Test Bank for Genetics Essentials Concepts and Connections 3rd Edition by Pierce ISBN 1464190755
9781464190759
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1. What is the role of the SRY gene in humans?
A) It initiates the X inactivation process in females.
B) It is located on the X chromosome and causes the X to pair with the Y chromosome during male meiosis.
C) It is located on the Y chromosome and initiates the developmental pathway toward the male phenotype.
D) It is located on an autosomal chromosome and represses expression of autosomal genes in order to balance their expression level with genes on the X chromosome.
E) None of the answers is correct.
2. What is the expected outcome for a human embryo with the XXXY chromosome constitution?
A) It would likely develop into a female who will not respond to the hormone testosterone.
B) It would likely develop into a sterile male with reduced testes.
C) It will always abort early in development before birth.
D) It would likely develop into a tall female who may be slightly cognitively impaired.
E) It would likely develop into a fertile man with a completely normal male phenotype.
3. Which of the following chromosome constitutions would never lead to a viable human baby being born?
A) XXX
B) XYY
C) XO (O = the absence of a second chromosome)
D) YY
E) XXY
4. In which of the following organisms is gender/sex determined by the temperature during embryonic development?
A) Humans
B) Mice
C) Fruit flies
D) Many snakes and all birds
E) Many turtles and alligators
5. In species of birds, males are the homogametic sex and females the heterogametic sex. Which if the following is TRUE in this system of sex determination?
A) The gender of the offspring is determined by the female parent.
B) Male offspring have a ZW chromosome constitution.
C) The gender of the offspring is determined by the male parent.
D) Female offspring have a ZZ chromosome constitution.
E) Female and male offspring have the same chromosome constitution.
6. In a germ-line cell from a female grasshopper (XX-XO sex determination system), when do the homologous X chromosomes segregate?
A) During mitosis
B) During meiosis
C) Immediately after fertilization
D) They do not segregate; gametes contain a copy of X and a copy of Y.
E) Both answers a and c are correct
7. In a germ-line cell from a human male that is dividing, when do the X and Y chromosomes segregate?
A) During mitosis
B) During meiosis
C) They do not segregate; gametes contain a copy of X and a copy of Y.
D) Immediately after fertilzation.
E) None of the answers is correct.
8. Which of the following human genotypes is associated with Klinefelter syndrome?
A) XXY
B) XXXY
C) XXXXY
D) All of the answers are correct.
E) None of the answers is correct.
9. What is the sex chromosome constitution of a male duck-billed platypus?
A) XX
B) XY
C) XO
D) ZZ
E) XXXXXYYYYY
10. An XXY chromosome constitution produces _____ development in humans and _____ development in diploid fruit flies.
A) female; female
B) male; male
C) female; male
D) male; female
E) male, intersex
11. With the XX-XO sex determination system, generally:
A) female offspring have one X chromosome, and it is inherited from their father.
B) male offspring have one X chromosome, and it is inherited from their mother.
C) male offspring have one X chromosome, and it is inherited from their father.
D) female offspring have one X chromosome, and it is inherited from their mother.
E) None of the statements is true.
12. Human males, with XY chromosomes are _____ and produce two different kinds of gametes, whereas females with XX chromosomes are _____ and produce only one kind.
A) homogametic; heterogametic
B) pleiotropic: epistatic
C) heterogametic; homogametic
D) epistatic; pleiotropic
E) epistatic; heterogametic
13. Red–green color blindness is X-linked recessive. A woman with normal color vision has a father who is color blind. The woman has a child with a man with normal color vision.Which phenotype is NOT expected?
A) A color-blind female
B) A color-blind male
C) A noncolor-blind female
D) A noncolor-blind male
14. A Barr body is a(n):
A) gene on the X chromosome that is responsible for female development.
B) patch of cells that has a phenotype different from surrounding cells because of variable X inactivation.
C) inactivated X chromosome, visible in the nucleus of a cell that is normally from a female mammal.
D) extra X chromosome in a cell that is the result of nondisjunction.
E) extra Y chromosome in a cell that is the result of nondisjunction.
15. Which of the following statements about X inactivation in mammalian females is FALSE?
A) Females that are heterozygous for an X-linked gene have patches of cells that express one allele and patches of cells that express the other.
B) Some genes on the inactive X continue to be expressed after the chromosome is inactivated.
C) X inactivation is random as to which X is inactivated and takes place early in embryonic development.
D) In an individual with three X chromosomes, one of the three will become inactivated.
E) Once an X chromosome first becomes inactivated in a cell, that same X will remain inactivated in somatic cells that are descendants of this cell.
16. What is the apparent purpose for X inactivation in humans and other mammals?
A) It allows for the levels of expression of genes on the X chromosome to be similar in males and females.
B) It allows for the levels of expression of genes on the autosomes to be similar to the levels of genes on the X chromosome.
C) It suppresses the expression of genes on the Y chromosome in males.
D) It reduces the amount of nondisjunction during meiosis in females.
E) It enhances the level of pairing between the two X chromosomes during meiosis in females.
17. The Lyon hypothesis helps us to understand which phenomenon in mammals?
A) X-linked inheritance
B) Evolution of the Y chromosome
C) Dosage compensation between males and females
D) Development of male and female secondary sexual characteristics
E) Sex determination
18. Women with Turner syndrome (XO) and normal women (XX) are clearly different phenotypically. In addition, the vast majority of XO conceptions abort before birth. However, both XO and XX women have one active X chromosome since the X in XO women remains active and one might expect that they would therefore have similar phenotypes. What is the MOST reasonable explanation for their different phenotypes?
A) XO women do not have a copy of the SRY gene.
B) Some genes remain active on the inactive X chromosome, and XX women will have two copies of these genes expressed while XO women only have one copy expressed.
C) In XO women, the single X chromosome has no partner to pair with during mitosis so that each cell division is delayed by pairing problems with the single X not finding a pairing partner.
D) XO women have two copies of the SRY gene so that they are forced to develop partway along the male pathway during embryogenesis.
E) None of these are reasonable explanations.
19. In which of the following individuals would you expect to find two Barr bodies in their somatic cells?
A) XX
B) XO
C) XXY
D) XXYY
E) XXX
20. Normal males (XY) and Klinefelter males (XXY) both possess only one active X chromosome. Nonetheless there are clearly phenotypic differences between the two. What is the MOST reasonable explanation as to why such differences exist?
A) The Y chromosome has higher gene expression levels when two X chromosomes are present compared to one X.
B) The Y chromosome has lower gene expression levels when two X chromosomes are present compared to one X.
C) Some genes remain active on inactive X chromosomes, so XXY males would produce higher expression levels for these genes compared to XY males.
D) XXY males exhibit a higher rate of problems during mitotic divisions than XY males.
E) XXY males don't have a copy of the SRY gene.
21. How many Barr bodies (condensed X chromosomes) would you predict in an XXY male with Klinefelter syndrome?
A) One per somatic cell
B) Two per somatic cell
C) Three per somatic cell
D) None
E) Either two or three depending on the tissue type
22. Women known to be heterozygous or carriers for the sex-linked recessive condition of hemophilia A were studied to determine the time required for their blood to clot. It was found that the time required for their blood to clot varied from individual to individual. The values obtained ranged from normal clotting at one extreme to clinical hemophilia at the other extreme. What is the most probable CORRECT explanation for these findings?
A) Some women had only one X chromosome, and it is inactive.
B) Some women had three copies of the X chromosome which allowed them to make extra amounts of gene products for their X-linked genes.
C) The women with normal clotting times probably had a mother with hemophilia while those with abnormal clotting times probably had fathers with hemophilia.
D) Random X inactivation probably results in individuals with different proportions of cells in their bodies expressing the normal allele at the hemophilia locus.
E) In women with abnormal clotting times, there was probably an epistatic interaction between an allele on the X chromosome and an allele on an autosomal gene that reduced the expression of the X-linked gene.
23. If a female Drosophila that is heterozygous for a recessive X-linked mutation is crossed to a wild-type male, what proportion of female progeny will have the mutant phenotype?
A) 100%
B) 0%
C) 33%
D) 25%
24. A woman is phenotypically normal, but her father had the sex-linked recessive condition of red–green color blindness. If she has children with a man with normal vision, what is the probability that their first child will have normal vision and their second child will be color blind?
A) 1/16
B) 3/8
C) 3/16
D) 3/6
E) 8/27
25. A eukaryotic diploid cell from an organism with the ZZ-ZW sex determination system has two pairs of autosomes and a pair of sex chromosomes, Z and W, shown below.
From what type of individual is this cell?
A) Male
B) Female
C) Hermaphrodite
D) Homogametic
E) Pseudoautosomal
26. A eukaryotic diploid cell from an organism with the ZZ-ZW sex determination system has two pairs of autosomes and a pair of sex chromosomes, Z and W, shown below.
A diploid cell from this individual begins to go through meiosis. After the completion of meiosis I, it becomes two cells. One of these two cells now undergoes meiosis II. Which of the following is a possible normal combination of chromosomes in one of the subsequent two cells after the completion of meiosis II?
A) One chromosome with the A allele, one with the B allele, one Z, one W
B) One chromosome with the A allele, one with the a allele, one with the B allele, one with the b allele, one Z, one W
C) A pair of chromosomes with A alleles, a pair of chromosomes with B alleles, a pair of Z chromosomes
D) One chromosome with an a allele, one chromosome with a B allele, one W
E) None of these is possible.
27. A eukaryotic diploid cell from an organism with the ZZ-ZW sex determination system has two pairs of autosomes and a pair of sex chromosomes, Z and W, shown below.
What is the probability of a gamete from this individual that has the following genotype: alleles A and b, chromosome Z?
A) 1/2
B) 1/4
C) 1/6
D) 1/8
E) 1/16
28. A eukaryotic diploid cell from an organism with the ZZ-ZW sex determination system has two pairs of autosomes and a pair of sex chromosomes, Z and W, shown below.
Assume A and B are dominant alleles. If this individual were crossed to an individual of genotype Aa Bb, what is the probability of a female offspring with the two dominant traits given by alleles A and B?
A) 1/8
B) 1/16
C) 9/16
D) 9/32
E) 3/32
29. A eukaryotic diploid cell from an organism with the XX-XO sex determination system has two pairs of autosomes and one X chromosome, shown below.
From what type of individual is this cell?
A) Male
B) Female
C) Hermaphrodite
D) Pseudoautosmal
E) Intersex
30. A eukaryotic diploid cell from an organism with the XX-XO sex determination system has two pairs of autosomes and one X chromosome, shown below.
A diploid cell from this individual begins to go through meiosis. After the completion of meiosis I, it becomes two cells. One of these two cells now undergoes meiosis II. Which of the following is a possible normal combination of chromosomes in one of the subsequent two cells after the completion of meiosis II?
A) One chromosome with the A allele, one with the B allele, and two X chromosomes
B) One chromosome with the A allele, one with the a allele, one with B allele, one with b allele, and two X chromosomes
C) One chromosome with the A allele, one chromosome with the B allele
D) One chromosome with the a allele, one chromosome with the B allele, one X chromosome
E) One chromosome with the A allele and one chromosome with the B allele AND one chromosome with the a allele, one chromosome with the B allele, and one X chromosome are BOTH possible.
31. A eukaryotic diploid cell from an organism with the XX-XO sex determination system has two pairs of autosomes and one X chromosome, shown below.
What is the probability of a gamete from this individual that has the following genotype: alleles A and b, chromosome X?
A) 1/2
B) 1/4
C) 1/6
D) 1/8
E) 1/16
32. The trait shown below is expressed only in males in the family. What is the BEST explanation for the inheritance of this trait?
A) Y-linked inheritance
B) Sex-linked (X-linked) recessive inheritance
C) Autosomal recessive inheritance
D) X-linked dominant inheritance
E) None of these explanations fit for this trait.
33. If a male bird that is heterozygous for a recessive Z-linked mutation is crossed to a wild type female, what proportion of the progeny will be mutant males?
A) 0% B) 100%
C) 75%
D) 50%
E) 25%
34. A woman has normal vision although her maternal grandfather (her mother's father) had red–green color blindness, a sex-linked recessive trait. Her maternal grandmother and the woman's own father are assumed to not possess a copy of the mutant allele. The woman marries a man with normal vision although his father was color blind. What is the probability that the first child of this couple will be color blind?
A) 1/2
B) 1/4
C) 1/8
D) 1/16
E) 1/12
35. Joan is phenotypically normal but had a child with the autosomal recessive disease cystic fibrosis (CF) from a previous marriage. Joan's father has hemophila A, a sexlinked recessive condition where the blood fails to clot properly. Her father has survived due to recent treatment advances. Joan now intends to marry Bill, who is also phenotypically normal but who has a sister, Jill, with CF. Bill's parents are phenotypically normal, and there is no history of hemophilia A in his family. Assume that Joan and Bill do marry and have a child. What is the probability that this child will have CF, but will not have hemophilia A ? (Hint: This problem requires that you utilize concepts from Chapter 3 as well as Chapter 4.)
A) 1/8
B) 1/12
C) 1/24
D) 3/32
E) 5/32
36. In humans there is a genetic disorder that results from a dominant mutation present in a gene located in the pseudoautosomal region of the Y chromosome and on the X chromosome.Which of the following statements is CORRECT?
A) All affected men marrying normal women will have no affected daughters.
B) All affected women marrying normal men will have affected daughters and no affected sons.
C) All affected men marrying normal women will have affected daughters, but all the sons will be normal.
D) All affected women marrying nomal men will have only normal sons and daughters.
E) None of the statements is correct.
37. Familial vitamin-D-resistant rickets is an X-linked dominant condition in humans. If a man is afflicted with this condition and his wife is normal, it is expected that among their children, all the daughters would be affected and all the sons would be normal. In families where the husband is affected and the wife is normal, this is almost always the outcome among their children when such families have been studied. Very rarely an unexpected result occurs in such families where a boy is born with the disorder. If the chromosomes of such unusual boys are examined, what might be expected to be found?
A) Some of the boys are XYY.
B) Some of the boys are XY but have lost the SRY gene from their Y chromosome.
C) Some of the boys are YY.
D) Some of the boys are XXY.
E) Some of the boys are XXX.
38. The R locus determines flower color in a new plant species. Plants that are genotype RR have red flowers, and plants that are rr have white flowers. However, Rr plants have pink flowers. What type of inheritance does this demonstrate for flower color in these plants?
A) Complete dominance
B) Incomplete dominance
C) Codominance
D) Complementation
E) Lethal alleles
39. Interactions among the human ABO blood group alleles involve _____ and _____.
A) codominance; complete dominance
B) codominance; incomplete dominance
C) complete dominance; incomplete dominance
D) epistasis; complementation
E) continuous variation; environmental variation
40. In the endangered African watchamakallit, the offspring of a true-breeding black parent and a true-breeding white parent are all gray. When the gray offspring are crossed among themselves, their offspring occur in a ratio of 1 black:2 gray:1 white. Upon close examination of the coats, each hair of a gray animal is gray. What is the mode of inheritance?
A) One gene pair with black dominant to white
B) One gene pair with codominance
C) One gene pair with incomplete dominance
D) Two gene pairs with recessive epistasis
E) Two gene pairs with duplicate genes
41. Suppose that extra fingers and toes are caused by a recessive trait, but it appears in only 60% of homozygous recessive individuals. Two heterozygotes conceive a child. What is the probability that this child will have extra fingers and toes?
A) 0.05
B) 0.10
C) 0.15
D) 0.25
E) 0.33
42. Polydactyly is the condition of having extra fingers or toes. Some polydactylous persons possess extra fingers or toes that are fully functional, whereas others possess only a small tag of extra skin. This is an example of:
A) variable expressivity.
B) complete dominance.
C) independent assortment.
D) complementation.
E) cytoplasmic inheritance.
43. Achondroplasia is a common cause of dwarfism in humans. All individuals with achondroplasia are thought to be heterozygous at the locus that controls this trait. When two individuals with achondroplasia mate, the offspring occur in a ratio of 2 achondroplasia:1 normal. What is the MOST likely explanation for these observations?
A) Achondroplasia is incompletely dominant to the normal condition.
B) Achondroplasia is codominant to the normal condition.
C) The allele that causes achondroplasia is a dominant lethal allele.
D) The allele that causes achondroplasia is a recessive lethal allele.
E) The allele that causes achondroplasia is a late-onset lethal allele.
44. Crossing two yellow mice results in 2/3 yellow offspring and 1/3 nonyellow offspring. What percentage of offspring would you expect to be nonyellow if you crossed two nonyellow mice?
A) 25%
B) 33%
C) 66%
D) 75%
E) 100%
45. In humans, blood types A and B are codominant to each other and each is dominant to O. What blood types are possible among the offspring of a couple of blood types AB and A?
A) A, B, AB, and O
B) A, B, and AB only
C) A and B only
D) A, B, and O only
E) A and AB only
46. A mother of blood type A gives birth to a child with blood type O. Which of the following could NOT be the blood type of the father?
A) A
B) B
C) O
D) AB
E) Any of the above is a possible blood type of the father.
47. You are studying body color in an African spider and have found that it is controlled by a single autosomal gene with four alleles: B (brown), br (red), bg (green), and by (yellow). B is dominant to all the other alleles, and by is recessive to all the other alleles. The bg allele is dominant to by but recessive to br. You cross a pure-breeding brown spider with a pure-breeding green spider. Predict the genotype of the progeny.
A) B/bg
B) Br/bg
C) br/by
D) by/bg
E) B/by
48. You are studying body color in an African spider and have found that it is controlled by a single gene with four alleles: B (brown), br (red), bg (green), and by (yellow). B is dominant to all the other alleles, and by is recessive to all the other alleles. The bg allele is dominant to by but recessive to br. You cross a pure-breeding brown spider with a pure-breeding green spider. Predict the phenotype of the progeny.
A) Half brown, half green
B) Three-fourths brown, one-fourth green
C) All brown
D) All green
E) All yellow
49. In rabbits, an allelic series helps to determine coat color: C (full color), cch (chinchilla; gray color), ch (Himalayan; white with black extremities), and c (albino; all white). The C allele is dominant to all others, cch is dominant to ch and c, ch is dominant to c, and c is recessive to all the other alleles. This dominance hierarchy can be summarized as C >
cch > ch > c. Indicate the phenotypic ratios expected in rabbits with the cross Ccch × Cch .
A) 1 full color:1 chinchilla
B) 1 full color:1 Himalayan
C) 1 chinchilla:1 Himalayan
D) 3 full color:1 chinchilla
E) 2 full color:1 Himalayan:1 albino
50. In rabbits, an allelic series helps to determine coat color: C (full color), cch (chinchilla; gray color), ch (Himalayan; white with black extremities), and c (albino; all white). The C allele is dominant to all others, cch is dominant to ch and c, ch is dominant to c, and c is recessive to all the other alleles. This dominance hierarchy can be summarized as C >
cch > ch > c. Indicate the phenotypic ratios expected in rabbits with the cross Cch × chc
A) 1 full color:1 chinchilla
B) 1 full color:1 Himalayan
C) 1 chinchilla:1 Himalayan
D) 3 full color:1 chinchilla
E) 2 full color:1 Himalayan:1 albino
51. In rabbits, an allelic series helps to determine coat color: C (full color), cch (chinchilla; gray color), ch (Himalayan; white with black extremities), and c (albino; all white). The C allele is dominant to all others, cch is dominant to ch and c, ch is dominant to c, and c is recessive to all the other alleles. This dominance hierarchy can be summarized as C >
cch > ch > c. Indicate the phenotypic ratios expected in rabbits with the cross Cch × cc.
A) 1 full color:1 chinchilla
B) 1 full color:1 Himalayan
C) 1 chinchilla:1 Himalayan
D) 3 full color:1 chinchilla
E) 2 full color:1 Himalayan:1 albino
52. In rabbits, an allelic series helps to determine coat color: C (full color), cch (chinchilla; gray color), ch (Himalayan; white with black extremities), and c (albino; all white). The C allele is dominant to all others, cch is dominant to ch and c, ch is dominant to c, and c is recessive to all the other alleles. This dominance hierarchy can be summarized as C >
cch > ch > c. Indicate the phenotypic ratios expected in rabbits with the cross cchch × chc.
A) 1 full color:1 chinchilla
B) 1 full color:1 Himalayan
C) 1 chinchilla:1 Himalayan
D) 3 full color:1 chinchilla
E) 2 full color:1 Himalayan:1 albino
53. In rabbits, an allelic series helps to determine coat color: C (full color), cch (chinchilla; gray color), ch (Himalayan; white with black extremities), and c (albino; all white). The C allele is dominant to all others, cch is dominant to ch and c, ch is dominant to c, and c is recessive to all the other alleles. This dominance hierarchy can be summarized as C >
cch > ch > c. Indicate the phenotypic ratios expected in rabbits with the cross Cc × chc
A) 1 full color:1 chinchilla
B) 1 full color:1 Himalayan
C) 1 chinchilla:1 Himalayan
D) 3 full color:1 chinchilla
E) 2 full color:1 Himalayan:1 albino
54. A mother with blood type A has a child with blood type A. Give all possible blood types for the father of this child.
A) O
B) B, AB
C) A, AB
D) A, B, O
E) A, B, AB, O
55. A mother with blood type B has a child with blood type O. Give all possible blood types for the father of this child.
A) O
B) B, AB
C) A, AB
D) A, B, O
E) A, B, AB, O
56. A mother with blood type A has a child with blood type AB. Give all possible blood types for the father of this child.
A) O
B) B, AB
C) A, AB
D) A, B, O
E) A, B, AB, O
57. A mother with blood type AB has a child with blood type B. Give all possible blood types for the father of this child.
A) O
B) B, AB
C) A, AB
D) A, B, O
E) A, B, AB, O
58. A blood-type B woman is married to a man whose blood type is unknown. They have three children whose blood types are B, O, and AB. What is the husband's genotype?
A) IAIA
B) IAIO (IO = i)
C) IBIO
D) IBIB
E) IOIO
59. You are studying a gene that controls ossicone (horn) length in giraffes. The wild-type long-ossicone allele (L) is dominant to the mutant short-ossicone (l) allele. However the L allele is only 60% penetrant. You cross two heterozygous giraffes. What proportion of offspring would you expect to exhibit the long ossicone phenotype? Assume the penetrance of L applies equally to both homozygotes and heterozygotes.
A) 0.40
B) 0.45
C) 0.55
D) 0.60
E) 0.75
60. You are studying a gene that controls ossicone (horn) length in giraffes. The wild-type long-ossicone allele (L) is dominant to the mutant short-ossicone (l) allele. However, the L allele is only 60% penetrant. You cross two heterozygous giraffes. What proportion of offspring would you expect to exhibit the short ossicone phenotype? Assume the penetrance of L applies equally to both homozygotes and heterozygotes.
A) 0.25
B) 0.40
C) 0.45
D) 0.55
E) 0.60
61. Hair color is determined in Labrador retrievers by alleles at the B and E loci. A dominant allele B encodes black pigment, whereas a recessive allele b encodes brown pigment. Alleles at a second locus affect the deposition of the pigment in the shaft of the hair; dominant allele E allows dark pigment (black or brown) to be deposited, whereas recessive allele e prevents the deposition of dark pigment, causing the hair to be yellow. What type of gene interaction does this represent?
A) Recessive epistasis
B) Dominant epistasis
C) Duplicate recessive epistasis
D) Duplicate dominant epistasis
E) Dominant and recessive epistasis
62. Hair color is determined in Labrador retrievers by alleles at the B and E loci. A dominant allele B encodes black pigment, whereas a recessive allele b encodes brown pigment. Alleles at a second locus affect the deposition of the pigment in the shaft of the hair; dominant allele E allows dark pigment (black or brown) to be deposited, whereas recessive allele e prevents the deposition of dark pigment, causing the hair to be yellow. A black female Labrador retriever was mated with a yellow male. Half of the puppies were black and half were yellow. If the genotype of the black female parent was Bb Ee, then what was the genotype of the other parent?
A) bb ee
B) bb EE
C) Bb ee
D) BB ee
E) BB EE
63. Suppose that the “fabulous” phenotype is controlled by two genes, A and B, as shown in the diagram below. Allele A produces enough enzyme 1 to convert “plain” to “smashing.” Allele a produces no enzyme 1. Allele B produces enough enzyme 2 to convert “smashing” to “fabulous.” Allele b produces no enzyme 2. The A and B genes are both autosomal and assort independently.
What will be the phenotype(s) of the F1 offspring of a true-breeding “fabulous” father and a true-breeding “plain” mother (aa bb)?
A) All “plain”
B) All “smashing”
C) All “fabulous”
D) “Plain” females and “fabulous” males
E) “Fabulous” females and “smashing” males
64. Coat color is determined by two loci in large cats. Two pink panthers fall in love and produce a large litter of baby panthers with the following phenotypic ratios: 12/16 pink; 3/16 black; and 1/16 white. What might be the genotype of the pink progeny?
A) A_ B_
B) A_ bb
C) aa B_
D) aa bb
E) A_ B_ and A_ bb
65. Coat color is determined by two loci in large cats. Two pink panthers fall in love and produce a large litter of baby panthers with the following phenotypic ratios: 12/16 pink; 3/16 black; and 1/16 white. What might be the genotype of the black progeny?
A) A_ B_
B) AA Bb
C) aa B_
D) aa bb
E) A_ B_ and A_ bb
66. Coat color is determined by two loci in large cats. Two pink panthers fall in love and produce a large litter of baby panthers with the following phenotypic ratios: 12/16 pink; 3/16 black; and 1/16 white. What is the genotype of the white progeny?
A) A_ B_
B) A_ bb
C) aa B_
D) aa bb
E) A_ B_ and A_ bb
67. Coat color is determined by two loci in large cats. Two pink panthers fall in love and produce a large litter of baby panthers with the following phenotypic ratios: 12/16 pink; 3/16 black; and 1/16 white. What kind of gene interaction is this?
A) Recessive epistasis
B) Dominant epistasis
C) Duplicate recessive epistasis
D) Duplicate dominant epistasis
E) Dominant and recessive epistasis
68. Two loci control body color in beetles. In a cross between a black beetle and a white beetle you obtain a ratio of 9 black to 7 white beetles. What kind of gene interaction is this?
A) Recessive epistasis
B) Dominant epistasis
C) Duplicate recessive epistasis
D) Duplicate dominant epistasis
E) Dominant and recessive epistasis
69. In order to determine if mutations from different organisms that exhibit the same phenotype are allelic, which test would you perform?
A) Testcross
B) Epistasis test
C) Complementation test
D) Allelic series test
E) Biochemical test
70. In purple people eaters, purple is dominant to white. A true-breeding white mutant is mated with a different true-breeding white mutant. All of the F1 are purple. When the purple F1 offspring mate with each other, their offspring occur in the ratio of 9 purple:7 white. Which phenomenon explains the purple F1 offspring?
A) Recessive epistasis
B) Dominant epistasis
C) Complementation
D) Mutation
E) Suppression
71. In the yawncat (a rare hypothetical animal), the dominant allele R causes solid tail color, and the recessive allele r results in white spots on a colored background. The black coat color allele B is dominant to the brown allele b, but these genes can only be expressed if the animal has an mm genotype at a third gene locus. Animals that are M_ are yellow regardless of which allele from the B locus is present. A mating between a solid yellowtailed male yawncat and a solid brown-tailed female yawncat produces 16 offspring with the following tail phenotypes: six solid yellow, two spotted yellow, three solid black, one spotted black, three solid brown, and one spotted brown. What is the MOST likely genotype of the male parent?
A) MM BB RR
B) MM Bb RR
C) Mm Bb RR
D) Mm BB Rr
E) Mm Bb Rr
72. In the yawncat (a rare hypothetical animal), the dominant allele R causes solid tail color, and the recessive allele r results in white spots on a colored background. The black coat color allele B is dominant to the brown allele b, but these genes can only be expressed if the animal has an mm genotype at a third gene locus. Animals that are M_ are yellow regardless of which allele from the B locus is present. A mating between a solid yellowtailed male yawncat and a solid brown-tailed female yawncat produces 16 offspring with the following tail phenotypes: six solid yellow, two spotted yellow, three solid black, one spotted black, three solid brown, and one spotted brown. What is the MOST likely genotype of the female parent?
A) mm bb rr
B) Mm bb rr
C) mm Bb rr
D) mm bb Rr
E) mm Bb Rr
73. In the yawncat (a rare hypothetical animal), the dominant allele R causes solid tail color, and the recessive allele r results in white spots on a colored background. The black coat color allele B is dominant to the brown allele b, but these genes can only be expressed if the animal has an mm genotype at a third gene locus. Animals that are M_ are yellow regardless of which allele from the B locus is present. A mating between a solid yellowtailed male yawncat and a solid brown-tailed female yawncat produces 16 offspring with the following tail phenotypes: six solid yellow, two spotted yellow, three solid black, one spotted black, three solid brown, and one spotted brown. What is the probability of the next offspring from these same two parents having a spotted brown tail?
A) 1/2
B) 3/16
C) 1/4
D) 1/16
E) 9/16
74. In chickens, comb shape is determined by genes at two loci (R, r and P, p). A walnut comb is produced when at least one dominant gene R is present at one locus and at least one dominant gene P is present at a second locus (genotype R_ P_). A rose comb is produced when at least one dominant gene is present at the first locus and two recessive genes are present at the second locus (genotype R_ pp). A pea comb is produced when two recessive genes are present at the first locus and at least one dominant gene is present at the second (genotype rr P_). If two recessive genes are present at the first and the second locus (rr pp), a single comb is produced. Give genotypes for comb shape of the parents in the following cross: Walnut crossed with single produces one walnut, one rose, one pea, and one single offspring.
A) RR PP × rr pp
B) RR Pp × rr pp
C) Rr PP × rr pp
D) Rr Pp × rr pp
E) Rr pp × rr pp
75. In chickens, comb shape is determined by genes at two loci (R, r and P, p). A walnut comb is produced when at least one dominant gene R is present at one locus and at least one dominant gene P is present at a second locus (genotype R_ P_). A rose comb is produced when at least one dominant gene is present at the first locus and two recessive genes are present at the second locus (genotype R_ pp). A pea comb is produced when two recessive genes are present at the first locus and at least one dominant gene is present at the second (genotype rr P_). If two recessive genes are present at the first and the second locus (rr pp), a single comb is produced. Give genotypes for comb shape of the parents in the following cross: rose crossed with pea produces 20 walnut offspring.
A) RR pp × rr PP
B) Rr pp × rr Pp
C) Rr pp × rr PP
D) RR pp × rr Pp
E) Rr pp × Rr Pp
76. A woman is deaf because of being homozygous recessive at an autosomal locus. She marries a deaf man who is also homozygous recessive at an autosomal locus. They have two children who have normal hearing. Assuming that this couple is the biological parents of these children, how is this situation MOST reasonably explained?
A) Imprinting the allele from the father is active, and the allele from the mother is inactive in the children.
B) Epistasis two genes are involved with the alleles at one locus suppressing gene activity at a second locus.
C) Sex-limited the trait is only expressed in one gender but not in both.
D) Imprinting the allele from the mother is active, and the allele from the father is inactive in the children.
E) Complementation the parents are homozygous recessive at different loci, and the children are heterozygous at both.
77. In chickens, comb shape is determined by genes at two loci (R, r and P, p). A walnut comb is produced when at least one dominant gene R is present at one locus and at least one dominant gene P is present at a second locus (genotype R_ P_). A rose comb is produced when at least one dominant gene is present at the first locus and two recessive genes are present at the second locus (genotype R_ pp). A pea comb is produced when two recessive genes are present at the first locus and at least one dominant gene is present at the second (genotype rr P_). If two recessive genes are present at the first and the second locus (rr pp), a single comb is produced. Give genotypes for comb shape of the parents in the following cross: pea crossed with single produces one single offspring.
A) rr PP × rr pp
B) RR Pp × rr pp
C) Rr PP × rr pp
D) Rr Pp × rr pp
E) rr Pp × rr pp
78. In chickens, comb shape is determined by genes at two loci (R, r and P, p). A walnut comb is produced when at least one dominant gene R is present at one locus and at least one dominant gene P is present at a second locus (genotype R_ P_). A rose comb is produced when at least one dominant gene is present at the first locus and two recessive genes are present at the second locus (genotype R_ pp). A pea comb is produced when two recessive genes are present at the first locus and at least one dominant gene is present at the second (genotype rr P_). If two recessive genes are present at the first and the second locus (rr pp), a single comb is produced. Give genotypes for comb shape of the parents in the following cross: rose crossed with pea produces two walnut, one single, and one pea offspring.
A) RR pp × rr PP
B) Rr pp × rr Pp
C) Rr pp × rr PP
D) RR pp × rr Pp
E) Rr pp × Rr Pp
79. In chickens, comb shape is determined by genes at two loci (R, r and P, p). A walnut comb is produced when at least one dominant gene R is present at one locus and at least one dominant gene P is present at a second locus (genotype R_ P_). A rose comb is produced when at least one dominant gene is present at the first locus and two recessive genes are present at the second locus (genotype R_ pp). A pea comb is produced when two recessive genes are present at the first locus and at least one dominant gene is present at the second (genotype rr P_). If two recessive genes are present at the first and the second locus (rr pp), a single comb is produced. Give genotypes for comb shape of the parents in the following cross: rose crossed with single produces 31 rose offspring.
A) RR PP × rr pp
B) RR pp × rr pp
C) Rr PP × rr pp
D) Rr Pp × rr pp
E) Rr pp × rr pp
80. A 13:3 ratio is obtained from the cross Aa Bb ×Aa Bb. What phenotypic ratio is expected if an Aa Bb individual is testcrossed (Aa Bb × aa bb)?
A) 1:1:1:1
B) 9:7
C) 2:1
D) 3:1
E) 13:3
81. In chickens, comb shape is determined by genes at two loci (R, r and P, p). A walnut comb is produced when at least one dominant gene R is present at one locus and at least one dominant gene P is present at a second locus (genotype R_ P_). A rose comb is produced when at least one dominant gene is present at the first locus and two recessive genes are present at the second locus (genotype R_ pp). A pea comb is produced when two recessive genes are present at the first locus and at least one dominant gene is present at the second (genotype rr P_). If two recessive genes are present at the first and the second locus (rr pp), a single comb is produced. Give genotypes for comb shape of the parents in the following cross: rose crossed with single produces 10 rose and 11 single offspring.
A) RR PP × rr pp
B) RR Pp × rr pp
C) Rr PP × rr pp
D) Rr Pp × rr pp
E) Rr pp × rr pp
82. In a certain species of plant, flowers occur in three colors: blue, pink, and white. A purebreeding pink plant is mated with a pure-breeding white plant. All of the F1 are blue. When the blue F1 plants are selfed, the F2 occur in the ratio 9 blue:3 pink:4 white. How many gene pairs control the flower color phenotype?
83. In a certain species of plant, flowers occur in three colors: blue, pink, and white. A purebreeding pink plant is mated with a pure-breeding white plant. All of the F1 are blue. When the blue F1 plants are selfed, the F2 occur in the ratio 9 blue:3 pink:4 white. What is the name for this type of interaction?
A) Recessive epistasis
B) Dominant epistasis
C) Duplicate recessive epistasis
D) Duplicate dominant epistasis
E) Dominant and recessive epistasis
84. In a certain species of plant, flowers occur in three colors: blue, pink, and white. A purebreeding pink plant is mated with a pure-breeding white plant. All of the F1 are blue. When the blue F1 plants are selfed, the F2 occur in the ratio 9 blue:3 pink:4 white. The genotype that produces white is ww. The presence of one W allele allows pink or blue color to occur. The alleles at the hypostatic locus are B (blue) and b (pink). What is the genotype of the pink parent?
A) bb WW
B) bb Ww
C) Bb Ww
D) Bb ww
E) BB ww
85. In a certain species of plant, flowers occur in three colors: blue, pink, and white. A purebreeding pink plant is mated with a pure-breeding white plant. All of the F1 are blue. When the blue F1 plants are selfed, the F2 occur in the ratio 9 blue:3 pink:4 white. The genotype that produces white is ww. The presence of one W allele allows pink or blue color to occur. The alleles at the hypostatic locus are B (blue) and b (pink). What is the genotype of the white parent?
A) bb WW
B) bb Ww
C) Bb Ww
D) Bb ww
E) BB ww
86. In a certain species of plant, flowers occur in three colors: blue, pink, and white. A purebreeding pink plant is mated with a pure-breeding white plant. All of the F1 are blue. When the blue F1 plants are selfed, the F2 occur in the ratio 9 blue:3 pink:4 white. The genotype that produces white is ww. The presence of one W allele allows pink or blue color to occur. The alleles at the hypostatic locus are B (blue) and b (pink). What is the genotype of the F1 plants?
A) bb WW
B) bb Ww
C) Bb Ww
D) Bb ww
E) BB ww
87. The presence of a beard on some goats is determined by an autosomal gene that is dominant in males and recessive in females. Heterozygous males are bearded, while heterozygous females are beardless. What type of inheritance is exhibited by this trait?
A) Sex-linked
B) Sex-limited
C) Sex-influenced
D) Autosomal recessive
E) Autosomal dominant
88. In domestic chickens, some males display a plumage pattern called cock feathering. Other males and all females display a pattern called hen feathering. Cock feathering is an autosomal recessive trait that is exhibited in males only. What type of inheritance is exhibited by this trait?
A) Sex-linked
B) Sex-limited
C) Sex-influenced
D) Autosomal recessive
E) Autosomal dominant
89. In domestic chickens, some males display a plumage pattern called cock feathering. Other males and all females display a pattern called hen feathering. Cock feathering is an autosomal recessive trait that is exhibited in males only. Two birds heterozygous for cock feathering are mated. What are the phenotypes of the parents?
A) Male with cock feathering, female with hen feathering
B) Male with hen feathering, female with cock feathering
C) Male with cock feathering, female with cock feathering
D) Male with hen feathering, female with hen feathering
E) Cannot be determined from the information given
90. In domestic chickens, some males display a plumage pattern called cock feathering. Other males and all females display a pattern called hen feathering. Cock feathering is an autosomal recessive trait that is exhibited in males only. Two heterozygous birds are mated. What fraction of the total offspring is expected to exhibit cock feathering?
A) 0
B) 1/8
C) 1/4
D) 1/2
E) 3/4
91. In domestic chickens, some males display a plumage pattern called cock feathering. Other males and all females display a pattern called hen feathering. Cock feathering is an autosomal recessive trait that is exhibited in males only. Two heterozygous birds are mated. What fraction of the male offspring is expected to exhibit cock feathering?
A) 0
B) 1/8
C) 1/4
D) 1/2
E) 3/4
92. In domestic chickens, some males display a plumage pattern called cock feathering. Other males and all females display a pattern called hen feathering. Cock feathering is an autosomal recessive trait that is exhibited in males only. Two heterozygous birds are mated. What fraction of the female offspring is expected to exhibit cock feathering?
A) 0
B) 1/8
C) 1/4
D) 1/2
E) 3/4
93. Male-limited precocious puberty results from a rare autosomal allele (P) that is dominant over the allele for normal puberty (p) and is expressed only in males. A male and female that both went through normal puberty have two sons. The first son undergoes precocious puberty but the second undergoes normal puberty. What is the genotype of the mother?
A) PP
B) Pp
C) pp
D) PP or Pp
E) Pp or pp
94. Male-limited precocious puberty results from a rare autosomal allele (P) that is dominant over the allele for normal puberty (p) and is expressed only in males. A male and female that both went through normal puberty have two sons. The first son undergoes precocious puberty but the second undergoes normal puberty. What is the genotype of the father?
A) PP
B) Pp
C) pp
D) PP or Pp
E) Pp or pp
95. Which organelle in an animal cell, in addition to the nucleus, contains genes?
A) Lysosome
B) Ribosome
C) Mitochondrion
D) Golgi body
E) Vesicle
96. Which of the following is a characteristic exhibited by cytoplasmically inherited traits?
A) Present in both males and females
B) Usually inherited from one parent, typically the maternal parent
C) Reciprocal crosses give different results
D) Exhibit extensive phenotypic variation, even within a single family
E) All of the above answers are correct.
97. Leber hereditary optic neuropathy (LHON) is a human disease that exhibits cytoplasmic inheritance. It is characterized by rapid loss of vision in both eyes, resulting from the death of cells in the optic nerve. A teenager loses vision in both eyes and is later diagnosed with LHON. How did this individual MOST likely inherit the mutant DNA responsible for this condition?
A) A nuclear gene from the father
B) A nuclear gene from the mother
C) A mitochondrial gene from the father
D) A mitochondrial gene from the mother
E) Any of the answers is possible.
98. The bicoid mutation (bcd–) in fruit flies is inherited as a maternal effect recessive allele. What is the expected ratio of phenotypes in the offspring of a cross between a bcd+/bcd
female and a bcd+/bcd– male?
A) 1 normal:1 mutant
B) 3 normal:1 mutant
C) 3 mutant:1 normal
D) All normal
E) All mutant
99. The phenomenon in which a gene's expression is determined by its parental origin is called:
A) sex-influenced.
B) sex-limited.
C) genomic imprinting.
D) maternal effect.
E) paternal effect.
100. A deletion of a small region on the long arm of chromosome 15 causes a developmental disorder in children called Prader-Willi syndrome when the deletion is inherited from the father. However, the deletion of this same region of chromosome 15 can also be inherited from the mother, but this inheritance results in a completely different set of symptoms, called Angelman syndrome. What type of genetic phenomenon does this represent?
A) Sex-influenced
B) Genomic imprinting
C) Cytoplasmic inheritance
D) Maternal effect
E) Paternal effect
101. The Himalayan allele in rabbits produces dark fur at the extremities of the body on the nose, ears, and feet. The dark pigment develops, however, only when a rabbit is reared at a temperature of 25°C or lower; if a Himalayan rabbit is reared at 30°C, no dark patches develop. What does this exemplify?
A) Dominance
B) Discontinuous characteristic
C) Genetic imprinting
D) Phenocopy
E) Temperature-sensitive allele
102. The SRY gene is located on the Y chromosome. This single gene encodes a protein called a transcription factor which binds to DNA and stimulates the transcription of other genes that lead to the development of male sex characteristics, including physical, biochemical, and behavioral phenotypes. What concept in genetics best describes this example?
A) Dominance
B) Discontinuous characteristic
C) Polygenic characteristic
D) Phenocopy
E) Pleiotropy
103. Multi-factorial traits are influenced by _____ and _____.
A) dominance; codominance
B) epistasis; pleiotropy
C) age; sex
D) genetic imprinting; reduced penetrance
E) polygenes; environment
104. List five different sex determination systems and a representative organism for each.
105. Explain the genders of human and diploid Drosophila XXY individuals.
106. Explain the genders of human and diploid Drosophila XO individuals.
107. While doing summer field work on a remote Indonesian island, you discover a new genus of lizard closely related to komodo dragons. You attempt to discover what sex determination system it uses by performing a series of controlled crosses on the island, using an isolated pair of lizards. Initially, all your crosses yield only males (in significant numbers). As fall begins and you prepare to leave the island, you find that your last cross yielded only females (in significant numbers). Suggest a mode of sex determination that explains this data.
108. Predict the sexual phenotype of a person who is XY but whose Y chromosome carries a deletion of the SRY gene. Explain your prediction.
109. Some organisms have multiple X and Y chromosomes and even different numbers of X and Y chromosomes. You discovered such a species. Females have eight X chromosomes, whereas males have four X and two Y. Describe the X and Y constitution of the gametes produced by this species both male and female that allows these chromosome numbers to be stably maintained.
110. A man and a woman are trying to have children but are unsuccessful. The man's autosomes appear normal, but his sex chromosomes, shown in the following diagram, are not. The diagram also shows a normal male's sex chromosomes for reference. In two to three sentences, explain the man's situation, including the type of chromosome mutation he carries, the specific regions of specific chromosomes involved, and why he is male.
111. A man and a woman are trying to have children but are unsuccessful. The man's autosomes appear normal, but his sex chromosomes, shown in the following diagram, are not. The diagram also shows a normal male's sex chromosomes for reference. Can you tell if the mutation came from the man's mother or the man's father? Explain how you can tell.
112. List three dosage compensation strategies for equalizing the amount of sex chromosome gene products.
113. Explain how dosage balance is achieved between X-linked genes and autosomal genes in mammals.
114. You are trying to develop a new species of newt as an experimental model system. You know that in other species of newt, green (G) is dominant to brown (g) skin color and is determined by a sex-linked gene. You cross brown males to green females and see that in the F1 all the males are green and all the females are brown. Which is the heterogametic sex in your species of newt?
115. Compare and contrast the patterns of inheritance expected for Y-linked and X-linked recessive inheritance in humans.
116. Red–green color blindness is an X-linked recessive condition. Juliet has a bit of difficulty passing the red-green color distinction test when she tries to get her driver's license. Her husband is not color blind, and neither is her son, Henry, nor her daughter, Roxanne. Roxanne has a son who is color blind. What is Juliet's genotype for the colorblind allele? How would you explain her partial color blindness?
117. You cross a female rat with pink toe pads (T) and pointy ears (Xe) to a male rat with black toe pads (t) and round ears (XE). The t and e alleles are both recessive, and the ear-shaped gene is X-linked, whereas the toe pad color gene is autosomal. The F1 progeny all have pink toe pads. What is the genotype of parental generation? What is the genotype of the F1 progeny? If the F1 are crossed to produce F2 progeny, what proportion of the F2 will be black-padded, pointy-eared males?
118. Marsupials, like cats, achieve dosage compensation by X inactivation. You are working in a lab that has discovered a mutation on the X chromosome in marsupials in the same gene that causes the tortoiseshell fur color phenotype in cats. You cross an X+Y blackfurred male with an XOXO orange-furred female. You expect that the X+XO female progeny will have tortoiseshell fur (like cats). Surprisingly, you find that all the females (n = 25) have solid orange fur. Offer a hypothesis to explain these results and describe a genetic test to support your hypothesis.
119. A geneticist is examining a culture of fruit flies and discovers a single female with strange spots on her legs. The new mutation is named melanotic. When a female melanotic fly is crossed with a normal male, the following progeny are produced: 123 normal females, 125 melanotic females, and 124 normal males. In subsequent crosses with normal males, melanotic females are frequently obtained, but never any melanotic males. Provide a possible explanation for the inheritance of the melanotic mutation. (Hint: The cross produces twice as many female progeny as male progeny.)
120. How do incomplete and codominance differ?
121. Describe the differences between dominance, codominance, and incomplete dominance.
122. How do incomplete penetrance and variable expressivity differ?
123. How does epistasis differ from Mendel's principle of dominance?
124. What is a dominant epistatic gene?
125. A homozygous strain of corn that produces yellow kernels is crossed with another homozygous strain that produces purple kernels. When the F1 are interbred, 280 of the F2 are yellow and 70 are purple.
a. If kernel color is controlled by a single gene pair with yellow dominant to purple, what would be the expected ratio of yellow to purple in the F2?
b. Do the observed data differ significantly from that expected in (a)? Explain your answer.
c. Provide an alternative explanation for the inheritance of kernel color and evaluate it by comparing observed data to that expected from your alternative hypothesis.
126. A yeast geneticist isolates two different haploid mutant yeast strains, Strain A and Strain B, which cannot grow unless the amino acid leucine is added to the growth media. Wild-type yeast strains can make their own leucine and do not require that it be added to the growth media. The geneticist discovers that each mutant yeast strain contains a single recessive mutation that leads to the observed leucine-requiring phenotype. When she crosses the two mutant strains together, she observes that the resulting diploid can grow without leucine added to the growth media. Explain the allelic relationship between the mutations in these two strains.
127. Discuss the difference between “cytoplasmic inheritance” and “genetic maternal effect.”
128. In some plant species, a single pair of alleles is involved in both flower color and stem color. For example, a plant with red flowers may also have red stems, whereas whiteflowered varieties of the same species have green stems. How would you explain this observation?
129. Two mice of the same species have different ear shapes. You find that one mouse, having normal shaped ears, was caught in a field in Kenya. The other mouse, with curled ears, was caught in the frozen tundra of Greenland. You determined that both mice have identical genotypes at the known gene loci controlling ear shape. How would you explain the differences in ear shape?
130. Two mice of the same species have different ear shapes. You find that one mouse, having normal shaped ears, was caught in a field in Kenya. The other mouse, with curled ears, was caught in the frozen tundra of Greenland. You determined that both mice have identical genotypes at the known gene loci controlling ear shape. How could you test the relative importance of environmental and genetic factors in determing ear shape?
This question as originally written is going to be confusing to some students because the question states that the two types of mice are identical at the genes that control ear shape so the answer appears obvious that the difference must be due to one or more environmental factors. Why would more testing be needed? I changed the wording of the question to “known gene loci” so that there may be genetic differences at other unknown loci and the answer would involve more analysis. To be consistent I made this change in question #129 also
131. Explain the differences between incomplete dominance and continuous variation.
132. You observe continuous variation in tail length in a wild population of rats. How would you determine whether this variation is an example of variable expressivity or polygenic inheritance?
133. You are studying a coat color gene (B, brown) in Mexican bats. You isolated a recessive allele (b) that causes yellow coat color, but you suspect that the phenotype may be sensitive to environmental conditions. To test your hypothesis, you examine the segregation ratio of phenotypes in F1 progeny from a cross between two heterozygotes. You do this once at normal laboratory temperatures (28°C) and once at temperatures closer to their native habitat (34°C) and record the following data:
a. What ratio do you expect in each experiment if temperature does not affect the phenotype?
b. What test can you use to determine if the ratio you observed is significantly different from the expected ratio?
c. Using that statistical test, is either observed ratio more different from the expected ratio than one would expect from chance alone? If so, suggest a biological explanation.
134. List at least four phenomena that can alter expected Mendelian phenotypic ratios in genetic crosses.
135. Cloning is a procedure by which exact genetic duplicates are made. Using cloning techniques, you produced 10 cloned cows. However, the fur color of each of the calves looks very different from one another. Explain why this might have occurred.
136. Explain how a phenotype like height in a tree can be due to the influence of both genes and environment.
Answer Key
104. (1) XX-XO; grasshoppers
(2) XX-XY; humans
(3) ZZ-ZW; birds
(4) X:A; Drosophila
(5) Environmental mollusks; many turtles, crocodiles, and alligators
105. An XXY human is male. The SRY on the Y chromosome determines maleness, in spite of the two X chromosomes. An XXY Drosophila is female. Drosophila is diploid, so there are two of each autosome. If there are two X chromosomes, the X:A ratio is 1.0, which is a female.
106. An XO human is female. In the absence of SRY from a Y chromosome, the person will develop as a female, in spite of having only one X chromosome. An XO Drosophila is male. Drosophila is normally diploid, so there are two of each autosome. If there is a single X chromosome, the X:A ratio is one-half, or 0.5, which is a male.
107. The crosses yielded all males or all females from the same parents. Male and female progeny were correlated with climatic conditions (summer versus fall). Environmental sex determination that is dependent on temperature is a likely explanation.
108. The SRY gene is the primary determinant of maleness in humans. If it is deleted, the gonads will not be induced to differentiate as testes, and the individual would likely follow the female developmental pathway. (Actually, this condition exists as Swyer syndrome. It turns out that the gonads do not develop into functional ovaries, indicating that genes other than the SRY also have roles in sexual differentiation. However, these individuals develop as apparently normal females until the lack of female hormones from the ovaries causes puberty not to be induced without hormone therapy.)
109. Females must produce one type of gamete, with four X chromosomes, whereas males produce two types of gametes, with four X chromosomes or with two Y chromosomes. Fertilization using a four X male gamete produces an eight X female, whereas fertilization with a two Y gamete produces a male.
110. He has an X with a translocation, meaning part of one chromosome has been moved to another. The translocated part is from Y and carries SRY, which determines maleness. He is missing the rest of Y, including the genes required for male fertility.
111. He inherited the translocation chromosome from his father because his mother could not have carried the SRY-containing chromosome.
112. (1) Inactivation of one sex chromosome in the homogametic sex
(2) Halving the activity of genes on both sex chromosomes in the homogametic sex
(3) Increasing the activity of genes on the sex chromosome in the heterogametic sex
113. In both males and females only one X chromosome is fully active, potentially creating an imbalance in gene expression between genes on the X chromosome and autosomes. The imbalance is prevented by upregulation of genes on the X chromosome.
114. Because the F1 females have the recessive brown phenotype, they must be hemizygous (i.e., they inherited a brown allele from their father and no allele from their mother). Therefore, the females of this species are the heterogametic sex. We use the ZZ-ZW nomenclature for species with heterogametic females. Therefore, your F1 females and males are ZgW and ZGZg, respectively.
115. Y-linked inheritance is easy to recognize because the trait is passed from a father to all of his sons and none of his daughters and continues to pass from fathers to sons in every generation. The trait does not affect females. An X-linked recessive trait is more common in males but can affect females as well. A son inherits the trait from his mother, who is typically a phenotypically normal carrier. Often, the trait will also be present in the mother's father or brothers or other male relatives. An affected female normally occurs from an affected father and a carrier mother.
116. Juliet is heterozygous. So every diploid cell in her eyes contains both X chromosomes, one with the defective red-green color vision allele, and one with the normal color vision allele. She is not completely color blind because some of her cells are inactivating the X chromosome with the recessive, defective red-green color vision allele and expressing the normal color vision allele, allowing her to see color. She has some difficulty with color vision because some of her cells are inactivating the X with the normal red-green color vision allele and expressing red-green color blindness. Her eyes are actually a mosaic of cells with two different phenotypes.
117. The parental generation was T/T; Xe/Xe and t/t; XE/Y. The F1 were T/t;XEXe and T/t;XeY. Black-padded, pointy-eared males are tt and Xe/Xe. One-quarter of the F2 progeny will be t/t, 1 4 will be XeY. Therefore, 1/16 of the progeny will be blackpadded, pointy-eared males.
118. It appears that the X+ chromosome from the male was inactivated in every female offspring. So perhaps in marsupials, unlike in cats, X inactivation is not random. Instead, in marsupials only the paternal X chromosome is inactivated. If this is true (and, in fact, it is), then it may be tested genetically. Crossing an orange-furred XOY male to a X+X+ black-furred female should produce only black-furred female progeny even though their genotype is X+XO, the same as the orange-furred female progeny from the first cross.
119. These observations can be explained by a single X-linked locus with two segregating alleles. The skewed sex ratio (2 female:1 male) in the F2 suggests a recessive lethal allele on the X chromosome that kills males that carry the lethal allele in one copy on their one X chromosome. The phenotype of the female parent also suggests that the allele is dominant for the melanotic trait. We will represent the mutant allele as M and the normal allele as +.
120. Incompletely dominant traits show an intermediate phenotype in the heterozygote, while codominant traits show both phenotypes in the heterozygote (e.g., AB alleles of blood type).
121. (1) Dominance is the condition in which one allele of a gene pair completely masks or inhibits phenotypic expression of the other allele.
(2) Codominance is the condition in which the complete expression of both alleles of a given gene pair is observed in heterozygotes; that is, the expression of neither allele influences the expression of the other.
(3) Incomplete (or partial) dominance is the condition in which one allele only partially inhibits the expression of the other allele in the phenotype. Heterozygotes exhibit phenotypes that are intermediate between those of the two homozygotes.
122. (1) If some individuals in a population don't express a trait, even though they have the corresponding genotype, the trait is said to exhibit incomplete penetrance in that population. When describing penetrance, think of populations. Polydactyly (extra fingers and toes) exhibits incomplete penetrance in human populations.
(2) A trait exhibiting variable expressivity is not expressed at the same degree among all individuals expressing it. Male pattern baldness in humans is an example of a trait that exhibits variable expressivity.
Note: For incomplete penetrance, not everyone with the genotype will express the phenotype. For variable expressivity, everyone with the genotype expresses the phenotype to some degree. Of course, some traits may (and often do) exhibit both incomplete penetrance and variable expressivity.
123. Phenotypic expression is often the result of products produced by multi-step metabolic pathways involving several different genes; each gene encodes an enzyme that regulates a specific biochemical step or event. Epistasis refers to the interaction among two or more genes that control a common pathway. For example, a mutation in any single gene contributing to a metabolic pathway can affect the expression of other genes in the pathway, and, of course, the final phenotype, depending on which biochemical step that gene controls.
Epistasis thus involves interaction among alleles located at different gene loci. This is in contrast to dominance, which involves interaction between alleles located at the same gene locus.
124. A dominant allele that, if present, determines the phenotype of a given trait regardless of which alleles at other loci are present
125. a. Because there are only two progeny classes, the simplest explanation is monohybrid inheritance with expected ratio of 3 yellow:1 purple.
b. A chi-square test of the observed numbers using an expected 3:1 ratio suggests rejection of this hypothesis (2 = 4.7, 0.025 < p < 0.05).
c. So, the next hypothesis to test is dihybrid inheritance. However, the progeny clearly don't segregate classic Mendelian dihybrid ratio (9:3:3:1) there are only two phenotypic classes. Therefore, some epistasis is likely. There are three epistatic ratios with two phenotypic classes to test: 9:7, 15:1, and Dividing each phenotypic class by 16 suggests that the 13:3 ratio is the closest. The 13:3 ratio is standard for the kind of epistasis called “dominant and recessive” interaction. In this particular kind of epistasis, only two F2 phenotypes are generated because a dominant genotype (e.g., A_) present at one locus recessive genotype at the other locus (bb) produce identical phenotypes in a 13:3 ratio (e.g., A _ B_, and aa bb produce one phenotype, and aa B_ produces another phenotype). To further substantiate is the correct ratio, a chi-square test can be done. In fact, among the alternatives only the 13:3 ratio accompanying genetic hypothesis should not be rejected (2 = 0.35 with 0.9 < p < 0.5).
126. The mutations in strains A and B are NOT allelic because complementation was observed. Strain A contains a mutation at gene A, which is recessive (a), and strain B contains a mutation at a separate genetic locus, gene B, which is also recessive. Strain A contains a wild-type B gene and strain B contains a wild-type A gene. These wild-type genes complement the corresponding mutant alleles in the diploid.
127. (1) In cytoplasmic inheritance, the genes controlling a given trait are inherited exclusively from the mother (through cytoplasmic organelles such as mitochondria) and can be expressed in both male and female progeny.
(2) In the genetic maternal effect, each individual's phenotype is determined by the genotype of the mother. Typically, the offspring's phenotype is determined by mRNA or protein factors loaded into the oocyte and encoded by the mother's genome. So while genes related to the trait are inherited from both parents (not so for the cytoplasmic inheritance), in a given generation, phenotype is determined exclusively by the mother's, not the offspring's, genotype.
128. This phenomenon, called pleiotropy, is the condition where a single gene affects multiple, apparently unrelated, phenotypic traits. In many other cases of pleiotropy, a single gene affects more than two phenotypic traits. For example, a mutant white-eye gene in Drosophila (fruit fly) also affects the structure and color of internal organs, causes reduced fertility, and decreases life expectancy. Another example involves sickle-cell anemia in humans (caused by a single nucleotide change in a hemoglobin gene), which has adverse effects on different organs and tissues.
129. Because both mice have the same genotype at the relevant loci controlling ear shape, there is most likely an effect of environment on phenotype. Because Kenya and Greenland have quite different climates (but also different food sources, humidity, sunlight intensities, etc.), it is possible that the very different temperature ranges within each region resulted in differential expression of identical genotypes for ear shape in each mouse for example, differential expression of temperature-sensitive allele(s) involved in ear development. However, the phenotypic differences could also be a result of differences in diet, light conditions, exposure to chemicals, nutrition, or a range of other nongenetic factors.
130. Develop true-breeding strains of mice for normal and curled ears in their original habitats. Then, rear and observe one pair of each (one control pair and one test pair) true-breeding strain in each habitat in Kenya and Greenland, mating them with only their fellow Greenland or Kenyan siblings and raising all offspring under the same conditions as originally present for the native species (e.g., the original location outside). If the phenotypes of the experimental and control offspring reared in Kenya are all the same (normal) and the phenotypes of both sets of mice in Greenland are all the same (curled), then the two-ear phenotypes are caused by environment. For example, they may result from temperature-sensitive alleles. Note that strict temperature sensitivity could be tested under laboratory conditions without the need to transport mice to different countries. If the phenotypes of the two sets of mice at any one geographic location are not all the same, then there is a genetic component involved in ear shape. Note that there may be both genetic and environmental components.
131. (1) In incomplete dominance, there will be three distinct phenotypes because the phenotype of a heterozygote is intermediate in appearance between the phenotypes of the two homozygotes. Incomplete dominance involves a single gene locus.
(2) Continuous variation refers to phenotypic variation exhibited by quantitative traits that are overlapping and distributed from one extreme to another. The continuous variation of quantitative traits is usually controlled by several genes whose alleles have an additive effect on the phenotype.
132. Take male and female rats from each phenotypic extreme (shortest and longest tails). Interbreed short with short and long with long under controlled laboratory conditions for several generations. If this is polygenic inheritance, then you will be able to develop different homozygous lines for short and long tails. But, if after several generations each line continues to produce progeny classes exhibiting significant variance in tail length, you could assume variable expressivity is the primary basis for the variation because the genotypes for each extreme line are (theoretically) homozygous and isogenic. Therefore, variances in tail length observed within each line cannot be the result of variable polygenic genotypes.
133. a. This is a simple monohybrid cross with brown dominant to yellow, so expect 3 brown:1 yellow.
b. The chi-square test
c. The chi-square test for the treatment at 34°C yields a value of 2 = 10.67, indicating a significant difference from the expected ratio of 3:1. This suggests that elevated temperatures reduce the penetrance of the yellow phenotype. The chi-square test for the treatment at 28°C yields a value of 0.08, indicating that these data fit the expected ratio of 3:1.
134.
(2)
(3)
135. A given phenotype arises from a genotype that develops within a particular environment. How the phenotype develops is determined by the effects of genes and environmental factors, and the balance between these influences varies from character to character. Since we are told that the calves are genetically identical, there must be environmental variation that explains the phenotypic differences. Even within the “constant” environment of a cow's womb, there is environmental variation!
136. The height reached by a tree at maturity is a phenotype that is strongly influenced by environmental factors, such as the availability of water, sunlight, and nutrients. Nevertheless, the tree's genotype still imposes some limits on its height: an oak tree will never grow to be 300 meters tall no matter how much sunlight, water, and fertilizer are provided.