final exam (a) name genome 371, spring 08faculty.washington.edu/pallanck/course materials...page 5...

Genome 371, Spring, 2008, Final Exam Print Name: _________________________

Final Exam (A) Name ___________________________________ Genome 371, Spring 08 Eichler/Pallanck TA/Quiz Section __________________________ 1. Two homologous chromosomes (human chromosome 2) are depicted below prior to the

beginning of cell division. The location of five loci each with two different alleles, the position of centromeres (oval) and direct repeats (arrows) are indicated. For the following questions, clear diagrams are essential to your earning credit for this question!

a) Draw how these chromosomes would look in each of the four gametes at the end

of Meiosis II as a result from a equal crossover between loci B and C ? (8 points)

A1B1C1

E1

F1

A2B2C2

E2

F2

A1B1

C1

E1

F1A

2B2

C2

E2

F2

A1B1C1

E1

F1

A2B2C2

E2

F2

A1B1

C1

E1

F1A

2B2

C2

E2

F2

b) Draw how these chromosomes would look in each of the four gametes at the end

of Meiosis II as a result from an unequal crossover event at the direct repeats between the homologous chromosomes? (8 points)

A1B1C1

E1

F1

A2B2C2

E2

F2

A2B2C2

E2

F1A

1B1C1

E1

F2

A1B1C1

E1

F1

A2B2C2

E2

F2

A2B2C2

E2

F1A

1B1C1

A2B2C2

E2

F1A

1B1C1

E1

F2


2. The pedigree shows the transmission of an autosomal dominant form of macular degeneration within a three generation family. You have genotyped a polymorphic marker that you believe to be linked to the disease trait. There are two different alleles for this marker (A and B).

a) In the space provided within the pedigree, provide the genotype of each individual with respect to the disease locus. Assume that the allele associated with macular degeneration (D) is completely penetrant. If the disease genotype for a specific allele can not be determined use a (?) symbol (4 points).

b) Based on the pedigree structure and transmission of the marker, indicate which of the children likely arose as a result of a recombination event in the gametes from II-1? Estimate the map distance between the marker and the disease locus (show your work)? (10 points)

2/12 X 100=16.7 cM

Non-parental= III-3 and III-6

c) Based on the number of recombinant and non-recombinant transmissions, calculate

the LOD score for a theta value of 10 (θ10)? Show your work. Is there evidence for or against linkage? (10 points)

θ10= log10 [(0.45)10 X (0.05)2/(0.25)12]=1.1547 or = log10[(0.0003405 X 0.0025)/0.00000006]=1.1519

There is no significant evidence for or against linkage. Inconclusive.

dd D?

Dd dd

dd Dd Dd Dd Dd Dd Dd Dd dd dd dd dd


BAC clone Markers

1 FAD

2 BIG

3 IGJ

4 DE

5 EBI

6 AD

7 ADE

8 GJ

9 JC

10 JCH

11 HK

3. You want to sequence the zebrafinch genome using the whole genome shotgun sequence assembly (WGSA) method by 454 pyrosequencing technology. The genome is 1.2 Gb in size and the average read length using this sequencing technology is 250 bp.

a) What is the minimum number reads that you will have to produce in order to sequence the entire genome at 10-fold coverage? (Hint: You can use the formula Cov=NL/G from problem set #6) (2 points) Cov=# reads X 250 bp/ 1.2 X 109 # or reads = (10 X 1.2 X 109 bp)/ 2.5 X 102 bp =4.8 X 107 or 48 million reads.

b) What portions of the zebrafinch genome will likely not be well assembled using

WGSA? Identify two different problematic regions and explain for each why WGSA will fail to resolve these. (4 points) (2-3 sentences maximum)

Centromeres, telomeres, common repeat regions and segmental duplications of high sequence identity; Whenever repeat-length exceeds the length of sequence read, sequences collapse during WGSA. In the case of telomeric and centromeric repeat sequences the higher order structure of over 10’s of kbp make the de facto repeat length greater than the read length. c) One of the regions identified in (b) contains a locus

important mating behaviour. You wish to obtain the complete sequence of the region using the clone-ordered based approach and subclone the entire genome in BACs. You previously identified 11 STS markers from your region of interest. You screen the library and identify 11 BACs that contain at least one of your markers as shown on the right. Draw how the BACs overlap in a map of the region based on the STS content of the BACs? What is the order of the markers along the chromosome (centromeric to telomeric)? The most telomeric BAC, its position along the chromosome and the location of the markers within clone 11 are shown as an example. (20 points)

centromere

H #11K#1

H KMarkers

#6

#4

#5

#2

#3

#8

#9

#10

F A D E B I G J C

#7

F A D

A D

A D E

D E

E B I

B I G

I G J

G J

J C

HJ C

centromere

H #11K#1

H KMarkers

#6#6

#4#4

#5#5

#2#2

#3#3

#8#8

#9#9

#10#10

F A D E B I G J C

#7#7

F A D

A D

A D E

D E

E B I

B I G

I G J

G J

J C

HJ C


d) What is the minimal tiling path (centromeric to telomeric) of clones needed to completely sequence the region of interest? (4 points) BAC clones #1, #4, #5, #3, #10,#11

e) You sequence the BACs, assemble the sequence and integrate it into the WGSA of

the zebrafinch genome. After six weeks, the sequence is annotated and you access your region through the USC genome browser (see below) (NOTE: A larger colored version of these snapshots is provided on the last page of the exam) (8 points)

How many RefSeq genes have been assigned to the region? __6__________ How large is the region that you subcloned and sequenced (to the nearest 50 kbp) _______650 kbp or 650,000 bp____ How many gene models are not confirmed based on alignment of chicken cDNAs (ESTs=cDNA) against the zebrafinch genome?__1_____ How many completely novel gene models are predicted for the zebrafinch genome based on the chicken cDNA/EST alignments?__4 ( 1 point for 5) _

f) Your are particularly interested in the PSMA3 gene as a candidate for mutations

associated with altered mating behaviour? Based on the zoomed in view of the gene (see next page), answer the following questions (12 points).

How many exons does this gene have? ______11_________ What is the approximate size of the gene (to the nearest kb)? ___8 or 9kbp______ Which strand (Watson or Crick) is the template strand?___Watson___________ Write out the mRNA sequence (5’ to 3’) encoding the first 3 amino acids of this protein?____5’AUGAGCUCC3’______________________________________________


4. The products of two genes (Gene C and Gene P) are necessary for the production of the

pigment anthocyanin in sweet pea flowers (pathway shown below, intermediate products are colorless). Colored flower alleles, C and P, are completely dominant to loss of function alleles (c and p). You breed a pure line pea plant with colored flowers with a homozygous recessive plant with white flowers (ccpp). (10 points)

a. What is the genotype of the P1 with colored flowers__PPCC___________ b. What is the genotype and phenotype of the F1? ____PpCc colored____________ c. If you self the F1, what is expected ratio of flower color in the F2? Show your

work by drawing a Punnett square and clearly labeling the phenotype of the different genotype classes? 9colored to 7 colorless Gametes PC pc Pc pC PC PPCC PpCc PPCc PpCC pc PpCc ppcc Ppcc ppCc Pc PPCc Ppcc PPcc PPCc pC PpCC ppCc PpCc ppCC Only P-C- are colored (9); All other genotype classes are colorless (7)


5. You are studying mismatch repair in yeast, which as you are aware from your undergraduate course in genetics plays a role in correcting mistakes made by DNA polymerase and thus reduces the mutation rate. In previous work you have discovered two mutant strains: one strain has a point mutation (allele mmr1pm) in a newly discovered mismatch repair gene and the other strain has a complete deletion of the MMR1 gene (mmr1Δ). The wild type allele is called MMR1. The mmr1pm, mmr1Δ and MMR1 WT strains are all in an ade2 genetic background (i.e., all three of these strains also have a missense allele of the ade2 gene and are thus unable to grow on media deficient in adenine). You test the following strains for growth on adenine-deficient plates and observe the following results:

Strain Genotype Result on adenine-deficient plate 1 α MMR1 ade2

2 a mmr1pm ade2

3 a mmr1Δ ade2

4 Diploid: α/a MMR1/mmr1pm ade2/ade2

5 Diploid: α/a MMR1/mmr1Δ ade2/ade2

a). Why is the wild type MMR1 strain (strain 1) able to form a colony on the adenine-deficient plate? (8 points) A rare spontaneous mutation reverted the ade2 strain to an ADE2 phenotype. b). Why do the two haploid mmr1 mutant strains (strains 2 & 3) form many more colonies than the wild type MMR1 strain (strain 1)? (8 points) The mismatch repair system corrects many of the errors that are made by DNA polymerase during DNA replication. In mutants that are defective in mismatch repair, none of the DNA polymerase errors would be corrected and this would result in a higher mutation rate. Thus, this higher mutation rate would increase the frequency of reversion mutations that revert the ade2 strain to an ADE2 phenotype. c). Suggest an explanation for why the two diploid strains (strains 4 & 5) are not behaving the same. (8 points) The heterozygote with the mmr1 deletion allele is showing the wild type phenotype, so the deletion allele is behaving as a recessive LOF allele-the wild type protein is still being made and is capable of mismatch repair. The heterozygote with the mmr1 point mutation is showing the mutant phenotype, so it is behaving as a dominant-negative mutation. The mutant protein must be acting as a poison subunit, interfering in some way with the activity of the wild type protein.


6. You are studying Spinobulbar Muscular Dystrophy (SBMA), a dominant sex-linked disease in humans that occurs when there are more than 40 repeats of a CAG trinucleotide in the androgen receptor gene. The unique sequence flanking the repeat region is as follows: 5' TTGACTTTCGCGACGAGTATG-----(CAG)n------ACGGTATTAGAGACGTACCAC 3' 3' AACTGAAAGCGCTGCTCATAC-----(GTC)n------TGCCATAATCTCTGCATGGTG 5' a). If you wanted to PCR-amplify the entire trinculeotide repeat region, which of the following primers would you use? Circle the primer(s) you choose. (6 points) Primers: #1) 5’-TGACTTTCGCGACGAGTATG-3’ #2) 5’-ACTGAAAGCGCTGCTCATAC-3’ #3) 5’-(CAG)12-3’ #4) 5’-GTGGTACGTCTCTAATACCGT-3’ #5) 5’-TGCCATAATCTCTGCATGGTG-3’ b). Shown on the right is the outline of a pedigree for the disease described above, and a representation of a gel showing PCR-amplified fragments detecting the number of CAG repeats. The DNA corresponding to each individual is directly below his or her place in the pedigree. Based on the information you have been given, fill in the pedigree to show the sex and phenotype (affected vs. unaffected) of all individuals except II-9 (save that one for the next question). (6 points) c). What is unusual about individual II-9? (4 points) The person looks like an XY male (only one PCR product), but appears to have inherited the father’s X chromosome (28 repeats). d). Suggest two distinct genetic mechanisms (two types of events) that could explain the genotype of II-9, stating whether that individual would be phenotypically female or male in each case. (8 points) Genetic mechanism: There are several possible explanations for this observation: -non-disjunction in the mom gave an egg with no X chromosome; the sole X chromosome is derived from the dad; the individual is an XO female.

= unaffected male

= unaffected female

= affected male

= affected female


-the person is an XY male, with a change in one of the alleles from the mom (either the 22-repeat allele changed to 28-repeats, or the >40-repeat allele shrank to 28 repeats). Alternatively, the person is an XX {28,28} female -- one of the alleles in the mom changed to a 28-repeat allele; both alleles are present, but since they are the same size, only one band is seen on the gel -the person is an XX female, but a mutation in the primer-binding site prevented the PCR primer from hybridizing to its target. -the person is XX female, but the disease allele expanded even further, to the point that the PCR product was so large that it was outside the range visualized on this gel


7. You have identified a diabetic mutant strain of mice. Your studies indicate that mice that are heterozygous or homozygous for this mutation both have an identical diabetic phenotype. From these findings, you propose the following three models to explain the effect of this mutation: Model 1: The mutation is haploinsufficient. Model 2: The mutation acts in a dominant-negative fashion. Model 3: The mutation acts through a gain-of-function mechanism. To distinguish your three models, you map and clone the gene responsible for this genetic form of diabetes, which you name DIA1, and then proceed to generate constructs that will allow you to study this gene further in mice. The following linear DNA fragments, all of which contain portions of the mouse DIA1 gene (exons are darkly shaded) and other DNA sequences that may be important in generating your desired mouse strains are available for this analysis:

You obtain homozygous C/C ; DIA1/DIA1 embryonic stem cells (ES cells; note that the tyrosinase genotype of the ES cells is the opposite of what was discussed in class) and now proceed to create two different genotypes of mice. Assume for this question that you have the appropriate culture medium and also have access to neomycin and gancyclovir. a). Which of the above versions of the gene would you use to create a knockout of the mouse DIA1 gene (dia1KO)? Briefly describe how you would select for cells that have a knockout, and why your strategy would work? (8 points) ES cells would be transformed with Construct E and selection would be performed in media containing neomycin and gancylovir. Cells that survive would have had to undergo a homologous recombination event in which the normal DIA1 gene had been replaced with a construct bearing the neor gene, but lacking the gans gene. This selection strategy would eliminate cells in which an intact construct E had inserted randomly in the genome, as such cells would be sensitive to gancyclovir. b). Which of the above versions of the gene would you use to create a mouse with three wild type copies of the mouse DIA1 gene (i.e., one extra copy)? Briefly describe how you would select for cells that have a third copy of the gene, and why your strategy would work? (8 points) Construct B would be used to transform ES cells and selection would be performed in media containing just neomycin. This would select for cells that had acquired the entire construct at a random location in the genome. Cells that undergo a homologous recombination event would lose the neor gene and would be sensitive to neomycin.

neoR: confers resistance to the drug neomycin ganS: confers sensitivity to the drug gancyclovir


Problem 7 continued: You follow standard procedures to make mice that are heterozygous (C/c ; DIA1/dia1KO) as described in the figure at right. You now proceed to cross these mice to one another to create homozygous knockout animals. e). What progeny would you expect (in terms of coat color and diabetes phenotype), and in what proportions given the three genetic models that you are considering? If a particular model does not predict a diabetes phenotype for the knockout animals, you may assume that the knockout animals behave like wild type mice. (12 points)

Genotype

Expected

ratios

Phenotype (MODEL 1)

haploinsufficient

Phenotype (MODEL 2) dominant-negative

Phenotype (MODEL 3)

Gain-of-function

C/C; dia1KO/dia1KO

1 Black; diabetic Black; diabetic Black

C/C; DIA1/dia1KO

2 Black; diabetic Black Black

C/C; DIA1/DIA1

1 Black Black Black

C/c; dia1KO/dia1KO

2 Black; diabetic Black; diabetic Black

C/c; DIA1/dia1KO

4 Black; diabetic Black Black

C/c; DIA1/DIA1

2 Black Black Black

c/c; dia1KO/dia1KO

1 White; diabetic White; diabetic White

c/c; DIA1/dia1KO

2 White; diabetic White White

c/c; DIA1/DIA1

1 white White White


8. You are using yeast as a model organism to study the frequency distribution of particular alleles in a population over time. You have generated several different diploid strains of yeast that are each heterozygous for a particular mutation. To perform your experiments, you inoculate growth media with diploid yeast cells of opposite mating type that are each heterozygous for the same mutation (so that they can randomly mate with one another over time). You then proceed to monitor the allele frequencies in the growth media over many generations. These experiments are then repeated multiple times for each combination of alleles tested. Show the results that you would expect to see from these analyses given the following effects of the mutant allele (in each case the Y-axis designates the frequency of the wildtype allele): Briefly (1 SENTENCE) justify your answers to the right of each diagram. a). Neutral (4 points)

b). Negative (4 points)

c). Positive (4 points)

Problem 8 continued:

In the case of neutral selection, the allele frequencies will drift randomly from their mean starting frequencies.

In the case of negative, or purifying selection, the allele with the negative phenotype (the mutant allele in this case) will decrease from its beginning frequency in the population until it gets to a very low frequency or reaches extinction.

In the case of positive, or Darwinian selection, the allele with the positive phenotype (the mutant allele in this case) will increase from its beginning frequency in the population until it gets to a very high frequency or becomes fixed.


Problem 8 continued: All three of the mutations that you studied in parts (a)-(c) above are missense alleles of the same gene. An alignment of a portion of the yeast protein sequence from this gene with the corresponding protein sequences from the fruit fly and human proteins is shown below: Yeast: …AGNNKGICPEPINLKIFSTHVVNLL… Drosophila: …AGNNKGINRQPINLKIFSPIDLEAR… Human: …AGNNKGILASPINLKIFSEWGNRKT… Assuming that the mutations responsible for your findings in parts (a)-(c) above change one amino acid in the portion of the yeast protein sequence shown in the above alignment, show where (by circling) the mutation is likely to reside in each case. Briefly justify your answers (NO MORE THAN 1-2 SENTENCES) d). Neutral: AGNNKGICPEPINLKIFSTHVVNLL (3 points) Sequences that are not evolutionarily conserved typically play less important roles in protein function and can often undergo change without an associated phenotype. e). Negative: AGNNKGICPEPINLKIFSTHVVNLL (3 points) Sequences that are evolutionarily conserved typically play important roles in protein function and when these sequences change they typically result in negative phenotypes. You have identified one additional yeast mutant and proceed to perform the same analysis with this mutant as you did in parts (a)-(c) above. The results of this analysis are shown below:

f). What do these findings tell you about this particular mutation? (6 points) This particular mutation appears to confer a negative phenotype when homozygous and/or as a haploid, but the mutation never reaches extinction. This is an indication of heterozygous advantage, or balancing selection: the heterozygotes have better fitness than the WT or mutant homozygotes, but the mutation homozygotes are least fit.

final exam (a) name genome 371, spring 08faculty.washington.edu/pallanck/course materials...page 5...

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