Bio 97 Fall 2001 Final Exam

SECTION 1: -(one scantron line/question)

 Correct answers are in bold and blue

1. Alice is a woman whose father suffers from hemophilia. He has been successfully treated with recombinant factor VIII protein. Alice’s mother’s family has no history of hemophilia. The likelihood that Alice is carrier of hemophilia A is:

  1. 100%
  2. 67% (that is, 2/3)
  3. 50%
  4. 25%
  5. 0%, Alice cannot be a carrier

2. Which of the following statements are true:

  1. All eggs produced by the same woman are genetically identical to each other
  2. All sperm produced by the same man are genetically identical to each other
  3. Both A and B
  4. All the gametes produced by a person are genetically identical to a somatic cell of that individual
  5. None of the statements (A-D) are true

3. A new drug is discovered that prevents spontaneous abortion. Certain religious groups lobby that the drug must be given to all pregnant women.

  1. This would save many lives and have almost no bad consequences
  2. This would lead to a substantial increase in kids born with birth defects
  3. This would lead to a substantial increase in kids born aneuploid
  4. This would lead to a substantial increase in kids born with a gross chromosomal rearrangement
  5. B and C and D

4. Which of the following statements about recessive alleles is the most correct:

  1. They are usually X-linked
  2. They usually encode a non-functional protein or a protein with diminished function
  3. They usually encode a protein that has gained a new function
  4. They rapidly eliminated from the gene pool if disadvantageous
  5. A person who is a carrier of one has no one to blame but themself (and their mate) if they have a kid affected by a genetic disease

5. Which of the following statements is true:

  1. Any time DNA is damaged, a mutation will result.
  2. The vast majority of DNA damage is caused by man-made chemicals and man-made radiation.
  3. Hundreds of new alleles are created in each gamete as a result of replication errors.
  4. Typical DNA polymerases can correct ("proofread") most of their own mistakes
  5. All of the statements (A-D) are true

 

The next six questions relate to the above figure. The pedigree shows the inheritance of the phenotypes cardiac arrhythmia (a) and partial deafness (d) in the W and F families. A study was undertaken to determine if the arrhythmia in these families was likely (1) the autosomal recessive form, Jervell Lang-Neilson syndrome, also characterized by partial deafness, (2) Ward-Romano syndrome, an autosomal dominant form, not associated with deafness, or (3) perhaps a third type, inherited in some other fashion.

6. The arrhythmia in the W family is most likely dominant, because it:

  1. affects both monozygotic twins in the third generation
  2. is sometimes associated with deafness
  3. is passed from generation to generation without skipping
  4. is transmitted by both sexes
  5. actually, it is most likely recessive

7. The deafness and the arrhythmia in the W family are the result of alleles at different loci because:

  1. non-disjunction occurs among the progeny of the first couple in generation II
  2. Multifactorial inheritence is suggested from comparing the concordance of the twins to their siblings
  3. independent assortment of these traits can be observed in generation III
  4. they exhibit epistasis
  5. actually, the deafness and arrhythmia are associated with the same locus, or with two tightly-linked loci 

8. Assuming that the arrhythmia in the F family is dominant, X-linked dominant inheritance can be ruled out because of the observation of:

  1. female to male transmission (that is, an affected woman and unaffected man have an affected son)
  2. male to male transmission
  3. lack of male to male transmission
  4. male to female transmission
  5. actually, the arrhythmia is most likely X-linked 

9. Assuming that the deafness in the W family is dominant, X-linked dominant inheritance can be ruled out because of the observation of:

  1. male to male transmission
  2. lack of male to male transmission
  3. two affected males have daughters whose hearing is normal
  4. an affected female has a daughter whose hearing in normal
  5. actually, the deafness is most likely X-linked

10. Strictly speaking, the only mode(s) of inheritance that can absolutely be ruled out for the transmission of the deafness characteristic is:

  1. Autosomal recessive
  2. X-linked dominant
  3. X-linked recessive
  4. B and C
  5. A and B and C

11. If arrhythmia without associated deafness was multigenic, then:

  1. one would have to consider the possibility that the allele segregating in the W family was of a different locus than the one segregating in the F family.
  2. none of the kids born from the marriage of the W and F families would be arrhythmic
  3. all of the kids born from the marriage of the W and F families would be arrhythmic
  4. environmental influences would be a major determinant of the arrhythmic phenotype.
  5. A and B

(end of questions dealing with the figure)

 

12. A segment of a chromosome has five genes (V, W, X, Y and Z) in the following order: VWXYZ

Shown below are three rearrangements (each occuring in a different individual) involving this region:

VWXYXYZ VWYZ VWYXZ

These rearrangements are (in no particular order):

  1. translocation, duplication, inversion
  2. deletion, inversion, translocation
  3. deletion, inversion, duplication
  4. deletion, translocation, duplication
  5. preconception, interception, misperception

13. Which of the following (A-D) could NOT lead to the development of a woman with Turner syndrome (45, X)?:

  1. Non-disjunction during meiosis I in the father
  2. Non-disjunction during meiosis I in the mother
  3. Non-disjunction of the Y chromatids during meiosis II in the father
  4. Non-disjunction during meiosis II in the mother
  5. All of the above (A-D) could result in Turner syndrome.

14. People afflicted with Turner syndrome (45, X) appear female, while those afflicted with Klinefelter syndrome (47, XXY) appear male. From this, and our knowledge of the karyotype of normal human males and females, we can tell that sex in humans is determined by:

  1. the number of X chromosomes
  2. the presence or absence of the Y chromosome
  3. by the egg, not the sperm
  4. any of the above (A-C) are consistent with the observations
  5. none of the above (A-C) are consistent with the observations

15. The only trisomies that cause disease in adults are those involving chromosome 21 or the sex chromosomes. This is because:

  1. Chromosome 21 and the sex chromosomes are more prone to non-disjunction
  2. Trisomies or monosomies of any other chromosomes have such detrimental effects that they do not allow development to adulthood
  3. There are no genes for fundamental life processes located on chromosome 21
  4. Chromosome 21 contains modifier genes for sex determination
  5. Aneuploidies of the remaining chromosomes exist but are not recognized because they cause have a normal phenotype

16. All of the following statements are FALSE, except:

  1. The polymerase chain reaction is a leading cause of cell suicide
  2. People carrying deleterious (that is, "bad") recessive alleles should probably not have children
  3. The widespread use of genetically engineered crops presents virtually no risk to farmland ecology
  4. The genetics of mammals is so different from the genetics of the fruit fly Drosophila melanogaster that there is little we can learn about the former from the latter
  5. It is possible to have a fetus tested for certain genetic disorders and to choose selective abortion on the basis of that information

17. Which of the following statements are FALSE?

  1. The recombination frequency between two linked genes is proportional to the distance between them
  2. It is possible to determine the linear order of linked genes on a chromosome
  3. Double cross-overs cause an overestimation of the distance between two loci
  4. Two genes on the same chromosome might not be linked
  5. None of the statements are false

SECTION II – fill in two entries on the scantron for each question

18&19. You want to clone a human gene. Which of the following will you NOT need:

  1. Human DNA
  2. An enucleated embryonic cell
  3. A restriction enzyme
  4. DNA ligase
  5. A plasmid vector

20&21. About 40 cases have been reported of women afflicted with quad-X syndrome (48, XXXX), and about 25 cases of penta-X syndrome (49, XXXXX). Both syndromes are characterized by multiple dysmorphisms, low IQ, developmental delay and incomplete sexual development. The karyotypes associated with these syndromes could result from:

  1. Simple non-disjunction (in meiosis I or II) in both the mother and the father
  2. Successive non-disjunctive meiotic divisions in the mother (that is, non-disjunction in meiosis I followed by non-disjunction(s) in meiosis II); normal meiosis in father
  3. Successive non-disjunctive meiotic divisions in the father; normal meiosis in mother
  4. Quad-X could result from A or B, but penta-X could only result from B
  5. Quad-X could result from B or C, but penta-X could only result from A

22&23. Consider the martian pea plant. It has a gene controlling seed color – yellow vs. green, and another gene controlling seed texture – smooth vs. wrinkled. A true-breeding line with smooth green seeds was crossed to a true-breeding line with wrinkled yellow seeds. All the progeny (the F1) had smooth yellow seeds. The F1 generation was allowed to self-fertilize. Assume that the two genes in question (seed color and seed texture) are completely linked. What is the expected ratio of phenotypes (smooth yellow: smooth green: wrinkled yellow: wrinkled green) in the F2 peas?

  1. 9:3:3:1
  2. 1:0:1:0
  3. 2:1:1:0
  4. 0:1:1:0
  5. 1:1:1:1

24&25. Scientists searching for a human autosomal dominant disease gene perform 110 Southern blots on the DNA from an extended family, some of whom are afflicted with the disease. They use as probes a panel of 110 markers spanning all the autosomes (at a density of 4 to 6 markers per chromosome). They find one marker on chromosome 2 that appears to be located about 40 cM from the disease gene. A logical next step would be to:

  1. Sequence the marker to see if they can find mutations in affected patients
  2. Try a different panel of markers spanning the autosomes, and try to find one linked to the disease gene
  3. Try a panel of markers specific for chromosome 2 and covering it with a higher density, in order to find a marker more closely linked to the disease gene.
  4. Give up, because less than 1% of the markers that they tried was associated with the disease gene
  5. Try to clone an extinct goat

26&27. Mutations in the K-ras gene are frequently found in colon cancer. These mutations are always missense mutations, primarily in codons 12 or 61. The mutations in the APC gene are also associated with colon cancer. In this case, however, nonsense and frameshift mutations are frequent, and are scattered throughout most of the coding region. From this it can be concluded that:

  1. K-ras is a tumor-suppressor gene and APC is an oncogene
  2. K-ras is an oncogene and APC is a tumor-suppressor gene
  3. Both K-ras and APC are oncogenes
  4. Both K-ras and APC are tumor-suppressor genes
  5. The tumor cells must have defective mismatch repair

28&29. Your sister suffers from a genetic disorder known as Marfan syndrome (incidence 1/10,000), but you and your parents are unaffected. What is the approximate chance that you are a carrier?

  1. 1/100, or 1%
  2. 1/4, or 25%
  3. 1/2, or 50%
  4. 2/3, or 67%
  5. 100%

30&31. DNA samples are taken from a married couple, and analyzed by restriction digestion, agarose gel electrophoresis and Southern blotting, using as a probe a DNA sequence that is part of the c-Myc gene. The sample from the husband shows a single band of 6.0 kb. The sample from the wife shows this band, and also two other bands: one of 2 and one of 4 kb. Their daughter is then typed with the same probe. The predicted banding pattern of the daughter is:

  1. The 6 kb band
  2. The 6 kb, 4 kb and 2 kb bands
  3. The 4 kb band and the 2 kb band
  4. Either the 6 kb band or the 2 kb and 4 kb bands
  5. Either A or B

32&33. A man smokes two packs of cigarettes per day from age 11, quits smoking at age 37, has his first child at age 39 and is diagnosed with malignant lung cancer at age 40. "I begged you not to smoke," says his wife, "and now you’ve probably given our baby your cancer".

Her statement is:

  1. reasonable, because cancer is a genetic disease
  2. reasonable, due to the inheritance of acquired characteristics
  3. reasonable, because the man probably had an inherited predisposition to cancer
  4. not reasonable, because the carcinogenic mutations arose somatically and were not passed to the gametes
  5. not reasonable, because the man stopped smoking before he got cancer, hence there is no connection between the smoking and the cancer

34&35. The test cross ab/ab x Ab/aB is performed. The following number of progeny of each genotype are obtained: 87 AaBb, 409 Aabb, 390 aaBb, 114 aabb.

What is the approximate distance (in map units) between the two genes in question?

  1. 25 centiMorgans (cM)
  2. 20 cM
  3. 15 cM
  4. 10 cM
  5. 5 cM

36&37. A deamination occurs on the cytosine residue in the following DNA sequence 5'-GAATCG-3'. (Note: the top strand is shown; this is the strand where the deamination occurs.) If the mutation is not repaired, and a round of DNA replication occurs, then the sequence of the newly-replicated complementary strand (i.e., the bottom strand) will be:

  1. 5'-CAATTC-3'
  2. 5'-CGATTC-3'
  3. 5'-GAATCG-3'
  4. 5'-GAATTC -3'
  5. None of the above

38&39. Sickle cell anemia is an autosomal recessive disease with an incidence of 1/400 among African-Americans. What is the approximate frequency of heterozygote carriers for this disease within this population? Assume Hardy-Weinberg conditions apply.

  1. 1/400
  2. 1/200
  3. 1/20
  4. 1/10
  5. 1/4

40&41. All of the following statements are FALSE except:

  1. Monozygotic twins arise from a single egg fertilized by two sperm
  2. The variance of a trait is determined either by environmental or genetic variance, and never by both together
  3. Athletic parents usually have athletic kids, hence the variation in athletic ability must be solely determined by genes
  4. Intelligent parents usually have intelligent kids, hence the variation in intelligence must be solely determined by the home environment
  5. Simple genetic mechanisms involving a small number of genes can lead to a quantitative trait having an approximately normal distribution

42&43. A certain population has a higher frequency of cystic fibrosis (an autosomal recessive genetic disease with a poor prognosis) than would be predicted based on Hardy-Weinberg assumptions. This could be due to:

  1. Heterozygote superiority
  2. Inbreeding
  3. Poor selection against rare deleterious recessive alleles in diploids
  4. A and B
  5. A and B and C (either answer scored correct because of ambiguous question)

44&45. You want to clone an extinct goat. Which of the following will you NOT need:

  1. An enucleated embryonic cell
  2. A donor nucleus isolated from frozen goat cells
  3. A surrogate mother
  4. A plasmid vector
  5. You will need all the above