name:______
student ID:______
Genetics L311 final exam
May 3, 2017
Directions: Please read each question carefully. Answer questions as concisely as possible. Excessively long answers, particularly if they include any inaccuracies, may result in deduction of points. You may use the back of the pages as work sheets, but please write your answer in the space allotted. However, you must show all your work. Clearly define your genetic symbols. We will not make guesses as to what a particular symbol is intended to mean. Also, don’t assume that strains are true-breeding unless this is stated in the question. Finally, show all your work. Good luck.
page 2 (30 points possible)
page 3 (26 points possible)
page 4 (26 points possible)
page 5 (24 points possible)
page 6 (36 points possible)
page 7 (18 points possible)
page 8 (24 points possible)
page 9 (16 points possible)
total (of 200 points possible)
1. Short answers (2 pts each, for total of 30 pts)
A. A diploid individual in which the two copies of a gene differ from one another is heterozygous.
B. Multifactorial traits are specified by multiple genes, often in conjunction with environmental inputs.
C. A distinct breeding group within a larger interbreeding group of individuals is referred to as a(n) subpopulation.
D. The migration of cancerous cells from one site in the body to another is called metastasis.
E. When only a single copy of a gene or segment of genome is present in otherwise diploid organisms, such as X-linked genes in human males, they are said to be hemizygous.
F. Genetic traits, such as height, cholesterol levels, etc. that can be measured are referred to as quantitative traits.
G. An organism into which a gene from a different species has been introduced is a(n) transgenic organism.
H. A single base pair difference in DNA sequence between individuals is referred to as a(n) single nucleotide polymorphism (SNP).
I. A(n) conservative transposable element is a DNA element that can excise itself and move to a new chromosomal position without increasing its number within the genome (be specific).
J. Apoptosis is the genetic program that is used to eliminate excess cells or cells that may become cancerous.
K. A heritable state of gene function not encoded within the DNA is referred to as epigenetic.
L. A point where the cell can halt its progress through the cell cycle, in the event of DNA damage or unattached chromosomes, for example, is called a(n) checkpoint.
Please provide a brief definition of each of the following:
M. equivalence group: Multiple cells with the same developmental potential.
N. morphogen: A chemical that can induce different fates depending on its concentration.
O. heteroplasmy: A mixture of mutant and wild type organelles, such as mitochondria.
2. Lesch-Nyhan Syndrome results from a mutation in the gene that encodes the enzyme hypoxanthine-guanine phosphoribosyltransferase (HGPRT). The HGPRT gene is located on the X-chromosome. Lesch-Nyhan affects about 1/400,000 boys. Assuming that Hardy-Weinberg applies, please answer the following.
A. What are the frequencies of the mutant and wild type hgprt alleles (6 pts)?
q = 1/400,000 or 0.0000025
p = 0.9999975
B. What is the frequency of Lesch-Nyhan carriers (i.e. heterozygotes) in this population (6 pts)?
2pq = 2(0.0000025)(0.9999975) = 0.0000049
2(0.0000025) = 0.0000050 is OK
C. The pedigree to the right shows autosomal recessive Gaucher disease. This disease affects about 1/50,000 individuals in North America. The frequency of heterozygotes is roughly 0.01. Symptoms appear from birth. What is the probability that the individual indicated by the question mark will be affected by Gaucher disease (6 points)?
0.01(2/3)(1/4) = 0.001667
3. The control of magical ability in humans is under the control of a simple, linear genetic pathway. Key genes have been identified and are found to operate in the pathway as shown.
ron herm hagrd vld hry magical ability
What phenotype would be expected for the following mutants? Note: Think of the gain of function as being continuously active independent of upstream genetic inputs (8 points).
Loss of function herm: no magic (= muggle)
Gain of function vld: no magic (= muggle)
Gain of function hry: magic
Loss of function vld and loss of function herm: magic
4. The brilliant poison dart frog E. qinae, is diploid with two pairs of chromosomes, one long and one short. The long chromosome contains two genes, sho and red. The short chromosome contains a single gene, grn. In a particular individual (a pet frog, named Cheryl), you demonstrate that the mutations on the long chromosome are in trans. Cheryl is homozygous mutant for grn. A. Show the cells produced by a mitotic division at the start of G1 in Cheryl. Diagram the chromosomes and show the location of each
allele (6 pts)
B. Imagine we’ve found a E. qinae individual with tetraploid chromosome content. Please diagram one of this individual’s cells during metaphase I of meiosis. Please draw a circle around one set of homologues. Draw a box around one pair of sister chromatids. (You do not need to show genes. 6 pts).
5. Drosophila anterior-posterior body axis is patterned by a series of sequentially acting classes of genes. A. Please provide the names and function of each class of gene, the order in which they act, and provide an example of one gene in each class (10 points).
morphogens provide A-P information (bcd, hb or nos)
gap genes begin to divide embryo into regions (knirps or kruppel)
pair rule genes establish homeotic genes specify segment
segments(ftz or eve) identity (antp or btx)
segment polarity genes define A-P within segment
and reinforce segment boundaries(wg or en)
B. You find a mutation in a gene that you call oddball, in which animals are missing every other body segment What class of genes is oddball likely to belong to (4 points)?
This is a pair rule gene.
6A. A key regulator of sex determination in Drosophila is the sxl gene. A generic form of the pathway is shown below.
A. Please explain how the production of sxl is initiated and maintained in animals with an X:A ratio of 1.0 (8 points).
XX activates Sxl PE because it makes activating dimer (because of having two copies of activating protein genes on the X chromosome), which producesprotein because otherwise default splice is skipped. Later Sxl PL turns on. Sxl protein made from early promoter binds mRNA and prevents default splice from occurring. Therefore protein continues to be made.
B. What phenotypes do you expect to result from the following mutations? Please provide a brief explanation of your rationale (6 points).
mutation phenotype rationale
loss of sxl male DSX undergoes default male splice
mutation that gives unregulated female DSX undergoes female splice
(i.e. always hi) tra activity
loss of sxl + unregulated female tra is downstream
tra activity
7. In C. elegans, cell-cell signaling specifies vulval cell fates, causing the Pn.p cells to adopt primary, secondary or tertiary fates. Give the predicted effect of the following mutations in each of the cell fates listed (i.e. increase, decrease or no change, 2 pts each, 10 pts total):
primary rationale
A. loss of adapter (sem-5) decrease can’t activate MPK to inhibit
transcription factors
B. gain-of-function mpk-1 (kinase increase transcription factors inhibited in
downstream of Ras) all 6 cells
C. gain of EGFR plus decrease can’t inhibit Tx factors, Ras
loss of Ras downstream
D. transcription factors (lin-1 or lin-31) decrease Tx factors inhibit primary in all 6
that can’t be inactivated
E. loss of adapter and transcription decrease Tx factors downstream
factor that can’t be inactivated
8. In your studies of the unusual ant species M. raoae you find two true breeding strains. Strain 1 is yellow-skinned and has oval shaped abdomens. Strain 2 is blue-skinned with triangular abdomens. Crossing a strain 1 female with a strain 2 male results in the following F1s:
BbXTXt All females are yellow-skinned with triangle shaped abdomens
BbXtY All males are yellow-skinned with oval shaped abdomens
A. Please give the genotypes of the F1s on the lines above (6 points).
B. What is the probability of obtaining a female with blue skin and a triangular abdomen from a cross of F1 males X F1 females? What is the probability of obtaining males of the same phenotype (6 points)?
Cross is BbXTXt X BbXtY
Probability of blue skinned, triangle abdomen female is ¼ X ¼ = 1/16
Probability of blue skinned, triangle abdomen male is ¼ X ¼ = 1/16
C. What genotypes and phenotypes do you expect from a cross of strain 1 males X strain 2 females (6 points)?
BBXtY X bbXTXT gives ½ BbXTXt and ½ BbXTY, which all have yellow skin and triangular abdomens.
9. While exploring the highlands of the Amazon basin you find a new species of resplendent hummingbird, which you name J. gavadae. Interestingly you notice that these birds are found in two strains. The first has golden wings and the second has iridescent green wings. You cross the two strains and find that the offspring all have iridescent green wings. You cross these F1’s and get the following:
903 golden A–B– and aaB–
227 iridescent green A–bb
73 brown aabb
A. Please give the genotypes of the F2s shown above (8 points).
B. What fraction of F2 golden-winged hummingbirds are homozygous for all genes present (6 points)?
AABB and aaBB are homozygous. These will be ¼(1/3) + ¼(1/3) = 1/6
C. A few hummingbirds are found to have short beaks. You cross short beaked males with wild type females and find the F1’s listed below. You cross the short beaked F1’s to produce the F2’s shown.
F1s: F2s
¼ short beaked male 1/3 short beaked male
¼ short beaked female 1/3 short beaked female
¼ long beaked male 1/6 long beaked male
¼ long beaked female 1/6 long beaked female
How do you explain these results (4 points)?
Short is autosomal dominant but recessive lethal.
10. Progression through the cell cycle is tightly regulated. In particular, we looked at how the cell regulates entry into S phase.
A. Please illustrate the steps involved in a cell’s decision to enter S phase. Include the relevant proteins, including the proteins involved in the DNA damage checkpoint that we discussed. You may use a diagram if you wish (6 points).
B. Cells have checkpoints at several points during the cell cycle that they can implement if needed. Briefly explain the purpose of these checkpoints (4 points).
The purpose of checkpoints is for the cell to verify that it is safe to move on through the cell cycle and that there are no significant problems. For example, one verifies that there is no DNA damage before replicating the DNA. Another confirms that chromosomes are properly attached to the spindle before moving through mitosis.
C. Using the pathway discussed in lecture, first determine whether the following mutations would increase or decrease the probability of cancer, or have no effect. Please briefly state your rationale (8 points).
Mutation Probability Rationale
Loss of cyclin decrease Will reduce proliferation
Constitutively active increase Active Ras drives proliferation
Ras (i.e. always active)
Loss of p53 increase Prevents DNA damage checkpoint and apoptosis
Non-phosphorylatable but otherwise decrease Rb won’t release E2F, it’s
Functional Rb plus loss of p53 downstream of p53
11. Around the world, many animals have moved from areas with day and night into the perpetual darkness of caves. Usually these animals quickly lose their eyes because these are of no use in the dark. You are interested in the genetics of eye degeneration so you obtain related fish from different parts of the world. Assume that each strain has lost its eyes due to a single gene mutation. All mutations are recessive. Strains from different countries have lost eyes independent of one another. Please include the genotypes of parents and offspring.
A. You cross cavefish from Mexico X cavefish from France and find all the progeny have eyes. How do you explain this result (4 points)?
They have eyes because the different fish have mutations in different genes required for eye development. Cross is aaBB X AAbb =>AaBb
B. You cross cavefish from France (the same fish as in part A) X cavefish from Borneo and all the progeny lack eyes. How do you explain this (4 points)?
The two strains have mutations in the same gene. Cross is bb X bb => bb
C. When you cross cavefish from Mexico (those in part A) X cavefish from Borneo (those in part B), what outcome do you expect? Why (3 points)?
French and Bornean fish have mutations in the same gene, which I’ve called B. We know from A that Mexican fish have a mutation in another gene, which I called A. Therefore the cross is aaBB X AAbb => AaBb and the fish will have eyes.
D. What outcome will result from a cross of the F1’s in part C? Please give phenotypes and their relevant frequencies (3 points).
Cross isAaBb X AaBb, which will produce:
9/16 A-B- The first (i.e. A-B-) have eyes the rest do not. Therefore the
3/16 A-bb ratio of fish with eyes:eyeless is 9:7.
3/16 aaB-
1/16 aabb
12. A man is homozygous mutant for a paternally silenced gene. If he marries a woman who is homozygous wild type (Assume all who marry in are wild type, 10 points):
A. What is the probability that his sons will show the trait? 0
B. What is the probability that his daughters will show the trait? 0
C. What is the probability that his daughter's sons will show the trait? 1/2
D. What is the probability that his daughter's daughter will show the trait? 1/2