Bsc 1010L Ch 13

Front 1

1. A genetic _____ indicates the distances between gene loci measured in terms of the frequency of recombination.
A. map
B. profile
C. pedigree
D. clone
E. karyotype

Back 1

A. map

Front 2

2. Of the 23 pairs of human chromosomes, 22 pairs are homologous and are found in both males and females. These are called ________.
A. bivalents
B. autosomes
C. recombinant chromosomes
D. somatic chromosomes

Back 2

B. autosomes

Front 3

3. Traits that are controlled by genes located on the X chromosome are said to be ________________.
A. autosomal
B. gametal
C. sex-linked
D. pleiotropic

Back 3

C. sex-linked

Front 4

4. Allele pairs are most likely to assort independently of one another when
A. they control unrelated traits.
B. they control related traits.
C. they are on the same chromosome.
D. they are sex linked.
E. they are on different chromosomes.

Back 4

E. they are on different chromosomes.

Front 5

5. The number of allele pairs that assort independently in an organism is generally much higher than the number of chromosome pairs. This phenomenon is due to
A. independent assortment.
B. segregation.
C. crossing over.
D. sex-linkage.
E. chromosome inactivation.

Back 5

C. crossing over.

Front 6

6. The theory of chromosomal inheritance was first proposed by
A. Mendel.
B. Morgan.
C. Knight.
D. Sutton.
E. Stern.

Back 6

D. Sutton

Front 7

7. In Drosophila, the sex of an individual is determined by
A. 1 pair of alleles.
B. the number of X chromosomes.
C. the number of Y chromosomes.
D. 1 pair of autosomes.
E. 2 pairs of alleles.

Back 7

B. the number of X chromosomes.

Front 8

8. In Morgan's experiments, the white eye allele in Drosophila was shown to be
A. located on the X chromosome.
B. located on the Y chromosome.
C. dominant.
D. located on an autosome.
E. codominant.

Back 8

A. located on the X chromosome.

Front 9

9. The geneticist who discovered the white eye mutation in Drosophila and helped establish that genes are carried on chromosomes was
A. Mendel.
B. Sutton.
C. Sturtevant.
D. Janssens.
E. Morgan.

Back 9

E. Morgan.

Front 10

10. Genetic exchange between 2 homologous chromosomes is called
A. synapsis.
B. pleiotropy.
C. crossing over.
D. allelic exchange.
E. independent assortment.

Back 10

C. crossing over.

Front 11

11. Occasionally, chromosomes fail to separate during meiosis, leading to daughter cells that have an abnormal number of chromosomes. This phenomenon is called
A. epistasis.
B. nondisjunction.
C. crossing over.
D. pleiotropy.
E. chromosome inactivation.

Back 11

B. nondisjunction.

Front 12

12. Humans who have lost one copy of an autosome are called
A. haploid.
B. trisomic.
C. bisomic.
D. monosomic.
E. monoploid.

Back 12

D. monosomic.

Front 13

13. In humans, individuals with trisomy of the ______ chromosome are most likely to survive until adulthood.
A. X.
B. 13th.
C. 15th.
D. 18th.
E. 21st.

Back 13

E. 21st.

Front 14

14. If a human female has 2 Barr bodies per cell, it is almost certain that
A. her father had 1 Barr body per cell.
B. her mother also had 2 Barr bodies per cell.
C. she developed from a fertilized egg with 3 X chromosomes.
D. she is genetically a male with female characteristics.
E. she is genetically a normal fertile female.

Back 14

C. she developed from a fertilized egg with 3 X chromosomes.

Front 15

15. A human female with only one X chromosome is said to have a condition called
A. X chromosome inactivation.
B. Angelman syndrome.
C. Turner syndrome.
D. Klinefelter syndrome.
E. Down syndrome.

Back 15

C. Turner syndrome.

Front 16

16. The most common fatal genetic disorder of Caucasians is
A. cholera.
B. cystic fibrosis.
C. hemophilia.
D. sickle cell anemia.
E. muscular dystrophy.

Back 16

B. cystic fibrosis.

Front 17

17. In sickle cell anemia, the defective hemoglobin differs from the normal hemoglobin by
A. the color of the pigment.
B. the size of the molecule.
C. a single amino acid substitution.
D. the total number of amino acids.
E. the type of blood cell it is found in.

Back 17

C. a single amino acid substitution.

Front 18

18. Hemophilia is caused by a
A. recessive allele on the X chromosome.
B. dominant allele on the X chromosome.
C. codominant allele on the X chromosome.
D. recessive allele on an autosome.
E. dominant allele on an autosome.

Back 18

A. recessive allele on the X chromosome.

Front 19

19. _______________ is a human hereditary disease that is caused by a dominant allele but does not show up in affected individuals until they are in middle age.
A. Cystic fibrosis
B. Sickle cell anemia
C. Tay-Sachs disease
D. Huntington's disease
E. Hemophilia

Back 19

D. Huntington's disease

Front 20

20. Amniocentesis is a procedure that is normally used
A. to reduce the risk of genetic disease.
B. for gene therapy.
C. to change the sex of the fetus.
D. for diagnosis of genetic disorders.
E. for nourishing the fetus.

Back 20

D. for diagnosis of genetic disorders.

Front 21

21. Huntington's disease is caused by a single dominant allele. It is a lethal disease, yet it persists in the human population. Which of the following statements best describes why?
A. Huntington's disease is sex-linked and every human has at least one X chromosome; thus, the chances are extremely high for this allele to be maintained in the human population.
B. Huntington's disease can present symptoms so mild that they appear to lack dominant expression of the allele in some individuals; in those cases, the allele is passed on to the offspring.
C. While lethal to a parent, Huntington's disease will not be lethal to the offspring since it can skip a generation.
D. Huntington's disease presents symptoms in mid-life, after most people have already had offspring.
E. Even though Huntington's disease is lethal, it improves chances for reproduction before the person dies.

Back 21

D. Huntington's disease presents symptoms in mid-life, after most people have already had offspring.

Front 22

22. In humans, the male has an X and a Y sex chromosome. The human female has two X chromosomes. In birds, the female has a Z and a W sex chromosome while the male has two Z chromosomes. Which of the following statements is accurate about which parent determines the gender of the offspring?
A. In humans and birds, the male determines the gender of all the offspring.
B. In humans and birds, the female determines the gender of all the offspring.
C. In humans, the male determines the gender of the offspring, and in birds the female determines the gender.
D. In humans, the female determines the gender of the offspring, and in birds the male determines the gender.
E. Determination of the gender of any human or bird offspring is related to the environmental conditions at the time of conception.

Back 22

C. In humans, the male determines the gender of the offspring, and in birds the female determines the gender.

Front 23

23. Sickle cell anemia is caused by a defect in the
A. oxygen-carrying pigment hemoglobin.
B. ability of the blood to clot.
C. ability of red blood cells to fight infection.
D. chloride ion transport protein.

Back 23

A. oxygen-carrying pigment hemoglobin.

Front 24

24. How many Barr bodies does a normal human female contain in each diploid cell?
A. 0
B. 1
C. 2
D. 3

Back 24

B. 1

Front 25

25. A test cross can be used to do all of the following except
A. determine whether an individual that displays a dominant phenotype is homozygous for the trait.
B. determine whether an individual that displays a dominant phenotype is heterozygous for the trait.
C. gather genotype information from phenotype information.
D. identify the chromosome on which a gene is located.

Back 25

D. identify the chromosome on which a gene is located.

Front 26

26. Which of the following animals is a genetic male?
A. bird ZW
B. grasshopper XO
C. honeybee diploid
D. Drosophila XXY

Back 26

B. grasshopper XO

Front 27

27. In humans, if non-disjunction led to an individual with a genotype of XO, that person would
A. be female because each cell lacks a Y chromosome.
B. be male because each cell has only one X chromosome.
C. display both male and female characteristics.
D. not survive.

Back 27

A. be female because each cell lacks a Y chromosome.

Front 28

28. In humans, if non-disjunction led to an individual with a genotype of XXY, that person would
A. be female because each cell has two X chromosomes.
B. be male because each cell has one Y chromosome.
C. display both male and female characteristics.
D. not survive.

Back 28

B. be male because each cell has one Y chromosome.

Front 29

29. In some species, sex determination is influenced by environmental temperature during development. If you wanted to determine the temperature at which one would obtain a 1:1 sex ratio in a particular species of turtle, which of the following experiments would best address this question?
A. Grow the turtles in two different incubators at temperatures of 22ºC and 30ºC.
B. Grow the turtles in three different incubators at temperatures of 26ºC, 28ºC, and 30ºC.
C. Grow the turtles in four different incubators at temperatures of 24ºC, 26ºC, 28ºC, and 30ºC.
D. Grow the turtles in five different incubators at temperatures of 22ºC, 24ºC, 26ºC, 28ºC, and 30ºC.

Back 29

D. Grow the turtles in five different incubators at temperatures of 22ºC, 24ºC, 26ºC, 28ºC, and 30ºC.

Front 30

30. Any genetic differences between individuals in a population are called
A. markers.
B. alleles.
C. polymorphisms.
D. SNPs.

Back 30

C. polymorphisms.

Front 31

31. The classic experiments performed by Creighton and McClintock in Maize
A. provided the initial evidence for genetic recombination.
B. provided evidence that genes located on the same chromosome do not assort independently.
C. allowed for the establishment of the first genetic map.
D. provided evidence for the physical exchange of genetic material between homologues.

Back 31

D. provided evidence for the physical exchange of genetic material between homologues.

Front 32

32. In humans, if an XY individual had a deletion of the SYR gene, that person would
A. develop as a female.
B. have both male and female characteristics.
C. have ambiguous genitalia.
D. develop as a male.

Back 32

A. develop as a female.

Front 33

33. Which statement about calico cats is false?
A. Calico cats can be male or female.
B. The different colored fur is due to the inactivation of one X chromosome.
C. The variation in coat color is an example of an epistatic interaction.
D. Calico cats are genetic mosaics.

Back 33

A. Calico cats can be male or female.

Front 34

34. If an XY human had a genetic disorder that causes insensitivity to androgens, that person's genotype and phenotype would be
A. XX, female.
B. XX, male.
C. XY, female.
D. XY, male.

Back 34

C. XY, female.

Front 35

35. Which offspring inherit all their mitochondrial DNA from their mother and none from their father?
A. daughters
B. sons
C. both sons and daughters
D. neither sons nor daughters

Back 35

C. both sons and daughters

Front 36

36. Nondisjunction of a single pair of autosomes can lead to all of the following except
A. aneuploidy.
B. monosomy.
C. trisomy.
D. euploidy.

Back 36

D. euploidy.

Front 37

37. If you needed to determine the order of genes on a chromosome, you should perform
A. a test cross.
B. a two-point cross.
C. a three-point cross.
D. a SNP test.

Back 37

C. a three-point cross.

Front 38

38. A 39-year-old woman is in her sixth week of pregnancy. Due to her advanced age, she is at higher risk for having a baby with Down's syndrome than younger pregnant women. She would like to find out as early as possible whether or not her baby has Down's syndrome. Her doctor should suggest
A. amniocentesis.
B. genetic counseling.
C. chorionic villi sampling.
D. a pedigree analysis.

Back 38

C. chorionic villi sampling.

Front 39

39. In Drosophila, dosage compensation is controlled by the male-specific lethal (MSL) complex consisting of MSL proteins and roX RNAs. Based on what you know about dosage compensation, the role of the MSL complex in males would be to
A. double the level of expression of genes on the X chromosome.
B. increase the level of expression of genes on the X chromosome by 50%.
C. decrease the level of expression of genes on the X chromosome by 50%.
D. decrease the level of expression of genes on the X chromosome by 100%.
E. double the level of expression of genes on the Y chromosome.

Back 39

A. double the level of expression of genes on the X chromosome.

Front 40

40. Given that these 2 gene loci are very closely linked, the genotypic ratio in the F2 generation should be closest to
A. 1:2:1
B. 1:1:1:1
C. 9:3:3:1
D. 3:1

Back 40

A. 1:2:1

Front 41

41. Given that these 2 gene loci are very closely linked, the phenotypic ratio in the F2 generation should be closest to
A. 1 brown, heavy: 2 red, thin: 1 red, thin
B. 1 brown, thin: 2 red, thin: 1 red, heavy
C. 3 red, thin: 1 brown heavy
D. 1 brown: 1 red: 1 heavy: 1 thin

Back 41

B. 1 brown, thin: 2 red, thin: 1 red, heavy

Front 42

42. What would be the results of a test cross with the F1 flies?
A. 1 brown, thin: 1 red, heavy
B. 1 brown, heavy: 1 red, thin
C. 1 brown, thin: 2 red, thin: 1 red, heavy
D. 1 brown, heavy: 2 red, thin: 1 red, thin

Back 42

A. 1 brown, thin: 1 red, heavy

Front 43

43. What is the relationship between recombination frequency and the actual physical distance on a chromosome?
A. As physical distance increases, the recombination frequency increases in a linear fashion.
B. As physical distance increases, the recombination frequency decreases in a linear fashion.
C. As physical distance increases, the recombination frequency first increases in a linear fashion, but gradually levels off to a frequency of 0.5.
D. As physical distance increases, the recombination frequency first increases, but then decreases.

Back 43

C. As physical distance increases, the recombination frequency first increases in a linear fashion, but gradually levels off to a frequency of 0.5.

Front 44

44. In a two-point cross to map genes A and B, you obtained 98 recombinant types and 902 parental types among the offspring. How far apart are these genes?
A. 9.8 cM
B. 0.98 cM
C. 90.2 cM
D. 9.02 cM
E. .098 cM

Back 44

A. 9.8 cM

Front 45

45. Morgan's student Sturtevant demonstrated that the recombination frequencies between a series of linked genes is additive. Examine the following recombination data from Sturtevant, and determine the proper order of the genes on the Drosophila X chromosome. Assume y is in the 0.0 position.

A. y m v w
B. y w v m
C. y m w v
D. y w m v

Back 45

B. y w v m

Front 46

46. Which of the following will help you distinguish between the two final suspects?
A. single nucleotide polymorphisms (SNPs)
B. human genetic map
C. linkage data
D. anonymous markers

Back 46

A. single nucleotide polymorphisms (SNPs)

Front 47

47. Why can't you use mitochondrial DNA to distinguish between these two suspects?
A. The sequence of mitochondrial DNA has not yet been determined.
B. The brothers share the same mitochondrial DNA.
C. There are no molecular techniques available that allow one to analyze mitochondrial DNA.
D. Because mitochondrial DNA is inherited in a paternal pattern.

Back 47

B. The brothers share the same mitochondrial DNA.

Front 48

48. In some human populations, the proportion of individuals who are heterozygous for the sickle cell allele is much higher than would be expected by chance alone. Why?
A. Individuals with two normal alleles have an advantage over heterozygous individuals.
B. Individuals with two harmful alleles have an advantage over heterozygous individuals.
C. Individuals with two harmful alleles have an advantage over individuals with two normal alleles.
D. Heterozygous individuals have an advantage over individuals with two normal alleles.

Back 48

D. Heterozygous individuals have an advantage over individuals with two normal alleles.

Front 49

49. A deletion of a particular stretch of chromosome 15 can cause either Prader-Willi syndrome or Angelman syndrome, depending on
A. the parental origin of the normal and deleted chromosome.
B. whether or not the region is methylated properly.
C. whether a translocation event has occurred.
D. whether a nondisjunction event has occurred.

Back 49

A. the parental origin of the normal and deleted chromosome.

Front 50

50. How did the development of anonymous markers aid in the production of a human genetic map?
A. Anonymous markers are genetic markers that do not cause a detectable phenotype, but can be detected by molecular techniques. The markers correspond to specific and unique chromosomal regions, thereby allowing for the identification and ordering of particular segments of DNA. Such information was essential to the generation of a human genetic map.
B. Anonymous markers are genetic markers that cause a detectable phenotype and can't be detected by molecular techniques. The markers correspond to specific and unique chromosomal regions, thereby allowing for the identification and ordering of particular segments of DNA. Such information was essential to the generation of a human genetic map.
C. Anonymous markers are genetic markers that do not cause a detectable phenotype, but can be detected by molecular techniques. The markers correspond to specific and unique genetic regions, thereby allowing for the identification and ordering of particular segments of the chromosome. Such information was essential to the generation of a human genetic map.

Back 50

A. Anonymous markers are genetic markers that do not cause a detectable phenotype, but can be detected by molecular techniques. The markers correspond to specific and unique chromosomal regions, thereby allowing for the identification and ordering of particular segments of DNA. Such information was essential to the generation of a human genetic map.

Front 51

51. Why isn't mitochondrial DNA a unique identifier?
A. Mitochondrial DNA is inherited through the paternal lineage. All offspring inherit their father's mitochondria, and therefore the same mitochondrial DNA. As a result, all family members that share a paternal lineage would have the same mitochondrial DNA. Mitochondrial DNA can therefore be used to confirm or eliminate a person's relationship within a paternal line, but cannot be used to identify a specific individual.
B. Mitochondrial DNA is inherited through the maternal lineage. All offspring inherit their mother's mitochondria, and therefore the same mitochondrial DNA. As a result, all family members that share a maternal lineage would have the same mitochondrial DNA. Mitochondrial DNA can therefore be used to confirm or eliminate a person's relationship within a maternal line, but cannot be used to identify a specific individual.
C. Mitochondrial DNA is inherited through the maternal lineage. All female offspring inherit their mother's mitochondria, and therefore the same mitochondrial DNA. As a result, all female family members that share a maternal lineage would have the same mitochondrial DNA. Mitochondrial DNA can therefore be used to confirm or eliminate a person's relationship within a maternal line, but cannot be used to identify a specific individual.

Back 51

B. Mitochondrial DNA is inherited through the maternal lineage. All offspring inherit their mother's mitochondria, and therefore the same mitochondrial DNA. As a result, all family members that share a maternal lineage would have the same mitochondrial DNA. Mitochondrial DNA can therefore be used to confirm or eliminate a person's relationship within a maternal line, but cannot be used to identify a specific individual.

Front 52

58. Suppose you are carrying out a series of crosses with an insect where the mechanism of sex determination is unknown. You discover a mutant female with short bristles and decide to cross it with a wild type male that has normal bristles. Half of the F1 progeny have short bristles but all of these short-bristled F1 progeny are males. Based on these results, a valid hypothesis would be
A. Males are ZW, females are ZZ, and short bristles are caused by a dominant allele on the Z chromosome
B. Males are ZZ, females are ZW, and short bristles are caused by a recessive allele on the Z chromosome
C. Males are ZZ, females are ZW, and short bristles are caused by a dominant allele on the W chromosome
D. Males are ZZ, females are ZW, and short bristles are caused by a dominant allele on the Z chromosome

Back 52

D. Males are ZZ, females are ZW, and short bristles are caused by a dominant allele on the Z chromosome

Front 53

59. In 1910, Morgan did a series of experiments with the fruit fly Drosophila, an organism where females are XX and males are XY. When a mutant male fly with white eyes was crossed with a wild type female with red eyes, none of the F1 progeny had white eyes but 18% of the F2 progeny had white eyes. Unexpectedly, all of these white-eyed F2 flies were males. Based on these results, Morgan concluded that white eyes is caused by a recessive X-linked allele. Suppose Morgan has found that half of the F1 progeny had white eyes but all of these white-eyed F1 flies were females. In this case, a valid hypothesis would be
A. White eyes is caused by a recessive Y-linked allele
B. White eyes is caused by a dominant Y-linked allele
C. White eyes is caused by a dominant X-linked allele
D. White eyes is caused by a dominant autosomal allele

Back 53

C. White eyes is caused by a dominant X-linked allele

Front 54

61. Genetic maps are based on recombination frequencies. Because both odd and even numbers of crossovers can occur between any 2 gene loci, as the physical distance between two loci increases, the maximum recombination frequency levels off at 50%. However, suppose you discovered a species where only an even number of crossovers can occur between any two gene loci. In this case, as the physical distance between two loci increases, you would expect the maximum recombination frequency to
A. remain at zero
B. increase with no limit
C. level off at 25%
D. level off at 75%
E. level off at 100%

Back 54

A. remain at zero