Explanations
1. E
In meiosis, sister chromatids separate during anaphase
II. During prophase I, sister chromatids are linked at their centromeres.
At metaphase I, each pair of sister chromatids lines up in the middle
of the cell together with its homologous pair of sister chromatids,
and only homologous chromosomes separate during anaphase I. During
metaphase II, each pair of sister chromatids again lines up along
the metaphase plate, but they do not separate until anaphase II.
2. D
Meiosis divides a single diploid cell into four haploid
cells. In the case of oogenesis, only one of those daughter cells
goes on to become a fully developed egg cell, leaving three “polar
bodies” to disintegrate; the original product, however, is still
four cells.
3. B
The first (F1)
generation of offspring from a hybrid cross will all look the same.
A hybrid cross involves the mating of a homozygous dominant (RR)
individual with a homozygous recessive (rr) individual, so the parents
do not look the same. In the F1 generation,
all offspring will have genotype Rr and will all show the dominant
trait. Other generations will contain some individuals with the
dominant phenotype (RR or Rr) and some with the recessive phenotype
(rr) and so will not all look the same.
4. D
Approximately 75% of the F2 generation
will exhibit the dominant phenotype. Genotypically, there will be
three types of offspring: 25% homozygous dominant, 50% heterozygous
dominant, and 25% homozygous recessive. Only the homozygous recessive
will show the recessive phenotype—therefore, a total of 75% will
have the dominant phenotype.
5. E
Because these genes are on separate chromosomes and
therefore act independently, we can use a Punnett square to work
out the results of this dihybrid cross.
For every sixteen offspring in the F2 generation,
approximately nine will exhibit the dominant phenotype of both genes
(genotype AABB, AABb, AaBB, or AaBb), three will be dominant on
gene A only (AAbb or Aabb); three will be dominant on gene B only
(aaBB or aaBb), and one will show a recessive phenotype on both
genes (aabb). The result is a phenotype ratio of 9:3:3:1.
6. D
In order to determine the genotype of a dominant-appearing
individual, the scientist would have to perform a test cross (also
known as a back cross) using a fully recessive individual. In this
case, that would be a short and green plant.
7. D
Given the fact that the expected 3:1 dominant-to-recessive
phenotype showed up on the same F2 offspring
plants, it looks like these genes are linked and were not acting
independently of each other. From this information, you can predict
that the two genes controlling these traits are on the same chromosome.
Since crossing-over limits the effects of linkage—and crossing-over
occurs more often when two genes are far apart on the same chromosome—
you can further determine that the two genes must be near each other
on the chromosome.
8. C
The genotype of a person who has an autosomal recessive
disease must be homozygous recessive. Therefore, if two parents
have the disease, 100% (not 50 or 75%) of their children should
get the disease, since neither parent has dominant alleles to contribute.
9. C
50% of the couple’s daughters would be expected to
have hemophilia, an X-linked recessive disease. Using a Punnett
square to do the cross of the father (X hemophilia/Y) and the mother
(X normal/X hemophilia), you should find that half of the female
children will be homozygous for the X hemophilia gene and have the
disease, while half will be heterozygous and be carriers of the
disease.
10. C
Although the father is color blind, the gene for this
condition is located on the X chromosome. The father could not have
given his son the condition, because he only contributes a Y chromosome
to his son; therefore, the gene that causes the disease must have
been inherited from the mother.
