Review the following concepts if needed:
Review Questions
- The following crossover frequencies were noted via experimentation for a set of five genes on a single chromosome:
Pick the answer that most likely represents the relative positions of the five genes.
- Imagine that in squirrels, gray color (G) is dominant over black color (g). A black squirrel has the genotype gg. Crossing a gray squirrel with which of the following would let you know with the most certainty the genotype of the gray squirrel?
- GG
- Gg
- gg
- Cannot be determined from the information given
- From a cross of AABbCC with AaBbCc, what is the probability that the offspring will display a genotype of AaBbCc?
- 1/2
- 1/3
- 1/4
- 1/8
- 1/16
Use the following pedigree of an autosomal recessive condition for questions 4–6.
- What is the genotype of person A?
- Bb
- BB
- bb
Cannot be determined from the given information
- What is the most likely genotype of person B?
- Bb
- BB
- bb
- Cannot be determined from the information given
- What is the probability that persons C and D would have a child with the condition?
- 1/ 2
- 1/4
- 1/6
- 1/8
- 1/10
- Which of the following disorders is X-linked?
- Tay-Sachs disease
- Cystic fibrosis
- Hemophilia
- Albinism
- Huntington disease
- A court case is trying to determine the father of a particular baby. The mother has type O blood, and the baby has type B blood. Which of the following blood types would mean that the man was definitely not the father of the baby?
- B and A
- AB and A
- O and B
- O and A
- None can prove conclusively
- Assume that gray squirrel color results from a dominant allele G. The father squirrel is black, the mother squirrel is gray, and their first baby is black. What is the probability that their second baby is also black?
- 1.00
- 0.75
- 0.50
- 0.25
- 0.00
- Imagine that tulips are either yellow or white. You start growing tulips and find out that if you want to get yellow tulips, then at least one of the parents must be yellow. Which color is dominant?
- White
- Yellow
- Neither; it is some form of intermediate inheritance
- Cannot be determined from the given Information
- Suppose that 200 red snapdragons were mated with 200 white snapdragons and they produced only pink snapdragons. The mating of two pink snapdragons would most likely result in offspring that are
- 50 percent pink, 25 percent red, 25 percent white
- 100 percent pink
- 25 percent pink, 50 percent red, 25 percent white
- 75 percent red, 25 percent white
- 100 percent red
- Which of the following represents the number of possible gametes produced from a genotype of RrBBCcDDEe?
- 2
- 4
- 8
- 16
- 32
- Which of the following diseases is not caused by trisomy nondisjunction?
- Down syndrome
- Klinefelter syndrome
- Turner syndrome
- Patau syndrome
- Edward syndrome
- The pedigree below is most likely a pedigree of a condition of which type of inheritance?
- Autosomal dominant
- Autosomal recessive
- Sex-linked dominant
- Sex-linked recessive
- A gene present only on the Y chromosome (holandric)
Answers and Explanations
- A—The crossover frequencies are an indication of the distance between the different genes on a chromosome. The farther apart they are, the greater chance there is that they will cross over during prophase I of meiosis. You are first told that A and B cross over with a frequency of 35 percent, so imagine that they are 35 units apart on a chromosome map.
I then tell you that B and C have a frequency of 15 percent. They are 15 units apart on the map, but you cannot yet be sure what side of gene A that C is on. Gene A and C cross with 20 percent frequency. This means that gene C must be in between A and B.
Gene A crosses over with D 10 percent of the time, and D crosses with B 25 percent of the time; therefore, D must also be in between A and B. It is closer to A than it is to B. You can use this knowledge to eliminate answer choices B and C.
Gene A crosses over with E with a frequency of 5 percent. You do not know which side of A gene E is on until you know its crossover frequency with B. Because the question tells you that it has a 40 percent frequency with B, you know that it must be on the left of A. This completes your map, leaving A as the correct answer.
- C—This is a test cross. To determine the genotype of an individual showing the dominant phenotype, you cross that individual with a homozygous recessive individual for the same trait. If they have no offspring with the recessive phenotype, then the individual displaying the dominant phenotype is most likely FF. If approximately one-half of the offspring have the recessive phenotype, you know the individual has the genotype Ff.
- D—The Punnett square shown below shows all the possible gamete combinations from this cross. Two-sixteenths or one-eighth of the possible gametes will be AaBbCc. A quick way to determine the number of possible gametes that an individual can produce given a certain genotype is to use the formula 2^{n}. For example, an individual who is AABbCc can have 2^{2} = 4 possible gametes because Bb and Cc are heterozygous.
- A—Person A must have genotype Bb because he has some children that have the recessive condition and some that do not. Because his wife is pure recessive, she can contribute only a b. The father must therefore be the one who contributes the B to the child who does not have the condition, and the second b to the one with the condition.
- B—Person B most likely has a genotype of BB. Because he does not have the condition, we know that his genotype is either BB or Bb. If it were Bb, then when crossed with his wife who has a genotype of bb, 50 percent of his children would be expected to have the recessive condition. None of the children have the condition, which leads you to believe that he is most likely BB. (This test is, of course, not 100 percent accurate. Answer choice B is not certain, but is the most probable conclusion.)
- C—We know that neither parent in the question has the recessive condition. We therefore need to calculate the probability that each of them is Bb. The probability that person C is Bb is 1. Because his mother has the condition, she must pass a b to him during gamete formation. So the only possible genotypes he can have are Bb and bb. Since he does not have the condition, he must be Bb with a probability of 1. The probability that person D is Bb is 0.67. Neither of her parents has the condition, but she has a brother who is bb. This means that each of her parents must be a carrier for the condition (Bb). You know that this woman is not rr, because she does not have the condition. As a result, there are only three possible genotypes from the cross remaining. Two of these three are Bb, giving her a probability of 2/3, or 0.67, of being Bb. The probability that both person C and person D are Bb is (1) × (0.67) = (0.67). Now it is necessary to calculate the probability that two Bb parents will produce a kid who is bb. The Punnett square says that there is a 0.25 chance of this result. To calculate the probability that they will have a child with the recessive condition, you multiply the probability that they are both Bb (0.67) times the probability that two individuals Bb will produce a bb child (0.25). Thus, the probability of an affected child being produced from these two parents is 1/6.
- C—Hemophilia is an X-linked condition. An XY male with hemophilia gets his Y chromosome from his father, and his X chromosome from his mother. All that is needed for the hemophilia condition to occur is a copy of the defective recessive allele from his mother.
- D—Types O and A would prove that he was not the father of this particular child. If the mother has type O blood, this means that her genotype is ii and she must pass along an i allele to her child. The baby has type B blood, and her genotype could be I^{B}i or I^{B}I^{B}. Since the mother must give an i, then the baby's genotype must be I^{B}i. It follows that the father must provide the I^{B} allele to the baby to complete the known genotype. If he is type O, he won't have an I^{B} to pass along since his genotype would be ii. This would also be the case if he were type A, because his genotype would be either I^{A}I^{A} or I^{A}i. Therefore, those two blood types would prove that he is not the father of this child.
- C—To figure out this problem, you need to know the genotype of the mother. The father is black, meaning that his genotype is gg. The
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