7.03 FALL 2009 PROBLEM SET 1 Due: September 23, 2008web.mit.edu/7.03/documents/2009FA/2009FA-PSets.pdfWhen you performed this cross, all the F1 progeny were black. These all have the

7.03 FALL 2009 PROBLEM SET 1

Due: September 23, 2008 1. (15 points) You have isolated a collection of yeast mutants that form dark tan colonies (wild type yeast are white). Dark mutants 1–5 are MATa and mutants 6–10 are MATα. Your analysis begins by pairwise mating of each mutant to a wild-type strain and to the mutants of the opposite mating type. The color of the colonies of the resulting diploids are shown in the table below.

a \ α Wildtype Mutant 6 Mutant 7 Mutant 8 Mutant 9 Mutant 10 Wildtype White White White Dark White White Mutant 1 White Dark White Dark White Dark Mutant 2 White White White Dark White White Mutant 3 White White White Dark White White Mutant 4 White Dark White Dark White Dark Mutant 5 White White Dark Dark Dark White a) Which of the mutants are dominant and which are recessive (2ps)? b) Organize the mutants into complementation groups (genes), indicating any ambiguities. (9 pts) c) Based on these complementation data, what is the minimum number of genes represented by this collection of dark mutants? What is the maximum number of genes? (4 pts) 2. (18 points) In a large-scale breeding population, two female flies with vestigial wings (short-winged flies) arise from different parents. You would like to know whether the two mutations that caused this vestigial wing phenotype are in the same gene or in different genes. Assume that you have an unlimited number of true-breeding flies with normal wings. Describe a set of crosses you would perform to make this determination and their outcomes. In your description, please indicate a set of circumstances that would prevent you from easily making this determination.

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3. (32 points) Say fur color in mice is determined by 2 genes. One gene determines what color pigment is produced, and another gene determines whether or not any pigment is produced at all. Suppose black fur is dominant to brown fur, but a mutation in the other gene prevents pigment of either color from being produced so the mice are white. You cross a true-breeding brown mouse with a true-breeding white mouse. a) Determine the ratio of each color mouse in the F2 generation if the F1 progeny are black. (4 pts) b) Determine the ratio in the F2 generation if the F1 progeny are white (Assume the parental white mice had only one mutation). (4 pts) c) When you performed this cross, all the F1 progeny were black. These all have the same genotype, but the black mice in the F2 generation have more than one genotype. You wish to determine the genotype of one of the black F2 mice, so you back cross to a heterozygous F1 mouse. For each possible genotype determine the expected ratio of fur colors in the progeny. (10 pts) d) You get 40 progeny from your test cross, of which 32 are black and 8 are brown. You think you know the genotype but your lab partner thinks it could still be two genotypes. Use the chi square test to determine if you can reject one of the remaining hypotheses. The table below gives chi square values for 1, 2 and 3 degrees of freedom. Use the convention that for p < 0.05 there is a statistically significant difference between the observed results and the results expected for a given model and therefore we can reject the model on the basis of the experimental data. (14 pts)

p value: .995 .975 0.9 0.5 0.1 0.05 0.025 0.01 0.005 df = 1 .000 .000 .016 .46 2.7 3.8 5.0 6.6 7.9 df = 2 .01 .05 .21 1.4 4.6 6.0 7.4 9.2 10.6 df = 3 .07 .22 .58 2.4 6.3 7.8 9.3 11.3 12.8

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4. (31 points) Consider a cross of two mice whose genotypes for five independently segregating traits are Strain 1: DdEEFFGgHH and Strain 2 DdEeffGgHH. Capital letters indicate the dominant alleles, and the all of the traits are independently scoreable.

(a) How many different types of gametes can each parent produce? (4 pts) (b) How many different phenotypes can result from this cross? How many different genotypes? (9 pts) (c) What fraction of the progeny will be phenotypically identical to the first parent? To the second parent? (4 pts) (d) i. What fraction of the progeny will be genotypically identical to the first parent? What fraction will be identical at four of the five loci? (4 pts) ii. What fraction of the progeny will be genotypically identical to the second parent? At four loci? (4 pts) (e) One of the progeny of the above cross has the phenotype "d" "E" "F" "g" "H". What cross should you perform to determine the genotype? (6 pts)

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5. (32 points) The following is a pedigree from a family with a rare, autosomal recessive disorder. In this question, R represents a dominant allele and r represents a recessive allele. Please note that this imaginary pedigree is from a society in which brother-sister marriages are common.

B A

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(a) What is the probability that individual A has genotype Rr? (2 pts)

(b) What is the probability that individual E has genotype Rr? (2 pts)

(c) i. What is the probability that G, when born, will be affected? (6 pts)

ii. What is the probability that G will have genotype Rr (9pts)?

Suppose that individual G is born, and does not have the disease.

(d) What is the probability that individual G has genotype Rr? (4 pts)

(e) What is the probability that individual E has genotype Rr, given that G has genotype Rr? (9 pts)

D C E F

Not affected

Affected

G?

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Due: October 2, 2009 1. (24 points) You are studying three autosomal mutations in the fruit fly Drosophila melanogaster. The cw– mutation gives the recessive “curly-wing” phenotype (wild-type flies (cw+) are straight-winged), the sh– mutation gives the recessive “short-bristled” phenotype (wild-type flies (sh+) are long-bristled), and the nh–mutation gives the recessive “no-bristled” phenotype (wild-type flies (nh+) are long-bristled). a) You mate a homozygous nh– cw+ sh– fly (which is no-bristled and straight-winged) to a true-breeding long-bristled and curly-winged fly to obtain an F1 generation. List all phenotypic categories of the flies in the F1 generation and the ratio in which these phenotypic categories are found in the F1 generation. The possible phenotypic classes are: short-bristled and curly-winged, long-bristled and straight-winged, short-bristled and straight-winged, long-bristled and curly-winged, no-bristled and curly-winged, no-bristled and straight-winged. b) To do a proper three-factor cross, you need to create an F2 generation by mating F1 females to a specific true-breeding male. What is the phenotype of the true-breeding male fly you would choose as the parent for the F2 generation?

You do a proper three-factor cross and, in the F2 generation, you obtain the following number of flies in each of the corresponding phenotypic classes:

short-bristled and curly-winged (230 flies) long-bristled and straight-winged (10 flies) short-bristled and straight-winged (18 flies) long-bristled and curly-winged (238 flies) no-bristled and curly-winged (28 flies) no-bristled and straight-winged (476 flies)

c) All of the long-bristled and straight-winged flies have the alleles sh+ nh+cw+ on the chromosome they inherited from their mother. Using this notation, list all of the possible chromosomes that the no-bristled and curly-winged flies could have inherited from their mother.

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d) Are sh– and nh– linked? If so, give the distance between them in cM. (HINT- for the no-bristled flies estimate the number of each possible genotype by matching to the reciprocal classes) e) At this point, can you conclude whether sh– and nh– are in the same gene or not? If so, state whether sh– and nh– are in the same gene or in different genes. f) Are sh– and cw– linked? If so, give the distance between them in cM. g) Are nh– and cw– linked? If so, give the distance between them in cM.

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2. (15 points) Wild-type yeast make white colonies. You have isolated two mutants that make red colonies, which you call red-1 and red-2. A red1 haploid mutant is crossed to a red2 haploid mutant. The resulting diploid is induced to sporulate, and 100 resulting tetrads are analyzed where all resemble one of the types below. A dark circle indicates a red colony, and a white circle indicates a white colony: Types 1 2 3

(a) Identify types 1, 2, and 3? (b) Of the 100 tetrads analyzed, 10 are of type 1, 5 are of type 2, and 85 are of type 3. Are the red1 and red2 loci linked? If so, how far apart are they in cM? One of the tetrads from above is selected for further testing. Each of the four spore clones is mated to a wild-type haploid yeast. The phenotypes of the resulting diploids are shown below:

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(c) When a diploid of type 3 is induced to sporulate, what will the tetrads look like with respect to red and white phenotypes? (d) A second tetrad from part (a) is chosen, and each of the four spore clones is again mated to a wild-type haploid yeast. In the diagram below, fill in the expected phenotypes of the resulting diploids. State any ambiguities that may exist.

3. (14 points) A Neurospora cross was made between a strain that carried the mating-type allele A and the mutant allele arg-1 and another strain that carried the mating-type allele a and the wild-type allele arg-1+. Four hundred linear octads were isolated, and they fell into the seven classes given in the following table:

1 2 3 4 5 6 7 A arg-1 A + A arg-1 A arg-1 A arg-1 A + A + A arg-1 A + A + a arg-1 a + a arg-1 a arg-1 a + a arg-1 a arg A + A arg-1 A + A arg-1 a + a arg-1 a + a + a + a arg-1 a +

127 125 100 36 2 4 6

A) Deduce the linkage arrangement of the mating-type locus and the arg-1 locus. Include the centromere or centromeres on any map that you draw. Label all intervals in map units. B) Diagram that meiotic divisions that led to class 6.

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4. (20 points) Wild-type Drosophila have red eyes, and white eyes is an X-linked recessive phenotype caused by a single mutation. A new single mutation that gives the recessive phenotype of apricot colored eyes is isolated. A female from a true-breeding apricot-eyed strain is crossed to a male from a true-breeding white-eyed strain. All of the resulting F1 flies have apricot eyes. (a) Are the white-eye and apricot-eye mutations in the same gene or in different genes? Explain your answer. A number of apricot-eyed F1 females from the cross, described above, are mated to males from a true-breeding white-eyed strain, and 1000 male progeny are examined. Among these progeny, only 6 flies have normal red eyes. (b) What is the measured distance between the white-eye and apricot-eye loci in cM? A new mutation is isolated that causes the recessive eye color “peach.” A female from a true-breeding peach-eyed strain is crossed to a male from a true-breeding white-eyed strain. All of the resulting F1 females have normal red eyes and all of the resulting F1 males have peach eyes. (c) Is the peach-eye mutation on an autosome or on the X-chromosome? Explain your answer.

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(d) Are the white-eye and peach-eye mutations in the same gene or in different genes? Explain your answer. A mutation that causes the recessive phenotype of crossveinless wings lies on the X-chromosome. A female from a true-breeding strain with apricot eyes and crossveinless wings is crossed to a male from a single mutant true-breeding strain with white eyes and normal wings. As expected, all of the F1 females from this cross have apricot eyes and normal wings. Many of these F1 females are crossed to wild-type males and 10,000 male progeny are examined. The observed phenotypes are as follows:

Phenotype Number Normal wings, white eyes 4,418 Crossveinless wings, apricot eyes 4,330 Normal wings, apricot eyes 610 Crossveinless wings, white eyes 590 Normal wings, red eyes 2 Crossveinless wings, red eyes 50

(e) Draw a genetic map showing the relative order of the crossveinless, apricot and white loci. Explain your reasoning.

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5. (12 points) Draw the meiotic segregation of these two chromosomes, starting with replication, then the two divisions. Note which division is reductional and which is equational.

A) Draw out the possible gametes that can result if chromosomal disjunction occurred during the reductional division for one of the chromosomes above.

B) Draw out the possible gametes that can result if the equational division was done reductionally by mistake for one of the chromosomes above.

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7.03 FALL 2009

PROBLEM SET 3 Due: October 19, 2009

1. (30 pts) Streptococcus pneumoniae secretes a short signaling peptide called ComC that is

thought to induce competence—the uptake of DNA from the environment—in bacterial

populations when the extracellular levels of this peptide reach a critical concentration. A

mutated version of this gene is shown below, beginning with the transcriptional start site and

ending with the transcriptional terminator sequence. S. pneumoniae populations that express this

mutated form of ComC are unable to uptake DNA.

5’ - ATCTTTCTCGGATGAAAAACACAGTTAAATTGGAACAGTTTGTAGCCTTGAAGGAAAAAGACTTGTAAAAGAT – 3’

3’ - TAGAAAGAGCCTACTTTTTGTGTCAATTTAACCTTGTCAAACATCGGAACTTCCTTTTTCTGAACATTTTCTA – 5’

a) Which strand—the upper or the lower—is used as the template in transcription? Why?

b) Write out the entire sequence of the mRNA made from this gene.

c) What is the peptide sequence encoded by this mRNA?

d) You learn that S. pneumoniae bacterial populations that can successfully uptake DNA

secrete a ComC peptide that is 41 amino acids long.

i) Briefly indicate a reason why the mutated peptide you translated above might

prevent S. pneumoniae populations that express this mutated gene from uptaking

DNA.

ii) If a single basepair substitution led to the mutated peptide, what kind of mutation

has occurred? Indicate on the sequence below where this mutation occurred.

5’ –ATCTTTCTCGGATGAAAAACACAGTTAAATTGGAACAGTTTGTAGCCTTGAAGGAAAAAGACTTGTAAAAGAT– 3’

3’ –TAGAAAGAGCCTACTTTTTGTGTCAATTTAACCTTGTCAAACATCGGAACTTCCTTTTTCTGAACATTTTCTA – 5’

ii) If the mutated base was originally a ‘C,’ what type of substitution has occurred?

Indicate all possibilities.

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iii) If a single basepair substitution did not occur, describe two other mutational

events that could have produced this mutated peptide sequence.

Upon further investigation you find that this mutation was caused by a single basepair

substitution. However, you discover a population of S. pneumoniae bacteria that contain this

mutated ComC sequence but can successfully uptake DNA. You suspect that nonsense

suppression may be at play, so you sequence the population’s tRNAs. Below is the sequence of

a wild type copy of the tRNA-Gln gene that recognizes the 5’ – CAA – 3’ codon on all mRNAs

in the bacteria. This sequence begins at the transcriptional start site and ends with the

transcriptional terminator sequence.

5’-CCCGTAGCTCAGATCTGGATATCCATCCTGCATGCATAGCTTGCTCATGCTGATACGCGCTAGGGT-3’

3’-GGGCATCGAGTCTAGACCTATAGGTAGGACGTACGTATCGAACGAGTACGACTATGCGCGATCCCA-5’

e) Which strand (the upper or the lower) is used as the template in transcription?

(Remember that tRNAs are directly transcribed from tRNA-encoding genes. There is no

mRNA intermediate in the production of a tRNA molecule from a tRNA gene!)

f) Put a box around the double-stranded DNA portion of the wild type tRNA gene that

encodes the anticodon portion of the tRNA on the sequence below: 5’-CCCGTAGCTCAGATCTGGATATCCATCCTGCATGCATAGCTTGCTCATGCTGATACGCGCTAGGGT-3’


You discover that the tRNA-Gln gene in the above population is mutated.

g) Given the information you know about the mutated ComC peptide, indicate on the

sequence below the most likely mutation that would result in nonsense suppression in this

population.

5’-CCCGTAGCTCAGATCTGGATATCCATCCTGCATGCATAGCTTGCTCATGCTGATACGCGCTAGGGT-3’


2. (14 points) Wild type Drosophila have red eyes. Your colleague is working with a pure

breeding stock of white-eyed Drosophila and obtained a revertant that had red eyes. She then

created a pure breeding stock of this revertant and sent it to you for analysis. She assures you

that the strain is homozygous at all loci. You want to distinguish between an intragenic revertant

(a reversion of the original white eye mutation back to wild type) or an extragenic suppressor (an

unlinked mutation that suppressed the original white eye mutation). In addition to the red-eye

revertant, you have stocks of wild type ( w+/ w+), and a pure breeding stock of the unreverted

(w/w) strain. In your answers below, use the following genetic symbols: w+ for wild type, w for

the mutation leading to white eye, s for wild type, and s+ for the mutation that leads to the

suppressor.

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a) What genotype would you expect if the red-eyed revertant strain is a pure breeding

intragenic revertant of the white-eyed strain?

b) What genotype would you expect if the red-eyed revertant strain still has the original

white mutation but is now homozygous for an unlinked extragenic dominant suppressor?

c) What two sequences of crosses would you perform to distinguish between an intragenic

and extragenic revertant? In your answer, provide the ratio of phenotypes and genotypes

you expect from each cross using the same genetic symbols as above.

3. (28 pts)

a) You would like to generate E. coli mutants that are auxotrophic for tryptophan. You have

Su+ and Su- E. coli strains and any type of media available to you. Explain how to use

Pam int-::Tn5 to generate Trp- auxotrophs.

b) You call one of your Trp- strains trp1- and you would like to map trp1- with respect to a

known gene gal1. You grow P1 phage on one of your Trp- strains and use the resulting

phage lysate to infect a gal1- strain, selecting for kanamycin resistance (Kanr). Among 100

Kanr transductants, you find that 68 are Gal+ and 32 are Gal–. What does this result tell you

about the relationship between the gal1– mutation and trp1?

c) What percentage of the Gal+ transductants will be Trp+?

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d) Using the same P1 lysate as in part (b) you infect another Gal– strain that you call gal2-,

selecting for Kanr

transductants. In this case, none of the 100 Kanr transductants are Gal+. Is

gal2 linked to trp1? What does this result tell you about the relationship between the gal1–

and gal2– mutations?

e) Preliminary P1 transduction experiments indicate that trp1 is linked to another Gal gene

gal3. Another mapping experiment shows that gal1 maps closer to gal3 than it does to trp1.

To find the relative order of trp1, gal1 and gal3 you set up two reciprocal crosses. In the first

cross you grow P1 on a strain that carries the Tn5 trp1 insertion and the gal1– mutation. You

then use this lysate to infect a gal3– mutant and select for Kanr. From 100 Kan

r transductants

examined, 96 are Gal– and 4 are Gal+. In the second cross you grow P1 on a strain that

carries the Tn5 trp1 insertion and the gal3– mutation. You then use this lysate to infect a

gal1– mutant, and select for Kanr. From 100 Kan

r transductants examined, 84 are Gal– and

16 are Gal+. Draw a genetic map showing the relative positions of the trp1- and the gal1–

and gal3– mutations. Express any measured distances as co-transduction frequencies.

f) Explain why it is necessary to carry out two reciprocal three-factor crosses in part (d) in

order to determine the relative positions of the gal1– and gal3– mutations.

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4. (14 pts) The diagram below shows the F factor and a portion of the E. coli chromosome that

has three different insertion sequences (IS) of the same type as is carried on F.

a) Starting with the F+ strain diagrammed you isolate an Hfr strain that can transfer the B

marker early, but transfers C very late. Show the recombination event that would produce

this Hfr and the final structure of the Hfr, showing all of the markers as well as the

position and orientation of each IS sequence and the origin of transfer (ori T).

b) Using the Hfr described in part a, you wish to isolate an F’ factor that carries the C gene.

Show the recombination event that would produce the desired F’ and the final structure of

the F’. Describe in general terms how you could select for the desired F’.

7.03 Problem Set 4 Due October 30, 2009

1. (24.5 points) You have discovered two mutations in the Lac operon and name them Lac-1 and Lac-2. You wish to characterize these mutations. For each mutant you apply a set of genetic tests that include i) testing the phenotype of the mutant itself, ii) the phenotype of the mutant in a merodiploid carrying an F’ with the wild type Lac operon (F’ Lac+), iii) the phenotype of each mutant in a merodiploid carrying an F’ with a Lac operon containing a LacZ- mutation (F’ LacZ-…), and iv) the phenotype of a wild type merodiploid carrying an F’ plasmid with Lac-1 or Lac-2 and LacZ-. Note: Neither lac-1 nor lac-2 are mutations in LacZ.

a) (4.5 points) The results are summarized in this chart. For the phenotype of each strain, fill in whether the strain has uninducible, constitutive, or inducible β-galactosidase activity

Strain - lactose + lactose Phenotype WT - + lac-1 + + lac-2 - - Lac-1/F' Lac+ + + Lac-2/F' Lac+ - + Lac-1/ F' LacZ- + + Lac-1 LacZ-/ F' Lac+ + + Lac-2/F' Lac-Z- - - Lac-2 LacZ-/ F' Lac+ - +

b) (2 points) What can you conclude about the Lac-1 mutation?

Recessive or dominant:

Cis- or trans-acting, with respect to LacZ (explain your reasoning):

c) (2 points) What type of mutation is Lac-1? Be specific, include what gene the mutation is in and what the consequences of the mutation are. Explain your reasoning.

d) (2 points) What can you conclude about the Lac-2 mutation:

Recessive or dominant:

Cis- or trans-acting, with respect to LacZ (explain your reasoning):

e) (2 points) What type of mutation is Lac-2? Include the same information as in part (c).

f) (4 points) From the information given, fill in the chart for phenotype of the merodiploid strains listed below. Explain your reasoning.

Strain -lactose +lactose Phenotype Lac-1 / F' Lac-2 Lac-1, Lac-2/ F' Lac+

g) (2 points) You discover a new mutation (Lac-3) and perform the same characterization as you did with Lac-1. Fill in the last column, as in part (a).

Strain -lactose +lactose Phenotype Lac-3 + + Lac-3/F'Lac-3 + + + Lac-3/F'LacZ- + + Lac-3 LacZ-/F'Lac+ - +

h) (2 points) What can you conclude about the lac-3 mutation?

Recessive or dominant: Cis- or trans-acting, with respect to LacZ (explain your reasoning):

i) (2 points) What type of mutation is Lac-3? Include the same information as in part (c).

j) (2 points) Fill in the following chart, giving the phenotypes of the merodiploid strains listed, using the information you have gathered about Lac-3 and Lac-2.

Strain -lactose +lactose Phenotype Lac-3 / F' Lac-2 Lac-2, Lac-3/ F' Lac+

2. (38 points) You have isolated an interesting mutation in Drosophila that you call sot1a— (short for seven-o-three1a—). You find that sot1a- mutant flies display behavior that indicates an increase in stress (excessive body shaking and chewing on limbs) as compared to wild-type flies. When you mate a pure-breeding sot1a— female with a pure-breeding wild-type male, you notice that all F1 progeny have the mutant phenotype (overly stressed). (a) (2 points) What is/are the possible mode(s) of inheritance of the sot1a — allele? Circle all possible choices. i. Autosomal dominant ii. X-linked recessive iii. Autosomal recessive iv. X-linked dominant v. Mitochondrial (b) (2 points) Next, you cross an F1 male with wild-type female. When you score the F2 progeny, half of them have the overly stressed mutant phenotype, regardless of gender. What is/are the possible mode(s) of inheritance of the sot1a— allele? i. Autosomal dominant ii. X-linked recessive iii. Autosomal recessive iv. X-linked dominant v. Mitochondrial To further characterize the pathway that leads to increased stress, you decide to test whether exposure to chemical compounds could lead to increased stress in wild-type flies, and found that the compound called “PSET” could induce stress. When wild-type flies are exposed to PSET they express a detectable protein called “stress factor”. The sot1a— flies express the “stress factor” even in the absence of PSET. Additionally, you generate a different mutation in the sot1 gene (sot1b—) and two other mutations (sot2— and sot3—) that are not in the sot1 gene, but may or may not be linked to it. When these three pure-breeding mutants were mated to wild-type flies, all F1 progeny showed normal response to PSET (no stress in the absence of PSET, and increased stress when exposed to PSET). Furthermore, when pure-breeding mutants were exposed to PSET, they showed the following behavior: Strain No PSET + PSET Wild-type Not stressed (no stress factor) Stressed (stress factor expressed)

sot1b—

sot1b—

Not stressed (no stress factor) Not stressed (no stress factor)

sot2—

sot2—


sot3—

sot3— Stressed (stress factor expressed) Stressed (stress factor expressed)

(c) (3 points) Characterize the phenotypes caused by the mutations sot1b—, sot2—, and sot3— in the following categories:

i. Dominant or recessive

ii. Uninducible or constitutive (with respect to expression of stress factor)

(d) (5 points) Next, you generate double mutants and determine the response of these doubly mutant strains to PSET. Specify the epistasis relationships between these mutations (i.e., for each strain observed below, specify which allele confers the epistatic phenotype). Remember that the sot1a— mutation was isolated because it caused flies to be stressed whether or not they were exposed to PSET. Strain No PSET + PSET sot1a— sot2—

sot1a— sot2—


sot1b— sot3—

sot1b— sot3—

Stressed (stress factor expressed)


sot2— sot3—

sot2— sot3—


(e) (5 points) Upon further investigation, you find that the sot2- mutation is in a gene (hereafter called sot2) that encodes “stress factor”. Stress factor directly causes stress when flies are exposed to PSET. You reason that the sot1a-, sot1b-,and sot3- mutations may be either in DNA sequences that directly regulate sot2 transcription or in genes that encode proteins that regulate sot2 transcription. To distinguish between these possibilities, you examine the following heterozygotes (note: alleles on the same side of the line belong on the same chromosome). Strain # Strain No PSET + PSET

1 sot1a— sot2—

+ +



2 sot1a— + + sot2—



3 sot1b— sot2—

+ +

Not stressed (no stress factor)


4 sot1b— + + sot2—



5 sot2— sot3—

+ +



6 sot2— + + sot3—



(i) Is sot1 cis- or trans-acting? Explain how you determined this from the given data (i.e. list which strain #’s put the relevant alleles in trans and which strain #’s put the relevant alleles in cis).

(ii) Is sot3 cis- or trans-acting? Explain how you determined this from the given data (i.e. list which strain #’s put the relevant alleles in trans and which strain #’s put the relevant alleles in cis).

(f) (9 points) Draw a genetic regulatory pathway for the genes defined by these mutations, using the genetic data gathered. Names the genes as follows: sot1 = the gene carrying the sot1a- and sot1b- mutations; sot2 = the gene carrying the sot2- mutation; and sot3 = the gene carrying the sot3- mutation. Include an explanation for the difference between the mutations sot1a— and sot1b—. (Note: assume that there’s only one linear pathway leading to the stress phenotype, such that relevant genes belong to a singly linear genetic regulatory pathway.)

(g) (12 points) Draw a gene regulatory model for the transcription of the sot2 gene that induces stress. Include in your model roles for the proteins SOT1 and SOT3, the molecule PSET, and any necessary regulatory DNA sequences upstream of sot2. Additionally, provide an explanation for differences in SOT1 protein function based on the differences between mutations sot1a— and sot1b—. 3. (25 points) You are interested in the regulation of a promoter of a carbohydrate-metabolizing enzyme, yummase, which digests the sugar yummose, in yeast P. seticus. You have developed an assay where you can gauge yummase activity. You know that two different compounds (yummose and maltose) impact whether the gene is turned on or off and aim to elucidate the mechanism of regulation. (a) (5 points) You begin by making a series of deletions in the region preceding the gene and test the response of yummase to yummose.

Enzyme Activity sugar present

Deleted region - yummose + yummose none 10 100

1 10 100 2 10 10 3 10 100 4 10 100 5 0 0

What can you conclude about the roles of regions 1-5 from this assay (i.e., specify a type of regulatory region where possible)? Be sure to specify if these mutations result in inducible, uninducible, or constitutive response to yummose in your explanation.

(b) (5 points) Knowing that maltose also plays a role in the transcription, and thus subsequent activity of the yummase enzyme, you extend your assay to include the presence and absence of maltose. The results are seen here:

Enzyme Activity sugars present

Deleted region none yummose only maltose only

maltose and yummose

none 10 100 1 1 1 10 100 1 1 2 10 10 1 1 3 10 100 1 1 4 10 100 10 100 5 0 0 0 0

Propose a model for regulation of expression of this gene that incorporates roles for regions 1-5 in gene regulation (if nothing can be concluded about a region, leave it out of the model). To do this, label the regions below with their appropriate roles, and include roles for yummose and maltose.

(c) (9 points) A friend in a different lab has stumbled across a series of mutations in a single gene, yumE, that affect yummose metabolism – seemingly by means of regulation. She names these alleles yumE-1, yumE-2, and yumE-3. yumE-1 is a deletion of the yumE gene. This gene maps to a chromosome that is different from yummase. You use the same assay as above and observe the following:

Enzyme Activity sugars present

mutation none yummose only maltose only

maltose and yummose

none 10 100 1 1 yumE-1 10 10 10 10 yumE-2 10 100 10 100 yumE-3 10 10 1 1

Propose a model that explains the effects the different mutations have on yumE activity. In your answer, explain the effects yummose and maltose have on yumE activity how yumE specifically regulates the promoter regions.

(d) (6 points) To test your hypothesis, you mate strains containing yumE-1, yumE-2, and yumE-3 to the deletion strains you constructed from parts A and B. You sporulate the diploids and confirm by PCR the presence of both a mutation, as well as a deletion. Fill in the values you would expect to see in the table below:

Enzyme Activity strain construction sugars present

Mutation Deletion none yummose only

maltose only

yummose and maltose

none none 10 100 1 1 yumE-1 2 yumE-1 5 yumE-2 2 yumE-2 4 yumE-3 2 yumE-3 4

(e) Analysis of Aus DNA in Japanese beetles revealed that of a large number of Wolbachia genes sought, only a few could be detected. Furthermore, more than half of the open reading frames identified in this Aus DNA contained stop codons or frame-shift mutations. Propose a brief hypothesis to account for these observations.


Due: November 16, 2009 1. (35 points) You are interested in a species of moth that comes in three colors – black (B), ivory (I) and grey (G). You are interested in the gene controlling this coloration, so you collect 1,000 of them from western Massachusetts. You collect 360 black, 550 grey and 90 ivory. A. i. (5 points) Calculate the allele frequencies if coloration is controlled by three distinct alleles in Hardy-Weinberg Equilibrium, where black is dominant to ivory and grey, and grey is dominant to ivory. A.ii. (5 points) Calculate the allele frequencies if coloration is controlled by two alleles, where the heterozygotes (grey) have an intermediate phenotype. A.iii. (5 points) Assuming random mating and absence of selection, what are the expected numbers of black, grey, and ivory moths you would have (out of 1000) given the two-allele model and frequencies you calculated in part ii? Is this population in Hardy-Weinberg Equilibrium? B.i. (2 points) You discover that there are only two alleles are responsible for color. You take a sample consisting of 500 black moths and 500 ivory moths to an island. What are the genotypic and phenotypic frequencies you expect to see after one generation of random mating?

B.ii. (6 points) You take the 1,000 moths, after equilibrium has been achieved, from the island in part B.i. and take them to another island where there are other animals. On this island, there are birds that are easily able to locate and predate the grey moths; the ivory moths are able to hide on lightly colored vegetation, and the black moths are able to hide on tree bark. Assume that the moths have the following fitness: WBB=WII=1, whereas WBI=0.3. What will be the allele frequencies in the adults of the next generation? What will the allele frequencies be at equilibrium? B.iii. (6 points) Suppose on this island that predators were more adept at finding the ivory moths, instead. Assume that the moths have the following fitness: WII= 0.3, WBI=0.3, and WBB=1.0. What will be the allele frequencies in the adults of the next generation? What will the allele frequencies be at equilibrium? B. iv. (6 points) Finally, suppose that on this island, the grey moths were best able to avoid predation. Predators most easily found the ivory moths, and the black moths had intermediate survival ability. Starting with the original population of moths from part B.i., assume that the moths have the following relative fitness: WII= 0.7, WBI=1.0, and WBB=0.8. What will be the allele frequencies in the adults of the next generation? What will the allele frequencies be at equilibrium (hint: at equilibrium the allele frequencies don’t change, so p’=p)?

2. (20 points) We are given the following family genotypic data where the father (155) and mother (154) have had six children (162, 156, 165, 158, 163, 160).

155 154 162 156 165 158 163 160 ------------------------------------------------------------------ rs726548 {g,a} {c,g}| {g,a} {a,c} {g,g} {a,g} {g,g} {c,g} rs1496328 {t,t} {t,g}| {t,t} {g,t} {t,t} {t,t} {t,t} {g,t} rs1016607 {c,a} {c,g}| {g,c} {c,c} {a,g} {g,c} {a,g} {c,a} rs896986 {g,c} {g,a}| {g,a} {g,g} {a,c} {a,c} {c,a} {c,g} rs1437747 {g,t} {t,t}| {t,t} {t,t} {t,g} {t,t} {t,g} {g,t} rs921115 {c,t} {c,a}| {c,c} {c,a} {c,t} {c,c} {t,a} {a,t} rs654135 {a,a} {g,c}| {a,c} {a,g} {a,c} {c,a} {c,a} {g,a}

a) (10 points) Compute the haplotypes of all of the individuals in this family. b) (6 points) Identify all of the potential points of meiotic recombination in each of the offspring by providing for each event what chromosome the recombination event occurred in (maternal or paternal) and the identity of the SNPs that bracket where the recombination event could have occurred. c) (4 points) The mother (154) has a rare dominant disorder, and you have linked this disorder to a haplotype where rs1016607 is A and rs896986 is C. Which of the offspring are likely to inherit this disorder?

3. (16 points) In a set of 700 gene expression experiments from S. Cerevisiae the expression levels of transcription factors are quantized into two levels (High and Low) for analysis. Your friend provides you with the following summary for the genes G1 and G2:

G2 High G2 Low G1 High 190 150 G1 Low 170 190

a) (4 points) If G1 and G2 are independent (but each occurring in a high or low

state with different probability) what would you have expected this table to be?

b) (4 points) How likely is it that the observed data occurred by chance

assuming that G1 and G2 are independent? (You can find a relevant Chi-Square Table via Google).

c) (4 points) Assuming the data did not occur by chance, what can you tell about the relationship between G1 and G2? Is one an activator or repressor of the other?

d) (4 points) You are now told that your friend examined all pairs of 10

transcription factors to identify G1 and G2 as having a potential regulatory relationship. How would you correct your estimate in (b) given this new information?

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DUE: November 30, 2009 1. (28 points) You are interested in studying a rare autosomal recessive genetic disorder.

You have identified two different families with this disorder and have constructed a pedigree for each family for your analysis. Preliminary findings suggest that the gene causing this disorder may be linked to a specific SSR – SSR2255. This SSR locus has four alleles, each with a different number of repeats. The pedigrees are listed below (affected individuals are darkened and the SSR genotype is listed below each individual).

Family 1 Family 2

a) In each family, which parent(s) is/are informative for calculating the LOD score? b) Calculate the LOD scores for θ=0.05, 0.1, 0.2, and 0.4 for Family 1. Do any of these LOD

scores suggest linkage between SSR2255 and the disease locus?

2, 4

1, 4 2, 3

1, 2 1, 3 1, 3 2, 4 1, 2 1, 4 1, 4 1, 4

2, 4 1, 1

2

c) Calculate LOD scores for θ=0.05, 0.1, 0.2, and 0.4 for Family 2. Do any of these LOD scores suggest linkage between SSR2255 and the disease locus?

d) Is it appropriate to combine LOD scores from Families 1 and 2? Explain. If yes, can you

conclude that SSR2255 and the disease locus from the combined LOD scores?

e) Disregarding information contributed by Family 2, what is the minimum number of independent families exactly like Family 1 (in terms of the disease and SSR2255 inheritance patterns) that you must analyze to be able to conclude that SSR2255 and the disease gene are linked? For which value(s) of θ can you make this conclusion?

f) You now learn that for Family 1, the father is unaffected and had an affected father with

genotype {5,4}. Recompute the LOD score for Family 1 given this new information for θ = 0.3.

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2. (17 points) You are studying the human skin cancer called melanoma. You obtain biopsy samples from 50 melanoma patients and 50 unaffected individuals and perform RNA-Seq with mRNA isolated from these samples. After analyzing this expression data, you find that a gene of uncharacterized function, which you call Human Melanoma-Associated 1 (HMA1), exhibits no expression in 48 of the 50 patients analyzed whereas it is normally expressed in unaffected individuals. Preliminary analysis of the gene’s sequence reveals the gene structure depicted below:

You find that the structure of the homologous mouse gene (MMA1) is highly similar to the structure of the human gene, so you decide to develop a knockout mouse model to study the effects of deleting the third exon of this gene. You want to employ a homologous recombination approach using embryonic stem cells derived from a true-breeding, white mouse strain.

a) Diagram a homologous recombination strategy that would allow you to delete exon 3 while keeping exons 1, 2, 4, and 5 intact. Assume that deleting exon 3 causes loss of gene function. Depict the features of your targeting construct that allow selection for integration of the construct by homologous rather than non-homologous integration.

b) You decide to use PCR to check that your method has worked. Describe the primers

you could use in a PCR reaction to confirm that your targeting vector has inserted into the correct location in the genome (i.e. describe which regions of DNA you would design your primers to hybridize).

c) You confirm that your strategy above successfully deleted Exon 3 by PCR. To generate a chimeric mouse, you inject the cells you isolated into a blastocyst derived from a true-breeding, black mouse and obtain the mouse below:

EXON1 EXON2 EXON3 EXON4 EXON5

Intron1 Intron2 Intron3 Intron4

4

i) Which region(s) indicated above are derived from your ES cells?

ii) Assuming that the mouse HMA1 gene (MMA1) acts in the same way as the human HMA1—loss of gene function causes cancer—do you anticipate that this mouse will have cancer? Why or why not?

To generate a mouse line that has a heterozygous deletion of Exon 3 of MMA1, you cross the mouse above to a wild type black mouse. In these mouse strains, black and white alleles are incompletely dominant, and so mice heterozygous at this locus are gray.

d) Which classes of progeny mice might you expect to see? For each class, indicate the genotype at the MMA1 locus.

e) Of the progeny you observe in part (d), which mice would you cross to obtain mice that are homozygous for the MMA1 exon 3 deletion, and at what frequency would they occur?

3. (18 points) In order to visualize different neurons in the mouse brain, researchers have utilized fluorescent proteins and the Cre-LoxP gene targeting system in designing the construct below, the “brainbow” construct:

Four genes encoding fluorescent proteins of different colors—Green FP, Yellow FP, Red FP, and Cyan FP—are flanked by three sets of incompatible LoxP sites (each LoxP site can recombine with its homologous LoxP site but not with the other two LoxP sites). At the end of the coding region for each fluorescent protein is a poly-adenylation sequence (pA). Translation of the mRNA produced from this construct will stop at the polyA sequence and thus only one fluorescent protein will be transcribed from this locus at a time.

GFP YFP RFP CFP

LoxP1

LoxP2

LoxP3

pApApApA

5

This construct is integrated into the genome of mouse cells, and controlled by a constitutively active promoter. Cre recombinase is also constitutively transcribed at a different site in the genome, and the Cre protein is present in an inducible form (called Cre-ER) such that when the inducer tamoxifen is present, Cre-mediated recombination will occur.

a) What color cells will you see in the absence of tamoxifen? Briefly explain your answer.

b) What color cells will you see in the presence of tamoxifen? Indicate all possibilities, and draw the constructs that will produce such colors.

c) You generate a mouse that has two brainbow constructs integrated into its genome. You find that each of these constructs is transcribed equally. The recombination event at each construct, mediated by Cre in the presence of tamoxifen, occurs independently. Therefore a given cell may express two different fluorescent proteins, and the color cells exhibit depends on the different combination of fluorophores expressed, as detailed in the table below. You record the colors observed in 1,000 cells.

i. What is the ratio of phenotypes (cell colors) you would observe in the presence of tamoxifen if p(no recombination) = p(recombination at loxP1) = p(recombination at loxP2) = p(recombination at loxP3) = 0.25? Use the color table below in formulating your answer.

+ Green Red Yellow Cyan

Green Green Yellow Lime Green Turquoise

Red Yellow Red Orange Purple

Yellow Lime Green Orange Yellow Green

Cyan Turquoise Purple Green Cyan

ii. What is the ratio of phenotypes you would observe in the presence of tamoxifen if p(no recombination) = p(recombination at loxP1) = 0.1 and p(recombination at loxP2) = p(recombination at loxP3) = 0.4?

Documents

7.03 FALL 2009 PROBLEM SET 1 Due: September 23, 2008web.mit.edu/7.03/documents/2009FA/2009FA-PSets.pdfWhen you performed this cross, all the F1 progeny were black. These all have the