Exam 1 questions

• In Mendel's peas, round seeds are dominant to wrinkled. A pure-breeding round seed plant is crossed with a pure-breeding wrinkled seed plant. All of the offspring are round. If one of these round offspring is crossed with a wrinkled plant, what will be the expected proportion of plants with wrinkled seeds in the next generation?

  a. 0%

  b. 25%

  c. 50%

  d. 75%

  e. 100%

  c. 50%
• Which of the following crosses would produce a 1:1 ratio of phenotypes in the next generation?

  a. AA x AA

  b. AA x aa

  c. Aa x Aa

  d. Aa x aa

  e. aa x aa

  d. Aa x aa
• In poodles, black is dominant to white. A black poodle is crossed with a white poodle. In a litter of four, all of the puppies are black. What is the best conclusion?

  a. The black poodle is definitely homozygous.

  b. The black poodle is probably homozygous.

  c. The black poodle is definitely heterozygous.

  d. The black poodle is probably heterozygous.


  b. The black poodle is probably homozygous.
• 4. In a cross between AaBbCc and AaBbcc, what proportion of the offspring would be expected to be A_BbCc? (A_ means AA or Aa.)

  a. 3/64

  b. 3/32

  c. 3/16

  d. 3/8

  e. 3/4

  c. 3/16
• Assume that attached earlobes are an autosomal recessive trait with 60% penetrance. If one parent is heterozygous (Aa) and the other homozygous recessive (aa) for the attached earlobe gene, what is the probability that their first child will have attached earlobes?

  a. 50%

  b. 18.75%

  c. 25%

  d. 30%

  e. 70%

  d. 30%See an expert-written answer!We have an expert-written solution to this problem!
• 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.

  d. The allele that causes achondroplasia is a recessive lethal allele.
• 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%

  a. 0%
• If a color-blind female has children, what do we know about all of her sons?

  a. Half of the sons will be color-blind.

  b. All sons will be color blind.

  c. All sons will have normal color vision.

  d. She would not be able to have a son.

  e. It is impossible to know without knowing the father's genotype.

  b. All sons will be color blind.
• Males and females express the same amount of X-chromosome genes despite the fact that females have more X chromosomes. Animals approach this problem in different ways. Which of the following is an example?

  a. Female Caenorhabditis elegansexpress X-chromosome genes at 50%.

  b. Male fruit flies double the expression of their X-chromosome genes.

  c. Female humans inactivate one of their X chromosomes.

  d. All of the above.

  e. None of the above.

  d. All of the above.
• Aneuploids of which chromosomes are less likely to result in a live birth?

  a. larger autosomes

  b. small autosomes

  c. sex chromosomes

  d. all of the above

  e. none of the above

  a. larger autosomes
• The following shows a boy's karyotype.

  How many Barr bodies (condensed X chromosomes) would you predict in his cells?

  a. One per cell

  b. Two per cell

  c. Three per cell

  d. None

  a. One per cell

  
• A phenotypically normal woman has phenotypically normal parents but she has a brother who has sickle cell disease (CSD) caused by a recessive mutant allele. What is the probability that the woman is heterozygous for the CSD allele?

  a. 1/4

  b. 1/2

  c. 3/4

  d. 2/3

  e. 1/3

  d. 2/3
• Freckles are caused by a dominant allele. A man has freckles but one of his parents does not have freckles. The man has fathered a child with a woman that does not have freckles. What is the probability that their child has freckles?

  a. 1/4

  b. 1/3

  c. 1/2

  d. 2/3

  e. 3/4

  c. 1/2
• A diploid somatic cell from a cat has a total of 38 chromosomes (2n=38). As in humans, sex chromosomes determine sex: XX in females and XY in males. What is the total number of chromosomes present in the cell during metaphase I of meiosis?

  a. 19

  b. 38

  c. 76

  d. 114

  e. 133

  b. 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

  b. incomplete dominance
• True/False: The phenotype of a dominant allele is never seen in the F1progeny of a monohybrid cross.

  False
• True/False: A Mendelian dihybrid cross involves one gene and two different alleles.

  False
• True/False: The principle of independent assortment can be demonstrated using a monohybrid cross.

  False
• True/False: The genotype is the physical appearance of a trait.

  False
• True/False: A testcross can be used to determine whether an individual is homozygous or heterozygous.

  True
• True/False: Mendel's experiments involved both discrete and continuous genetic traits.

  False
• True/False: Temperature during embryonic development determines sex in reptiles like turtles and alligators.


  True
• True/False: In humans SRY is the male determining gene.

  True
• True/False: Female mammals that are heterozygous for an X-linked gene have patches of cells that express one allele and patches of cells that express the other.

  True
• True/False: The cellular reason for X-inactivation is to reduce the amount of nondisjunction during meiosis.

  False
• Short Answer: In deer mice, red eyes (r) is recessive to normal black eyes (R). Two mice with black eyes are crossed. They produce two offspring, one with redeyes and one with black eyes. Give the genotypes of parents and offspring of this cross.

  For red eyes (recessive) to be expressed in the progeny (i.e., to segregate in the progeny), both parents must be heterozygotes (Rr).

  Rr ×Rr ---> 1 (RR), 2 (Rr), 1 (rr) = 3:1 (black eyes:red eyes).

  
  

  Note that even with only two progeny produced, you have determined the genotypes of the parents because you observe segregation of the recessive allele (red eyes).
• Short Answer: The cells illustrated below belong to a species with a diploid chromosome number of four. Each of the cells below is in which stage of mitosis or meiosis?

  
• Short Answer: Explain why mitosis does not produce genetic variation and how meiosis leads to the production of tremendous genetic variation.


  Mitosis produces cells that are genetically identical to the parent cell. Meiosis includes two distinct processes that contribute to the generation of genetic variation: crossing over shuffles alleles on the same chromosome into new combinations, whereas the random distribution of maternal and paternal chromosomes shuffles alleles on different chromosomes into new combinations.
• Short Answer: 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 tortoise shell fur color phenotype in cats. You cross an X+Y black-furred 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.


  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.
• Short Answer: A homozygous strain of corn that produces yellow kernels is crossed with another homozygous strain that produces purple kernels. When the F1are interbred, 280 of the F2 are yellow and 70 are purple.

  
  

  a. If kernel color is controlled by a singlegene pair with yellow dominant to purple, what would by the expected ratio of yellow to purple in the F2?

  Because there are only two progeny classes, the simplest explanation is monohybrid inheritance with an expected ratio of 3 yellow:1 purple.
• b. Do the observed data differ significantly from that expected in (a)? Explain your answer.

  A chi-square test of the observed numbers using an expected 3:1 ratio suggests observed data differ significantly from that expected (Χ2= 4.7, 0.025 < p < 0.05).