Study Guide Exam 2 

What is caused by faulty genetic imprinting?

1. Looking at the map of GAL genes below, the coding strands for GAL1  and GAL10 are:  

a. Upper (GAL 1) and lower (GAL 10) strands

b. Lower (GAL 1) and upper (GAL 10) strands

c. Upper (GAL 1) and upper (GAL 10) strands

d. Lower (GAL 1) and lower (GAL 10) strands

Answer: B

2. HP-1 binds H3-K9-methyl. If HP-1 is mutated, this suppresses position  effect variegation (fly eyes in flies with the w+ translocation have eyes that are mostly red-instead of variegated). HP-1 is a:

How many phenotypes predicted in the progeny of this cross, assuming dominance of purple (w) allele over white (w) allele?

a. Chromatin writer

b. Chromatin reader

c. Chromatin eraser

Answer: B

3. Defective genomic imprinting of the H19 and IGF2 genes leads to  major developmental defects.

Do you think that this is because:

a. Expression of 2 gene copies of H19 is required for proper development b. Expression of 2 gene copies of IGF2 or 2 gene copies of H19 is  detrimental to proper development

c. Expression of 1 copy of H19 and 1 copy of IGF2 is detrimental to proper development

Answer: B

4. How many PHENOTYPES PREDICTED in the progeny of this cross,  assuming dominance of purple (W) allele over white (w) allele?

What is meant by syntenic genes?

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a. One

b. Two

c. Three

d. Four

Answer: B

5. How many PHENOTYPES PREDICTED in the progeny of this cross,  assuming dominance of purple (W) allele over white (w) allele? a. One

b. Two

c. Three

d. Four

(ratio of phenotypes?)

a. 1:1:1:1 We also discuss several other topics like Does vibration help nerve pain?

b. 2:1

c. 3:1

d. 2:2

Answer: C

6. How many different GENOTYPES PREDICTED in the progeny of this  cross?

a. One  

b. Two

c. Three

d. Four

Answer: C

7. For round vs. wrinkled seeds, Mendel observed for (Rr x Rr) cross:  5474 round

1850 wrinkled

= 7324 seeds

What ration of round: wrinkled did Mendel predict? We also discuss several other topics like Why do stimulus cause an action potential?

a. 1:1

b. 2:1

c. 3:1

d. 2:2

Answer: C

Use the table below for question 8:

8. Where does X2 = 0.263 fall on the table: under 1 df?

a. 0.95 – 0.90

b. 0.70 – 0.50

c. 0.20 – 0.10

d. 0.05 – 0.01

Answer: B

a. Accept null hypothesis; Law of Segregation is correct

b. Reject null hypothesis; Law of Segregation is incorrect

Answer: A

9. You carry out a DNAse HS expt, using EcoRl and the DNA probe shown.

If you want to learn more check out Do abused parents become abusers?

If there is a DNase HS site where indicated, will you get the results shown in  1, 2, or 3?

Answer: C (3)

Use the image below for question 10:

10. What regulatory regions bind positive transcription factors in Cell 1A? Don't forget about the age old question of Notre dame in paris is a what church?
We also discuss several other topics like You perform a cross between a heterozygous tall pea plant and a homozygous short pea plant and obtain 30 tall plants and 20 short plants in the f1 generation. assuming standard mendelian inheritance of this character, how many tall and short plants would

a. Reg 1 and Reg 3 only

b. Reg 2 and Reg 4 only

c. Reg 1, Reg 2, Reg 3 only

d. Reg 2, Reg 3, Reg 4 only

Answer: C

Use the following image below for question 11:

11. Investigator wants to follow the fate of the 6 cells shown below

Which regulatory region, driving GAL 4 is needed to make sure these cells  (and perhaps more) express GFP?

a. 1

b. 2

c. 3

d. 4

Answer: D

Use the following image below for question 12:

12. Investigator wants to follow the fate of the 6 cells shown below:

What regulatory region, linked to GAL 80 can now make sure that column C  cells don’t express GFP?

a. 1

b. 2

c. 3

d. 4

Answer: C

13. Regions of human chromosomes 2 and 5 have been shown to  exhibit segmental aneuploidy, meaning that parts of the chromosomes  were deleted. If there are NO phenotypic consequences of these  chromosomal deletions, the genes are said to be __________. a. Codominant

b. Haplo-sufficient

c. Epistatic

d. Haplo-insufficient

Answer: B

14. Defects in the short stature homeobox (SHOX) genes result in a  wide range of phenotypic disorders. Some individuals have severe  skeletal defects, called Leri-Weill dyschondrosteosis (LWD), and others  have a milder defect, known as SHOX-related short stature. If both

copies of an allele must be present for a normal phenotype and a  defect in one copy of the allele is sufficient to cause a mutant  phenotype, the allele is said to be __________.

a. Codominant

b. Haplo-sufficient

c. Epistatic

d. Haplo-insufficient

Answer: D

15. A protein called suppressor of gene silencing 3 (SGS3) protects  RNA from degradation before it can be transformed into dsRNA during  plant posttranscriptional gene silencing pathways. Complete loss-of function sgs3 mutants lack mature siRNAs, and complete gain-of function mutations would exhibit high levels of both mature siRNAs and other cleavage products. A newly identified mutant overaccumulates  one type of RNA cleavage product but has fewer mature siRNAs. This  mutation that displays a novel phenotype is most likely what type of  mutation?

a. Neomorphic

b. Amorphic

c. Hypermorphic

d. Hypomorphic

Answer: A

16. If you were to cross a pure-breeding black cat with a pure breeding white cat and you got 100% gray offspring, what type of  inheritance would be demonstrated? Assume coat color is controlled by one gene with two alleles, B (black) and b (white).

a. Incomplete dominance

b. Recessiveness

c. Incomplete penetrance

d. Dominance

Answer: A

17. Individuals who are heterozygous for the sickle cell trait have  genotypes bAbS, and both wild-type beta-chains and mutant beta  chains of hemoglobin are found in their blood. The ability to detect  both allelic products in heterozygotes is an example of which type of  inheritance?

a. Incomplete dominance

b. Incomplete penetrance

c. Dominance

d. Codominance

Answer: D

18. When you cross two heterozygous black dogs you would expect a ratio of ¾ black : ¼ white in their offspring. However, you find that you get a ratio of ⅔ black : ⅓ white, with equal numbers of males and  females in both groups. What genetic inheritance pattern can explain  these results?

a. The allele for black coat is lethal.

b. The trait is sex-influenced.

c. The trait is sex-limited.

d. The alleles are codominant.

Answer: A

19. Hereditary hemochromatosis (HH) is a genetic condition that  causes "iron overload." Patients with severe forms of the disease must  donate blood regularly to remove excess iron from their bloodstreams.  Clinically, it is difficult to determine which HH patients will actually  develop increased iron levels and the adverse symptoms associated  with the disorder. Which type of inheritance pattern describes the  inability to predict whether an individual with the HH allele will develop the disorder based on his or her genotype?

a. Variable expressivity

b. Pleiotropy

c. Epistasis

d. Incomplete penetrance

Answer: D

20. Marfan syndrome is an autosomal dominant disorder associated  with a mutation in collagen and resulting in skeletal, optical, and  cardiovascular abnormalities. Each person with a dominant mutant  allele expresses at least one of the symptoms, with varying degrees of  severity. This is an example of __________.

a. Variable expressivity

b. Epistasis

c. Incomplete penetrance

d. A sex-influenced trait

Answer: A

21. In sweet pea, red flower color is controlled by two genes involved in the anthocyanin pathway, A and B. Both genes are required for  expression, so that a recessive pair of alleles at either locus would  result in a pink phenotype. If an AaBb plant were allowed to self-cross  and you got only red and pink flowers, what dihybrid ratio would you  expect in the offspring?

a. 9:6:1

b. 13:3

c. 9:7

d. 9:3:4

Answer: C

22. Squash shows dominant epistasis, meaning white masks the  presence of other color. If white (W) is dominant to color (w) and yellow (Y) is dominant to green (y), which of the following allelic ratios would  you expect if you crossed two WwYy plants?

a. 12 white : 3 yellow : 1 green

b. 9 white : 3 yellow : 4 green

c. 12 yellow : 3 green : 1 white

d. 9 green : 6 yellow : 1 white

Answer: A

23. Suppose you have a gene S that acts as a suppressor. The  presence of an S allele would suppress the expression of the dominant  allele at R. A dihybrid cross would result in a phenotypic ratio of 13:3.  Which genotypes would be included as part of the 3/16 in this ratio? a. RRSs and rrSs

b. RrSs and RRSs

c. rrss and Rrss

d. Rrss and RRss

Answer: D

24. Mutation of several different genes can result in albinism.  Suppose two parents both have albinism, but they have a child with  normal pigmentation. What is the possible genetic explanation for this  result?

a. Genetic complementation

b. Haploinsufficiency

c. Hypomorphic allele

d. Suppression

Answer: A

25. What is meant by syntenic genes?

a. Genes that produce more progeny with parental phenotypes b. Genes located on the same chromosome that always assort  independently

c. Genes located on the same chromosome that always exhibit genetic  linkage

d. Genes located on the same chromosome that may assort  independently or exhibit genetic linkage

Answer: D

26. What possible gametes will be produced from a parent who has  the genotype AaBBCc, assuming the three genes are NOT syntenic? a. A, a, B, C, c

b. ABC, ABc, aBC, aBc

c. Aa, BB, Cc

d. ABC, ABc, AbC, abC, Abc, abc, aBC, aBc

Answer: B

27. In the map of the Drosophila X chromosome, the recombination  frequency between vermillion and rudimentary is 26.9%, and the  distance between white and rudimentary is 45.2%. If you know that the order of the genes is w-v-r, how far apart are white and vermillion? a. 18.3 cM

b. 45.2 cM

c. 72.1 cM

d. 1830 cM

Answer: A

Recombination frequency is proportional to the percentage, and 45.2 − 26.9  = 18.3 cM.

28. For a given test cross with an expected ratio of 1:1:1:1, you  calculate the chi-square value and compare it to the chi-square table to determine your p value. What would a p value of p << 0.005 tell you  about the two genes?

a. This result indicates that the observed ratios are very close to the  expected results, suggesting that the genes are linked.

b. This result indicates that the observed ratios are very close to the  expected results, suggesting that the genes are not linked. c. This result indicates a significant deviation between observed and  expected results, suggesting that the genes are not linked. d. This result indicates a significant deviation between observed and  expected results, suggesting that the genes are linked.

Answer: D

29. The probability of one single crossover event is 20%, and that of  the other single crossover is 40%. If each crossover event is  independent, what is the expected number of double crossover  events?

a. 0.04, or 4%

b. 0.08, or 8%

c. 0.20, or 20%

d. 0.6, or 60%

Answer: B

30. You observe five double crossovers among 100 test-cross  progeny, or 5%. If the two single-crossover frequencies are 0.2 and  0.35, what is the coefficient of coincidence (c)?

a. 0.0035

b. 0.02

c. 0.12

d. 0.714

Answer: D

31. For a dihybrid organism AB/ab, the frequency of the Ab gamete is 10%. What is the frequency of the AB gamete?

a. 10%

b. 40%

c. 45%

d. 90%

Answer: B

32. For a trihybrid organism ABC/abc, the probability of NO  crossovers between genes a and b is 80% (0.8), and the probability of  NO crossovers between b and c is 70% (0.7). What is the frequency of  the Abc gamete?

a. 0.03

b. 0.07

c. 0.12

d. 0.28

Answer: B

33. Which of the following statements is true?

a. Only the sex chromosomes of the homogametic sex undergo higher  rates of recombination; the rates for autosomes are the same in each  sex.

b. Chromosomes of the homogametic sex undergo higher rates of  recombination than chromosomes of the heterogametic sex. c. Autosomes have higher rates of recombination in the heterogametic  sex than in the homogametic sex.

d. Autosomes show a higher recombination rate than do sex  chromosomes, in both the homogametic and heterogametic sex.

Answer: B

34. Which of the following statements is false?

a. Linkage disequilibrium in a population can be a result of recent  immigration into the population.

b. Linkage disequilibrium in a population can be a result of natural  selection.

c. Linkage disequilibrium can be used to study the specific function of a  gene.

d. Analysis of the patterns of linkage disequilibrium and of the different  haplotypes involved, such as in the case of the beta-globin gene, can  be used to understand the origin of alleles

Answer: C

35. What types of gametes result from each type (meiosis I and meiosis II) of  meiotic nondisjunction?

a. Meiosis I nondisjunction results in two aneuploid cells and two normal  diploid cells, whereas meiosis II nondisjunction results in four aneuploid cells.

b. Meiosis I nondisjunction results in four aneuploid cells, whereas meiosis II nondisjunction results in four normal diploid cells.

c. Meiosis I nondisjunction results in four normal diploid cells, whereas  meiosis II nondisjunction results in four aneuploid cells.

d. Meiosis I nondisjunction results in four aneuploid cells, whereas meiosis II nondisjunction results in two aneuploid cells and two normal diploid  cells.

Answer: D

35. When during the cell cycle would you expect to find  chromosomes that are highly condensed and easily visualized by  microscopy?

a. Prophase

b. Anaphase

c. Interphase

d. Metaphase

Answer: D

36. Will a genomic DNA library made from kidney contain genes  expressed only in pancreas?

a. Yes

b. No

Answer: B

37. Can you find a pancreas-specific cDNA in a cDNA library made  from kidney?

a. Yes

b. No

Answer: A

38. Bacteria transfer genetic material from one to another by three  processes. Which of the following processes involves the transfer of  DNA from a donor bacterium to a recipient bacterium by way of a viral  vector?

a. Transfection

b. Conjugation

c. Transduction

d. Transformation

Answer: C

39. Which type of bacterial DNA contains genes that promote  antibiotic resistance?

a. F plasmid

b. R plasmid

c. Bacteriophage

d. Bacterial chromosome

Answer: B

40. Exconjugants are identified by their genotypes, which are distinct from those of either the donor strain or the recipient strain. Which  method would you use to identify an exconjugant?

a. Isolation of the F plasmid from the cell's DNA

b. Microscopy to look for the presence of the conjugation pilus c. Growth on a selective growth medium

d. Growth on a complete medium containing no antibiotics Answer: C

41. What happens to linear DNA within a bacterial cell? a. Linear DNA remains in the cell indefinitely but cannot be incorporated  into the chromosome.

b. Linear DNA is degraded by nuclease enzymes.

c. Linear DNA becomes circularized, and the ends are sealed together to  form a new DNA plasmid.

d. Linear DNA is added to the large bacterial chromosomes by nicking the chromosome and resealing the nick with DNA ligase.

Answer: B

42. Which type of DNA transfer is unable to facilitate high-resolution  mapping of closely linked genes?

a. Transformation

b. Conjugation

c. Direct genomic sequencing

d. Transduction

Answer: B

43. After a transformed cell undergoes DNA replication and cell division cycle, what is the genetic composition of the resulting  daughter cells?

a. Two daughter cells that retain the recipient chromosome and are not  genetically altered

b. One transformant that retains the genetic alterations and one  nonaltered daughter cell

c. Two transformants, both of which retain the genetic alterations d. Some cells that retain the transforming DNA and others that do not  retain the new DNA and revert back to the recipient chromosome

Answer: B

44. What name describes a virus that infects bacteria and is capable  of integrating its DNA into the host chromosome?

a. Virulent phage

b. λ phage

c. Temperate phage

d. Prophage

Answer: C

45. Which step of the lytic cycle is commonly accompanied by  fragmentation of the host chromosome and, occasionally,  mispackaging of a fragment of the host chromosome into a phage  head?

a. Injection of the phage chromosome into the host cell

b. Packaging of phage chromosomes into phage heads

c. Transcription and translation of phage genes

d. Replication of phage DNA

Answer: B

46. If a gene is under negative transcriptional control, what would  you expect to find bound to the operator?

a. Inducer protein

b. Inducer–corepressor protein complex

c. Repressor protein

d. RNA polymerase

Answer: C

47. What compound can bind to the allosteric binding domain of an  activator protein, converting the activator to an inactive conformation? a. Inducer

b. Inhibitor

c. Repressor

d. Effector

Answer: B

48. The enzyme β-galactosidase catalyzes which of the following  reactions?

a. Glucose → lactose + galactose

b. Galactose → lactose + glucose

c. Lactose → 2 glucose

d. Lactose → galactose + glucose

Answer: D

49. Which gene in the lac operon can be mutated without affecting  the cell’s ability to transport and break down lactose for metabolism? a. lacA

b. lacY

c. lacZ

d. lacP

Answer: A

50. In response to heat stress, bacteria alter gene transcription by  activating heat stress response genes through expression of what  gene?

a. An alternative sigma factor

b. Antitermination stem loop

c. Antisense RNA

d. Translation repressor proteins

Answer: A

51. How do translation repressor proteins regulate translation in  bacteria?

a. Translation repressor proteins bind near the Shine–Dalgarno sequence  and interfere with the recognition of the 16S ribosomal subunit. b. Translation repressor proteins bind to the operator region of the gene,  which interferes with RNA polymerase binding.

c. Translation repressor proteins bind to the mRNA and activate  alternative sigma factors to change promoter-recognition capacity of  the RNA polymerase core enzyme.

d. Translation repressor proteins form an antitermination stem loop  structure, which allows RNA polymerase to transcribe the gene through the leader region and into the structural genes of the operon.

Answer: A

52. How does TPP work to regulate transcription in B. subtilis? a. TPP binds to the riboswitch and prevents RNA polymerase from  binding.

b. TPP binds to the mRNA transcript and recruits nucleases to degrade it. c. TPP binds to the riboswitch, which allows a stem loop structure to form  that is followed by a poly-U sequence and causes termination of  transcription before the thi operon is transcribed.

d. TPP binds to the operator region and interferes with RNA polymerase  binding.

Answer: C

53. How does TPP work to regulate translation in E. coli? a. When TPP levels are high, TPP binds to the small ribosomal subunit,  which decreases its recognition of the Shine Dalgarno sequence. b. When TPP levels are high, it binds to the riboswitch and basically stops  transcription.

c. When TPP levels are high, TPP binds to the riboswitch, which results in  a stem loop structure forming that blocks recognition of the Shine Dalgarno sequence.

d. When TPP levels are high, TPP binds to the Shine-Dalgarno sequence  and physically blocks rRNA from binding.

Answer: C

54. After the circularization of the λ phage chromosome, what  controls transcription of genes whose products determine whether a  phage enters the lytic or lysogenic cycle?

a. Inducers and repressors

b. Early promoters and early operators

c. Alternative sigma factors

d. Late genes

Answers: B

55. During the induction process, what does the RecA protein target  and cleave?

a. cI protein

b. cII/cIII protein complex c. Cro protein

d. λ repressor protein Answer: D