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BIOL105 Final Exam Spring, 2011

Name: ____________________________________ ID #: ____________________________________ Name of TA: ____________________________________ • Please fill in the requested information in the section above. • Read each question through COMPLETELY and CAREFULLY before starting to answer it. • Write legibly. You may work through problems and give your answers in pencil or pen. Completely erase or scratch out anything that you have corrected or replaced. • Most questions have a box provided for the final answer. That is where we will look for the answer you want graded. • Use correct genetic nomenclature, including "/" and ";" symbols where appropriate. You may use "-" to indicate any allele. maximum points points scored

Q1: 9 ____________

Q2: 6 ____________ Q3: 10 ____________

Q4: 8 ____________ Q5: 4 ____________

Q6: 10 ____________

Q7: 9 ____________ Q8: 12 ____________

Q9: 12 ____________ Q10: 6 ____________

Q11: 6 ____________

Q12: 6 ____________ Q13: 5 ____________

Q14: 6 ____________ Q15: 9 ____________

Q16: ____ 10 ____________

Q17: ____ 5 ____________ Q18: 7 ____________

Total: 140 ____________

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Question 1) (9 points) In Arabidopsis, the ragged leaf gene (R gene) and hairy roots gene (H gene) are 40 m.u. apart on one arm of chromosome 3. a) (2 points) You test cross R h / r H plants, and examine progeny plants. What % of the progeny do you expect to have an "R h" phenotype? b) (2 points) You cross R h / r H females with R H / r h males. What percentage of the progeny do you expect to be R h / R h? c) (2 points) The miniature flower gene (M gene) is on a different chromosome. You cross R h / r H; M/m plants with each other, and examine progeny plants. What % of the progeny plants do you expect to have the fully recessive “r h m” phenotype? d) (3 points) Challenge question: You cross R h / r H; M/m plants with each other, and examine progeny plants. What % of the progeny plants do you expect to have the fully dominant “R H M” phenotype?

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Question 2) (6 points) The frequencies of gametes of different genotypes, determined by test-crossing a triple heterozygote, are as follows: gamete genotype % A B D 13 a b d 14 A B d 7 a b D 6 A b d 26 a B D 27 a B d 3 A b D 4 a) Correctly write the genotype of the triple heterozygote? Show the proper gene order and proper cis-trans configuration of alleles. b) What is the accurate genetic distance between the outer genes?

c) If this cross had been performed without the middle gene, by how many m.u. would you have underestimated the distance between the outer genes?

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Question 3) (10 points) Below are shown a plasmid vector and a region of human chromosomal DNA. Erythromycin is an antibiotic. HRP encodes an enzyme that causes bacteria to turn brown when grown on substrate. All of the restriction sites shown yield sticky ends upon digestion with their respective enzymes. You want to clone the geneticase gene, so that you can express it in bacteria and make a million dollars by selling its protein product to students (it synthesizes correct answers to genetics exams).

a) (2 points) With which restriction enzyme would you cut the vector and the geneticase-containing genomic DNA? Circle the single best answer. BamHI

EcoRI

HindIII

PvuII

SalI

b) (4 points) After ligating cut vector and cut human DNA together, what starting genotype of bacteria would you introduce your recombinant plasmids into? Circle one. erythromycin-sensitive or erythromycin-resistant HRP+ or HRP- c) (4 points) To find the bacteria containing plasmids with human DNA inserts, you should do which of the following - circle one of the 2 choices in each set of parentheses (screen or select) for cells that (can or can't) grow on erythromycin (screen or select) for cells that are (white or brown)

geneticase

B = BamHIE = EcoRIH = HindIIIP = PvuIIS = SalIstreptomycin-

resistance genekanamycin-resistance gene

origin ofreplication

E BP

HS E

S E H B PH E S P B H

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Question 4) (8 points) Mutations in two genes, myc and puma, result in a form of pancreatic cancer. The wild-type allele of myc is a proto-oncogene and the wild-type allele of puma is a tumor suppressor gene. a) (4 points) What is the genotype of this cancer with regard to these two genes? Circle your answer. Choose one: myc(gf)/+ or myc(lf)/+ or myc(lf)/myc(lf) Choose one: puma(gf)/+ or puma(lf)/+ or puma(lf)/puma(lf) b) (4 points) Homozygous recessive mutations in a third gene, when combined with mutations in myc and puma, result in an aggressive form of cancer that is difficult to treat. This mutation could be in a (circle all that apply): a) gene that normally promotes cell division

b) gene that normally inhibits cell division

c) gene that normally promotes apoptosis

d) gene that normally inhibits apoptosis

Question 5) (4 points) Indicate beside each of the Retinoblastoma (Rb) statements below whether the statement is True (T) or False (F). ____ One known role of Rb is to bind E2F in order to promote entry into S. ____ Phosphorylation of Rb by Cdk-cyclin causes Rb to release E2F. ____ The hereditary mode and the non-hereditary mode of forming eye tumors both require

that both copies of the Rb gene be mutant. ____ In family pedigrees, the hereditary form of retinoblastoma behaves as a dominant

disease.

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Question 6) (10 points) Mutations in the mef gene are recessive and maternal-effect. Mutations in the yel gene are recessive and zygotic (i.e. not maternal-effect). Mutations in both genes cause visible phenotypes but are not lethal. Both genes are autosomal. Indicate below what percentage of the progeny from each cross you would expect to show a mutant phenotype (any mutant phenotype). If the cross is not possible, then indicate "NP" instead. What % of progeny will female fly male fly have a mutant phenotype?

yel -/ yel + yel -/ yel + __________

yel -/ yel - yel -/ yel + __________

mef -/ mef + mef -/ mef + __________

mef -/ mef - mef -/ mef + __________

mef +/ mef -; yel +/ yel - mef +/ mef -; yel +/ yel - __________

Question 7) (9 points) In marigolds, flower color is controlled by a biochemical pathway. The wild-type color is deep orange. Homozygous r/r mutants have red flowers, and homozygous g/g mutants have gold flowers. The R and G genes map to different chromosomes, both the R and G genes are haplo-sufficient, and the R gene is epistatic to the G gene. a) (4 points) Write out the biochemical pathway, showing the order of pigment conversions (the pigments are orange, red, and gold). Show which conversion is catalyzed by the R protein, and which by the G protein. b) (3 points) You mate r/r x g/g mutant plants. Correctly write the genotype(s) of the F1 plants and describe their flower color. c) (2 points) You mate F1 marigolds with each other. Among the F2 plants, what proportion would you expect to have gold flowers?

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Question 8) (12 points) A researcher treated haploid yeast cells with a mutagen and obtained 5 haploid arginine-requiring mutants (a, b, c, d, and e), all of which require that arg be added to the medium in order for them to grow. (Without arg, none of them can grow.) Complementation testing and mapping showed that: a and b are in the same gene a and c are in different genes right next to each other (i.e. no c.o.) on chromosome 2 a and d are in different genes that map 15 m.u. apart on chromosome 2 a and e are in different genes on different chromosomes Fill in the charts below with the %s that would have led to these conclusions. a) (4 points) Complementation test: The arg-requiring mutants were mated with each other in the pairwise combinations shown below (a x a, a x b, a x c, a x d, and a x e), and the resulting diploid cells were plated on medium lacking arg. Fill in the chart below with the % of diploid cells that grew on medium lacking arg. a b c d e a ___ ___ ___ ___ ___ b) (4 points) Mapping experiment: The diploids from the complementation test were induced to undergo meiosis, and the resulting haploid spores were plated on medium lacking arg. Fill in the chart below with the % of haploid cells that grew on medium lacking arg. a b c d e a ___ ___ ___ ___ ___ c) (2 points) Correctly write the genotype of the diploids formed by mating a and d. c) (2 points) Correctly write the genotype of the diploids formed by mating a and e.

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Question 9) (12 points) Mutations in the ApoE gene in mice result in a condition that mimics atherosclerosis, in which fatty deposits build up in arteries. To create a mouse model of atherosclerosis, you generate a disrupting construct, in which a gene conferring colchicine resistance has been inserted into the mouse ApoE gene. Colchicine normally kills mouse embryonic stem cells (ES) cells. Your disruption construct also includes the thymidine kinase. Ganciclovir kills cells expressing thymidine kinase. You introduce the disrupting construct into ES cells. a) (2 points) What type of ES cells do you want to generate by introducing the disruption construct?

a) colchicine sensitive, ganciclovir resistant cells

b) colchicine sensitive, ganciclovir sensitive cells

c) colchicine resistant, ganciclovir resistant cells

d) colchicine resistant, ganciclovir sensitive cells

e) none of the above

b) (2 points) You isolate ES cells with your disruption construct. What is their genotype with regard to the ApoE gene? Use A for the wild-type ApoE allele and a for the mutated ApoE allele. c) (2 points) The ES cells you generated in part a came from wild-type brown mice. You transplant the ES cells you generated into embryos from "golden" mice; the golden mutation (g allele) is recessive and causes mice to have golden fur. Some of the embryos develop into F1 adults with patches of brown fur and patches of golden fur. If you mate F1 mice with patches to golden mice, what genotype (at the A and G loci) of F2 mice do you want? Note: A and G are on different chromosomes. d) (2 points) You mate these selected F2 mice to generate your F3 mice. What fraction are brown and exhibit atherosclerosis?

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e) (4 points) Introduction of the colchicine resistance marker removes a BamHI restriction site that is present in the wild-type ApoE gene (the starred arrow in the drawing below). You isolate DNA from F3 mice, digest it with BamHI and perform a Southern blot with the probe depicted below. Draw the blot you would expect to see for the genotypes indicated. Circle the band that indicates the mutated ApoE gene. * probe A/A A/a a/a

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Question 10) (6 points) Indicate whether each statement is True (T) or False (F).

___ Recombination occurs between sister chromatids of homologous chromosomes. ___ In a two point cross between two genes, double crossovers do not affect the calculated genetic distance between the two genes. ___ Recombination does not occur if a plant or animal is homozygous. ___ Genes cannot lie more than 50 m.u. apart on a chromosome. ___ Mutations that alter a region that does not code for a protein may alter phenotype. ___ If a gene is haplo-insufficient, a loss-of-function mutant allele is dominant to the wild-type allele. Question 11) (6 points) Match the term with its definition: null mutation

neomorphic mutation

dominant negative mutation

hypermorphic mutation

a) mutation that reduces gene function b) mutation that produces a novel protein or causes inappropriate expression, resulting in a new phenotype c) mutation that abolishes gene function d) mutation that results in a hyperactive form of the protein e) mutation that produces a protein that associates with wild-type subunits, inactivating them Which of the choices below is an example of a hypermorphic mutation? a) mutation of Ras that results in the protein always being bound to GTP

b) inappropriate expression of the Antennapedia gene in the head of a fruit fly so that

legs develop where antennae should be

c) mutation in lin-3 that produces vulvaless worms

d) mutation in bicoid that affects embryonic patterning in flies

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Question 12) (6 points) The allele B gives mice a black coat and b gives a brown one; B is dominant to b. The genotype e/e of another, independently assorting, gene prevents the expression of B and b, making the coat color beige, whereas E/- permits the expression of B and b. Both genes are autosomal. In the following pedigree, black symbols indicate a black coat, crosshatched symbols indicate brown and white symbols indicate beige. I 1 2 II 1 2 3 4 5 6 III 1 2 3 4 5 6 7 Give the genotypes for the following individuals: I-1 II-3 III-5 Question 13) (5 points) For the pedigree to the right, what is the mode of inheritance? a) autosomal dominant

b) autosomal recessive

c) sex-linked dominant

d) sex-linked recessiv

e) none of the above

b) What is the probability that the individual indicated by the hexagon will be affected? Assume that the allele that results in the trait is rare, i.e. spouses that marry into the family do not carry a mutant allele.

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Question 14) (6 points) A mutation in gene R runs in families, and the pedigree below illustrates the syndrome caused by loss of gene R function. R syndrome is a rare trait; spouses can be assumed not to be carriers.

I-1 I-2

II-1 II-2 II-3 II-4 II-5 II-6

III-1 III-2 III-3 III-4 III-5

In 1997, investigators searching for the R gene identified a linked RFLP. The RFLPs associated with the mutant allele and the wild-type allele are shown below. The arrows indicate BamHI sites. mutant allele wild-type allele 1 2 3 4 a) (2 points) Which probe would you use to distinguish between the RFLPs? a) 1 b) 2 c) 3 d) 4 b) (4 points) You digest the genomic DNA of several members of the family depicted in the pedigree above with BamHI and perform a Southern blot with the appropriate probe. Indicate the banding pattern you would expect to see for the following individuals: I-1, II-1, and II-2. In the last lane indicate the banding pattern you would expect to see for individual III-4 if an observable cross-over occurred between the RFLP and the gene in one of his parents. Circle the RFLP associated with the mutant allele in all individuals who carry it. I-1 II-1 II-2 III-4 if cross-over in parent

top of gel bottom of gel

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Question 15) (9 points) Drosophila have compound eyes made up of repeating units called ommatidia. Each ommatidium is made up of 8 photoreceptor cells. One of these cells has a unique morphology and is called R7. Development of the R7 cell is controlled by a signaling pathway. There are gain-of-function (gf) alleles of some of the genes in this pathway, and they produce multiple R7 cells when there should only be one. There are loss-of-function (lf) alleles of most of the genes in this pathway, and they result in no R7 cells at all. The types of mutations and the phenotypes they produce are shown below: gene and allele phenotype boss(lf)/boss(lf) no R7 cells boss(gf)/+ multiple R7 cells ras(lf)/ras(lf) no R7 cells ras (gf)/+ multiple R7 cells rolled(lf)/ rolled(lf) no R7 cells To figure out the epistatic interactions between the genes, the following double mutant combinations were generated and the phenotypes recorded: boss(gf)/+; ras(lf)/ras(lf) no R7 cells boss(lf)/boss(lf); ras (gf)/+ multiple R7 cells ras (gf)/+; rolled(lf)/ rolled(lf) no R7 cells boss(gf)/+; rolled(lf)/ rolled(lf) no R7 cells a) (3 points) Draw the pathway, indicating the order of gene action: R7 fate b) (2 points) Which gene is epistatic to the other two? c) (4 points) You identify a new gene, sevenless, in this pathway. A loss of function allele of sevenless results in no R7 cells. You suspect that sevenless acts between boss and ras. Describe a double mutant combination that would allow you to test this prediction and the phenotype (no R7 cells or multiple R7 cells) you would expect if sevenless indeed acts between boss and ras? Double mutant combination Phenotype

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Question 16) (10 points) The R and H genes are on different chromosomes. In a male heterozygous at both the R locus and the H locus, a germ cell suffered nondisjunction of the R and r bearing chromosomes in meiosis I. a) Correctly write the R and H genotypes of the 4 resulting sperm. (You should ignore the effects of recombination.) b) Correctly write the genotype at the R and H locus of a monosomic embryo that could result from the union of one of the sperm from part a above with a normal oocyte/egg from a mother bearing all dominant alleles. Question 17) (5 points) The cell below is from a diploid (2n=6) animal. Draw the correct number of chromosomes and the correct arrangement for metaphase of mitosis. The B gene and E gene both map to one arm of chromosome 1. The A gene maps to a different chromosome. Add allele labels to your chromosomes. Here is the allele information you need. This organism is heterozygous at each locus. One parent of the heterozygote was homozygous for the dominant alleles, the other parent was homozygous for the recessive alleles. Show the allele arrangements in the absence of recombination effects.

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Question 18) (7 points) a) (2 points) The VNTR bands below contain variable numbers of the repeat AATCGATCA. Which of the sets of VNTR banding patterns shown below would be consistent with a father (F) and his child (C)? a) b) c) d) e)

b) (2 points) How many VNTR loci (defined by AATCGATCA repeats) are present in the genome? a) 2 b) 4 c) 8 d) 16 e) we can’t tell c) (3 points) The above blots were generated using a probe to the repeated sequence present in each VNTR. Now suppose that you have generated a probe to unique sequence adjacent to the repeat region at one particular VNTR locus on chromosome 7, as shown below. Using this probe on the blot below, create a banding pattern that would be consistent with a father (F) and his child (C). F C sites for restriction enzyme used repeat probe region