Question
A model showing the cells in anaphase I and anaphase II of meiosis during a nondisjunction event is shown in Figure 1
Figure 1. Model of a nondisjunction event
Which of the following best predicts the effect of the chromosomal segregation error shown in Figure 1?
A. All of the resulting gametes will have an extra chromosome.
B. All of the resulting gametes will be missing a chromosome.
C. Half of the resulting gametes will have an extra chromosome and the other half will be missing a chromosome.
D. Half of the resulting gametes will have the correct number of chromosomes, and the other half will have an incorrect number of chromosomes.
▶️Answer/Explanation
Ans: D
In this instance, nondisjunction of sister chromatids during meiosis II will produce two gametes with the correct number of chromosomes (n). The remaining two gametes that are produced will be n+1 and n−1, respectively.
Question
Sex chromosomes determine the phenotype of sex in humans. Embryos containing XX chromosomes develop into females, and embryos containing XY chromosomes develop into males. The sex chromosomes separate during meiosis, going to different gamete cells.
A woman is heterozygous for the X-linked recessive trait of hemophilia A. Her sex chromosomes can be designated as \(X^HX^h\). During meiosis the chromosomes separate as shown in Figure 1.
Figure 1. Transmission pattern for sex chromosomes of a woman heterozygous for hemophilia A
into gametes
If the woman and a man with normal clotting function have children, what is the probability of their children exhibiting hemophilia A
▶️Answer/Explanation
Ans: B
Half of the sons would inherit the \(X^H\) allele from the woman and a Y chromosome from the man. Without a \(X^H\) allele, these sons would develop hemophilia A.
Question
Huntington’s disease has been traced to the number of CAG repeats in the HTT gene, which is located on chromosome 4. The phenotypic influence of individual alleles with different numbers of repeats is shown in Table 1.
Which of the following is most likely the immediate cause of the first appearance of Huntington’s disease in a person?
A. A point mutation occurs in the HTT gene.
B. The first appearance of the CAG repeat occurs in the HTT gene.
C. An allele with more than 39 CAG repeats was inherited by the affected person.
D. The person inherited two alleles that each contained 20 CAG repeats.
▶️Answer/Explanation
Ans: C
At some point, all alleles for the HTT gene contain less than 39 CAG repeats. Only after this threshold is crossed will a person inherit an allele that will cause symptoms of Huntington’s disease.
Question
Figure 1 illustrates the X and Y chromosomes during meiosis I and meiosis II of normal spermatogenesis in a mammal species.
Figure 1. X and Y chromosomes during meiosis I and meiosis II
If the normal spermatogenesis is disrupted, the gametes can have different chromosomes than expected. Which of the following is the most likely cause of one of the four gametes having two X chromosomes and one having neither an X nor a Y chromosome?
A. Nondisjunction of the chromosomes during meiosis I
B. Nondisjunction of both the X and Y chromosomes during meiosis II
C. Nondisjunction of the Y chromosome during meiosis II
D. Nondisjunction of the X chromosome during meiosis II
▶️Answer/Explanation
Ans: D
A nondisjunction event of the X chromosome in meiosis II would result in both of the X chromatids going to one gamete and no X chromatids going to the other gamete.