: a. resistant cells were produced by phage infection

b. resistant cells were selected by phage infectionc. phage infection did not have any effect on resistant cellsd. resistance cells were produced by environmental mutagense. none of the above

13. The Nucleotide Excision Repair system relies on:a. the proof reading capabilities of DNA polymerase IIIb. a photolyasec. Transcription Factor IIH (TFIIH)d. DNA ligase IVe. all of the above

Page 5: Genetics 2009 3rd Exam

14. You have discovered a mutant Andalusian horse with a very smooth lateral four beat gait similar to the Paso Fino breed. This gait is much more comfortable to the rider and will be prized in the equestrian community. In breeding studies you realize that this smooth gait is caused by a single dominant mutation. You find that by placing this mutation over a deficiency of the locus, the gate becomes smoother. Horses with two wild type copies in addition to the mutation have a rougher gate. This mutation is most likely a:

a. Neomorphb. Antimorphc. Hypermorphd. Hypomorphe. Amorph

15. In maize, there are three different breeding lines, which have the following gene orders on chromosome 3:

 1: a b c i h g k j d e f l2: a b c i h g f e d j k l3: a b c d e f g h i j k l

Which of the below is true?

 

a. An inversion of part of the chromosome in line 1 gave rise to the version in line 2.b. An inversion in part of the chromosome in line 3 gave rise to the chromosome in strain 1.c. Unequal crossing-over led to the productions of lines 3 and 1.d. A deletion led to the production of line 1 from line 2.e. All of the above are false.

16. You have identified the following mutation:

Wild type: ATG CTC AGT CTT GTT TGG TAAAmino Acids: Met Leu Ser Leu Val Trp Stop

Mutation: ATG CTT AGT CTT GTT TGG TAAAmino Acids: Met Leu Ser Leu Val Trp Stop

This mutation is a:a. Insertionb. Nonsense mutationc. Frameshift mutationd. Transversione. Silent Mutation

Page 6: Genetics 2009 3rd Exam

17. To phenocopy is to:a. to produce a specific phenotype by extra-genetic meansb. characterize the phenotypic variance generated by different environmentsc. determine the number of loci that regulate a quantitative traitd. to generate a second mutation with a gene that reverses the phenotypee. generate a dominant phenotype due to the loss-of-function of a gene.

18. The norm of reaction refers to:a. the variation of phenotypes in a populationb. the role of the dominance genetic variance in broad sense heritabilityc. the movement of the population mean after selectiond. the phenotypic response of a given genotype to different environmentse. none of the above

19. A Holliday Junction:a. generates a four-stranded DNA moleculeb. is capable of branch migrationc. is a test to establish genetic dominanced. a & be. none of the above

20. How many homozygous genotypes are produced by 6 loci, each with 2 alleles?a. 81b. 243c. 16d. 64e. 256

21. If a clonal population (all individuals with the same genotype) displays large variation for height, then:

a. heritabilities in both broad and narrow senses for height in this population equal zero

b. the narrow sense heritability is positive, but the broad sense heritability is zeroc. the broad sense heritability is positive, but the narrow sense heritability is zerod. there is significant environmental variance for this populatione. a & d

22. Which of the following are necessary for evolution by natural selection to be effective?

Page 7: Genetics 2009 3rd Exam

I. Variation in a population so that some individuals are better adapted than others.II. Competition for resources so that not all organisms can survive.III. Useful traits will be inherited by offspring.IV. Neutral mutations must be common.V. The number of species present must increase over time.

A. I, II, IV.B. I, III, IV.C. I, III, V.D. I, II, III, IV, V.E. I, II, III.

23. What is the underlying design of a fitness (adaptive) landscape? They:A. connect average population genotype to the complete phenotype of each genotype.B. connect average population genotype of two populations – allowing a direct comparison between populations.C. connect population fitness to the average population phenotype. D. connect average population genotype to population fitness.E. All of the above.

24. What combination of factors combine such that neutral mutations fix at an approximately constant rate?I. Neutral mutations occur at an approximately constant rate, per population.II. Neutral mutations occur at an approximately constant rate, per individual.III. The ratio of beneficial to neutral mutations is approximately constant. IV. The probability that a given neutral mutation will reach a frequency of 1 (i.e. fix) in the population does not depend on population size. V. The probability that a given neutral mutation will reach a frequency of 1 (i.e. fix) in the population is proportional to population size.

A. I and II.B. I and IV.C. III only.D. II and V.E. IV and V

25. In a sequencing reaction containing everything required for normal DNA synthesis plus a mix of deoxy-ATP and dideoxy-ATP, where will reactions terminate?A. Opposite some A nucleotides on the template strand.B. Opposite some G nucleotides on the template strand.C. Opposite some T nucleotides on the template strand.D. Opposite some C nucleotides on the template strand.E. Dideoxy-ATP will not terminate a growing DNA strand.

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26. You are trying to clone a human gene into a bacterial vector so that you can use bacteria to produce large amounts of the protein the gene encodes. You have cloned the gene OK (you checked using Sanger sequencing), but the bacteria don’t seem to be expressing the protein. What could be wrong?A. The human gene may have introns that are not spliced out in bacteriaB. The human gene may have a promoter sequence not recognized by the bacterial transcription machinery.C. The human protein may need post-translational modification before being functionalD. The human gene may lack a Shine-Dalgarno sequence necessary for efficient translation in bacteria.E. All of the above

27. What is T-DNA?A. The region of DNA transferred from the Ti plasmid to a plant genome. B. The region of DNA acquired by the Ti plasmid from a plant genome. C. The region of DNA transferred from the Ti plasmid to another bacteria.D. Any region of DNA on the Ti plasmid is called T-DNA.E. Any region of DNA that would provide a benefit to Agrobacterium tumefaciens if transferred to a plant genome.

28. What color test do you have?A. BlueB. GreenC. YellowD. Pink