Download BIOS 1700 Dr. Tanda 15 November 2016 Week 13, Session 2 1. T/F

BIOS 1700
Dr. Tanda
15 November 2016
Week 13, Session 2
1. T/F Alleles are a different form of a given gene.
2. T/F Phenotypes are visible or measureable.
3. T/F Mendel’s firs law: independent assortment
4. T/F A test cross is a diagnostic cross of a heterozygote and a recessive homozygote.
5.T/F The PiZ allele makes a1AT gene less active.
6. Based on the bands observed for the VNTR (variable number of tandem repeats)
polymorphism in the accompanying gel diagram, which of the individuals M, H, K, or L could
not be siblings of individual X?
7. While doing a pedigree analysis of a European royal family, you notice a disease in a female
child of two healthy parents. There are also some cousins with the same genetic disease. If this
disease is controlled by a single gene, then the MOST likely explanation for these observations is
that the disease is:
a. a spontaneous mutation.
b. recessive and only seen in homozygous recessive individuals such as the daughter.
c. dominant and only seen in homozygous dominant individuals such as the daughter.
d. only seen in heterozygous individuals such as the daughter.
e. dominant and is seen in homozygous dominant or heterozygous individuals.
8. Sickle-cell anemia is due to a mutation of the b-globin gene on Chromosome 11. The wild-type
allele is designated as the A allele. One of mutant alleles of the gene is called the S allele. This
allele has one base (or nucleotide) substitution, A  T, and this substitution also changes its
amino acid sequence from glutamic acid to valine. Therefore, this substitution is classified as a
(________) mutation.
9. While doing a pedigree analysis of a royal family from the ancient Mayan empire, you notice
that a prince with the genetic disease married a person from outside the royal family without a
history of the disease in her family. Of their six children, two have the disease and four are not
affected. The prince's mother and father also had this disease. If this disease is controlled by a
single gene, then the MOST likely explanation for disease in the children would be that the
disease is:
a. a spontaneous mutation.
b. recessive and is only seen in homozygous recessive individuals such as the two
affected children.
c. dominant and is only seen in homozygous dominant individuals.
d. dominant and is seen in heterozygous or homozygous dominant individuals.
10. In humans, ability to roll the tongue (R) is dominant to being unable to roll (r). Having
freckles (F) is dominant to having no freckles (f). A man heterozygous for both traits marries a
woman heterozygous for both traits. What is the probability that they will have a child with
freckles? (Hint: make a Punnett square)
a. 9/16
b. 3/16
c. 1/16
d. 3/4
e. 1/4
11. Suppose that, in humans, ability to roll the tongue (R) is dominant to being unable to roll (r),
and having freckles (F) is dominant to having no freckles (f). If a woman heterozygous for both
traits married a man with no freckles who couldn't roll his tongue, what is the probability that
they would have a freckled, tongue-rolling child? (Hint: make a Punnett Square)
a. 9/16
b. 3/16
c. 1/16
d. 3/4
e. 1/4
12. Ultraviolet light is a mutagen, but humans need some exposure to it in order to synthesize
vitamin D3. The amount of ultraviolet light that penetrates the skin depends on the skin's
pigmentation: more melanin (skin pigment) means less penetration. Certain mutations result in
decreased melanin production. Such mutations:
a. are harmful only if inherited from both parents.
b. are beneficial only if inherited from both parents.
c. are neutral only if inherited from one parent, not the other.
d. may be harmful in one environment and beneficial in another.
13. During the formation of female gametes, nondisjunction of X chromosomes may occur during
meiosis I, resulting in two types of eggs with different compositions of sex chromosomes. If
normal sperm fertilize these two types of egg, which of the following are possible sex
chromosome complements in the resulting fertilized egg?
a. XX and XY
b. XYY and XO
c. XYY and YO
d. XXY and XO
e. XXX and XYY
14. Cancer can be caused by mutations. Genetic analysis of a tumor found in a patient we shall
call Anna shows that the cell proliferation was triggered by a somatic mutation in the MYC gene,
causing this gene to be inappropriately activated. Anna is concerned about passing this cancer on
to the children she plans to have in the future. Should she be concerned?
a. Yes, she should worry because tumor growth was triggered by a genetic change, and
mutations are passed on through cell divisions.
b. No, this is not something to worry about because mutations in cancer cells cannot be
passed on in cell division.
c. No, she should not worry about her children because this did not occur in a germ-line
cell.
The diagrams below depict the relative proportions of individuals affected with a certain
condition (shaded bar) and individuals not affected (open bar), in individuals carrying either the
A–T or the G–C allele of an SNP polymorphism. (Heterozygous genotypes carry both alleles and
are included in both categories.)
15. Which graph shows a pattern that suggests that the G–C allele is a risk factor for the disease?
a. graph M
b. graph H
c. graph K
d. graph L
e. graph Q
16. In a person with the genotype XXY, can we definitively tell that nondisjunction took place in
either the mother or the father? Explain.
Hairy cell leukemia is a cancer of white blood cells that responds to treatment with drugs that
inhibit DNA synthesis. A mutation in the BRAF gene, designated BRAF V600E, is associated
with hairy cell leukemia. BRAF V600E differs from the nonmutant BRAF gene in a single base
pair. Imagine that the single base-pair change added a HaeIII restriction site indicated by the
arrow on the gene below. You decide to identify hairy cell leukemia with PCR and the restriction
enzyme HaeIII.
17. Which lane on the gel represents DNA from cells heterozygous for the BRAF V600E allele?
a. lane 1
b. lane 2
c. lane 3
d. lane 4
18. In the gel from the previous question, which lane represents an individual who is homozygous
for the nonmutant allele?
a. lane 1
b. lane 2
c. lane 3
d. lane 4
19. In the gel from the previous question, which lane represents an individual who is homozygous
for the HaeIII restriction site?
a. lane 1
b. lane 2
c. lane 3
d. lane 4
20. The nonmutant allele of the BRCA1 gene helps to suppress tumor formation in women who
are heterozygous for the mutation. Women heterozygous for BRCA1 nevertheless have a 50% to
70% chance of developing breast cancer before age 70, and the usual reason is that the nonmutant
allele is lost or inactivated in a lineage of cells. One possible mechanism for such “loss of
heterozygosity” is:
a. germ cells in the affected individual develop a mutation in the nonmutant allele of
BRCA1.
b. a somatic mutation in a breast cell inactivates the nonmutant BRCA1 allele.
c. a silent mutation occurs in the nonmutant BRCA1 allele.
21. What would be a ratio of yellow and green seeds in the F2 generation if the F1 plant produced
A and a gametes in the 2/3 to 1/3 ratio in Fig. 16.7, respectively? In Fig. 16.7 in the textbook, the
A and a are produced in the 1/2 to 1/2 ratio, respectively. But, I am asking you to change this
ratio to the 2/3 to 1/3 ratio in this question.
Answer the question using a Punnett square and show your work using the multiplication
and the addition rules.