strains ofmating type a:

Problem set questions from Exam 1 Unit – Basic Genetic Tests, Setting up and AnalyzingCrosses, and Genetic Mapping

Basic genetic tests for complementation and/or dominance

1. You have isolated 20 new mutant yeast strains that are defective in synthesis ofthreonine, an amino acid. These Thr- mutants do not grow on minimal medium, butthey do grow on minimal medium supplemented with threonine. Ten of your Thr-mutants (numbered 1 through 10) were isolated in a strain of mating type alpha (MATα). The other 10 Thr- mutants (numbered 11 through 20) were isolated in a strain ofmating type a (MAT a). You cross each of the MAT a haploid strains to each of theMAT α haploid strains, and you include crosses to the appropriate wild-type haploidstrains. Your experimental observations are shown in the table below, where (-)indicates diploids that did not grow on minimal medium and (+) indicates diploids thatdid grow on minimal medium (that had NOT been supplemented with threonine).

(a) Unfortunately, when all the data were collected and the plates discarded, you lostsome of your data. From the data that remains, see if you can reconstruct the full table.

strains of mating type α

wild-type 1 2 3 4 5 6 7 8 9 10wild-type + + + + + - +

11 - + +12 + - + - -13 - + + + +14 - - - - -15 - + - +16 + - + - -17 + - + - -18 + + + - - +19 + + -20 + - + + -

(b) Which mutations give recessive phenotypes?

(c) Which mutations give dominant phenotypes?

(d) Which mutations do you know to be in the same gene?

(e) Based on these experiments, what is the minimum number of genes required forthreonine synthesis?

(f) What is the maximum number of genes that these 20 mutants could represent?

2. You have isolated a set of five yeast mutants that form dark red colonies instead ofthe usual white colonies of wild-type yeast. You cross each of the haploid mutants to awild-type haploid strain and obtain the resulting diploids shown below.

Mutant 1

Xwild-type diploid

Mutant 2

Xwild-type diploid

Mutant 5

Xwild-type diploid

Mutant 4

Xwild-type diploid

Mutant 3

Xwild-type diploid

(a) What do these results tell you about each of the mutants?

(b) Next you cross each haploid mutant strain to a different haploid mutant of theopposite mating type. From the results shown below deduce as much as you can abouthow many different colony color genes you have isolated mutations in, and whichspecific mutations lie in the same gene.

(c) Clearly state any remaining ambiguities.

Mutant 1

XdiploidMutant 2

Mutant 1

XdiploidMutant 3

Mutant 1

XdiploidMutant 4

Mutant 1

XdiploidMutant 5

Mutant 2

XdiploidMutant 3

Mutant 2

XdiploidMutant 4

Mutant 2

XdiploidMutant 5

Mutant 3

XdiploidMutant 4

Mutant 3

XdiploidMutant 5

Mutant 4

XdiploidMutant 5

3. You have isolated five new Arg– mutants in yeast. Each mutant cannot grownunless the amino acid arginine is provided in the growth medium. After obtainingversions of each mutant in mating type a and in mating type α, you perform all of thepossible pairwise matings shown in the table below. A "+" at the intersection of the twoparental strains indicates that the diploid can grow without arginine added to themedium, whereas a "–" indicates that the diploid can’t grow without arginine.

(a) Give as much information as you can about your new Arg– mutants. Indicate whichmutants have dominant Arg– phenotypes, and which have recessive Arg– phenotypes.Also state how many genes are represented by your collection of five mutants, andwhich mutations lie the same gene. Assume each strain carries only a single Arg–mutation.

(b) The amino acid permease that allows cells to take up arginine will also transportcanavanine, which is a toxic analog of arginine. Mutations that interfere with thispermease’s transporter activity can block canavanine uptake and will therefore allowcells to grow in the presence of an amount of canavanine that would kill a wild-typeyeast cell. You have isolated a set of five canavanine resistant mutants. As before, youobtain versions of each mutant in mating type a and mating type α, and you perform allof the possible pairwise matings shown in the table below. In this table a "+" at theintersection of the two parental strains indicates that the diploid can grow in thepresence of high levels of canavanine, whereas a "–" indicates that the diploid is assensitive to canavanine as wild-type.

1

2

3

4

5

+– – –+–

–

++

–

–+–

–

1mutants of mating type α

2 3 4 5

mutants ofmating type a

–

Give as much information as you can about your new CanR mutants. Indicate whichmutants have dominant CanR phenotypes, and which have recessive CanRphenotypes. Also state how many genes are represented by your collection of fivemutants, and which mutations lie the same gene. Assume each strain carries only asingle CanR mutation.

4. You have isolated twenty His– yeast mutants. Each single mutant cannot growunless the amino acid histidine is supplemented in the growth medium. Mutants 1 – 10are of mating type α and mutants 11 – 20 are of mating type a. You performedcomplementation tests by mating each α strain to each a strain. The table below showsthe results, where a "+" at the intersection of the two parental strains indicates that thediploid can grow without histidine added to the medium, and a "–" indicates that thediploid can’t grow without histidine. Unfortunately, after all of the data are collected,some of the data is lost.

(a) From the data that remains, see if you can reconstruct the full table that follows.

1

2

3

4

5

+ – –+–+

–+

1mutants of mating type α

2 3 4 5

mutants ofmating type a

++

++

+

+–

(b) Determine how many different genes are represented by your collection of His–

mutants. For your answer, you should indicate which mutations fall into the samecomplementation group. Any remaining ambiguities in the assignment of mutations tocomplementation groups should be explicitly stated.

Human Pedigrees and Probability

1. The following pedigree shows the segregation of two different rare recessive traits.Assume no new mutations and complete penetrance.

(a) Assuming that the two traits are due to unlinked autosomal genes, calculate theprobability that the indicated child will have both recessive traits.

(b) Assuming that the two traits are due to linked autosomal genes that are 10 cM apart,calculate the probability that the indicated child will have both recessive traits.

(c) Assuming that the two traits are due to X-linked genes that are 10 cM apart,calculate the probability that the indicated child will have both recessive traits, if thatchild is born male.

(d) Assuming that the two traits are due to X-linked genes that are 10 cM apart,calculate the probability that the indicated child will have both recessive traits, if thatchild is born female.

= female showing trait 1 only

= male showing trait 2 only

?

= male showing both traits

2. Each of the families below exhibits a different very rare genetic disorder whereindividuals expressing the disorder are shown by solid symbols. Assume completepenetrance and also assume that no new mutations have arisen in these families. Giveall possible modes of inheritance that are consistent with each pedigree (your choicesare: autosomal recessive, X-linked recessive, or autosomal dominant). Also indicatethe predicted genotypes of each individual in the pedigree using:A for autosomal alleles giving dominant phenotypes,a for autosomal alleles giving recessive phenotypes,XA for X-linked alleles giving dominant phenotypes,and Xa for X-linked alleles giving recessive phenotypes.Do this for each possible mode of inheritance. In ambiguous cases, give all possiblegenotypes.

(a)

?

(b)

?

(c)

?

3. Each of the following pedigrees contains individuals carrying two different rarerecessive traits indicated as follows:

Assume no new mutations and complete penetrance.The loci for the two traits are linked on the same autosome and lie 20 cM apart. Foreach pedigree, calculate the probabilities that the individual indicated by ? will have onlytrait 1.

(a) (b)

(c)

4. You have just been hired as a genetic counselor for a royal family that still engagesin a significant amount of inbreeding. As your first assignment, you are presented withthe following pedigree where the filled symbol represents a male in the royal family whohas a rare recessive disease.

Your job is to calculate the probability that the child indicated by ? will have the disease.To do this, assume that no new mutations arise within the pedigree and that nounrelated individual is a carrier (because this is a very rare disease). Also assumecomplete penetrance.

(a) If the disease is autosomal recessive, what is the probability that the child indicatedby the ? will have the disease?

(b) If the disease is autosomal recessive, and the first child born has the disease, whatis the probability that a second child will have the disease?

(c) If the disease is X-linked, what is the probability that a child will be born with thedisease if that child is born male?

(d) If the disease is X-linked, what is the probability that a child will be born with thedisease if that child is born female?

5. The producers of a soap opera have hired you as a consultant. The story lineincludes two families, each containing individuals that have a rare trait. The families arediagramed below — individuals are numbered, and those expressing the trait arerepresented by the filled symbols. The scriptwriters are contemplating a number ofdifferent couplings between individuals in the two families. Because they are concernedwith the genetic accuracy of the story, they want you to figure out what the offspringfrom each possible mating might be like. Assume no new mutations and completepenetrance.

1 2 5 6

3 4 7 8

(a) Assume that the rare trait is autosomal recessive. Consider the possible matingsdescribed below. For each, calculate the probability that the child will have the raretrait.

Female 2 and Male 5

Female 6 and Male 4

Female 7 and Male 4

Female 3 and Male 8

(b) Now assume that the rare trait is autosomal dominant. Again for each of thepossible matings given below, calculate the probability that the child will have the raretrait.

Female 2 and Male 5

Female 6 and Male 4

Female 7 and Male 4

Female 3 and Male 8

(c) Finally, assume that the rare trait is X-linked recessive. For each of the possiblematings given below, calculate the probability that the child will have the rare trait.Explicitly give a probability for sons and a probability for daughters in any case wherethe probabilities for a boy or a girl having the trait differ.

Female 2 and Male 5

Female 6 and Male 4

Female 7 and Male 4

Female 3 and Male 8

6. Each of the families below exhibits a different very rare genetic disorder whereindividuals expressing the disorder are shown by solid symbols. Assume completepenetrance and also assume that no new mutations have arisen in these families. Giveall possible modes of inheritance that are consistent with each pedigree (your choicesare: autosomal recessive, X-linked recessive, or autosomal dominant).Also calculate the probability that the next child indicated by a (?) will be affected giveneach mode of inheritance. In the case of X-linked recessive inheritance, calculateseparate probabilities for sons and daughters.

(a) (c)

? ?

(b) (d)

? ?

Mendelian Genetics and calculating statistical significance using Chi Square

1. Consider a hypothetical insect species that has red eyes. You isolate twomutations, each of causes the loss-of-function in an enzyme that normally leads tosynthesis of a red eye pigment and thus yield mutants with white eyes. You establishthat the white-eyed phenotype is recessive, which makes sense, because loss-of-function mutations normally yield recessive phenotypes. It is possible that bothmutations you have isolated are in the same gene. But it is also possible that you haveisolated mutations in two different genes, and that there is thus a pathway in which twodifferent enzymes each operate to produce red pigment. If there are two genes thatencode enzymes that produce red pigment, two different pathways for pigmentproduction can be drawn. The two genes might act in series such that a mutation ineither gene would block the formation of red pigment. Alternatively, the two genescould act in parallel such that mutations in both genes would be required to block theformation of red pigment.

If there are two genes, assume that they are unlinked. Further complexity arises fromthe possibility that mutations in either gene that lead to a block in enzymatic activitycould be either X-linked or autosomal.

(a) Such considerations yield the following six possibilities:1) A one-gene pathway, and the mutation is autosomal.2) A one-gene pathway, and the mutation is X-linked.3) A two-gene pathway in series, with both mutations being autosomal.4) A two-gene pathway in series, with one mutation being autosomal and the

other being X-linked.5) A two-gene pathway in parallel, with both mutations being autosomal.6) A two-gene pathway in parallel, with one being autosomal and the other being

X-linked.

Gene 1Series

Gene 2red pigment

Gene 1

Parallel Gene 2 red pigment

For each of the six possible cases outlined above, determine the expected phenotypicratios of the F1 progeny and the F2 progeny for the following cross:A cross between a wild-type female insect with red eyes and a true-breeding white-eyedmale insect (who carries mutations in both genes if it is a two-gene model).

(b) You obtain a true-breeding white-eyed mutant male (who carries mutations in bothgenes if it is a two-gene model), which you cross to a wild-type female. All of the F1progeny have normal red eyes. Crosses among these F1 insects yield 24 progeny -- 5with white eyes and 19 with red eyes. Determine whether this data is consistent witheach of the six possibilities outlined in part (a). Use the table below of chi-squaredprobabilities for your statistical tests.

For each chi square test you do, give the observed and expected phenotypic ratios, thedegrees of freedom, your calculated value for χ2, and a rough estimate of the p value.Also, state what your null hypothesis was, and whether or not you can reject your nullhypothesis.

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

2. Consider the following mouse breeding experiment involving two different raretraits. Assume that each rare trait is caused by a specific allele of a single gene. Amale mouse with both traits is crossed to a normal female, and all of the offspringappear normal. A female offspring from this cross is mated multiple times to a normalmale to produce several litters of offspring. A total of 32 offspring are scored as havingthe following characteristics:

16 normal females

6 normal males

2 males with Trait One only

1 male with Trait Two only

7 males with both traits

(a) What is the mode of inheritance of each of the two traits? Explain your reasoning.

(b) Use the chi-square test to determine whether the two traits appear to be linked.Note that you are trying to determine whether or not an expectation based on the nullhypothesis (that is, that the two traits are unlinked) differs significantly from theobserved data. There are a number of different ways to set up this test, but there is onebest way to test for linkage. Show your work and use the table below which gives pvalues as a function of chi-square values and degrees of freedom.

For your answer, give the observed and expected phenotypic ratios, the degrees offreedom, your calculated value for χ2, and a rough estimate of the p value. Also, statewhether or not you can reject your null hypothesis.

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

(c) Based on the data, give your best estimate of the distance between the genes that,when mutated, cause Trait One and Trait Two.

3. In a cross between a male mouse from a true-breeding black strain and a femalefrom a true-breeding tan strain, all of the F1 progeny are gray.

(a) Based on the information that you have at this stage, is it possible that a singlegene determines the differences in coat color among the two parental strains andprogeny? If so, what coat colors should appear in the F2 generation and at whatfrequencies?

(b) In fact, when F1 mice are crossed among themselves, the following F2 progeny areproduced: 30 gray mice, 10 black mice, 8 tan mice, and 2 dark brown mice.

Propose a genetic model to account for the existence of the dark brown F2 mice. Foryour answer give the genotypes of the parental mice, the F1 mice and each class of F2mice.

(c) Use the Chi-square test to show that the observed frequencies fit with the expectedfrequencies based on your model. For your answer, give the observed and expectedphenotypic ratios, the degrees of freedom, your calculated value for χ2, and a roughestimate of the p value. Also, state what your null hypothesis was, and whether or notyou can reject your null hypothesis.

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

(d) Returning to the original true-breeding parental strains, you do a different cross, andfind that things are even more complicated. A male from the tan strain is crossed to afemale from the black strain. As expected all of the female F1 mice are gray, but to yoursurprise, all of the male F1 mice are black.

Propose a genetic model to account for this new data. Give the genotypes of the maleand female parental mice and the male and female F1 mice. Finally, predict the types ofmice that will appear in the F2 generation. Specify the coat colors, sex, and expectedfrequency of each class.

(e) Given the genetic model that you have developed in part (d), return to the crossesdescribed in parts (a) and (b). For these crosses, give the genotypes of the male andfemale parental mice and the male and female F1 mice. Finally, predict the types ofmice that will appear in the F2 generation. Specify the coat colors, sex, and expectedfrequency of each class.

4. Being a mouse geneticist, you maintain a large colony of mice. One day you spot amouse in your colony with a novel and interesting phenotype: a kinked tail. You breedthe kinked-tail mouse (a male) with several wild-type females and observe that abouthalf of the offspring (both males and females) have kinked tails and half have normaltails.

(a) Is the kinked-tail phenotype dominant or recessive to wild-type?

(b) When two of the kinked-tail offspring from part (a) are crossed, what fraction of theresulting mice would you expect to have kinked tails?

(c) When you cross kinked-tail offspring from part (a), you find that one-third of theresulting kinked-tail males produce no sperm and thus are sterile. The other two-thirdsof the resulting kinked-tail males (and all of the normal-tail males and all of the females)are fertile. Propose a two-gene model to account for these findings.

(d) Your colleague informs you that he has isolated a pure-breeding mouse strain inwhich males produce no sperm but have normal tails, and in which females arephenotypically normal (fertile; normal tails). You explain to your colleague that this isimpossible. Why?

(e) Your colleague discovers a mutant that displays a dominant phenotype in bothsexes – “short tail.” Through extensive breeding, your colleague identifies a series ofshort-tail females that, when crossed with wild-type males, produce exclusively short-tailprogeny. You cross these short-tail females with fertile, kinked-tail males and observethe following offspring: 35 short-kinked-tail females, 32 short-non-kinked-tail females, 28short-kinked-tail males, 36 short-non-kinked-tail males. (All of these offspring arefertile.) Propose two different two-gene models to account for these ratios.

(f) To distinguish between your two models from part (e), you select, from among theoffspring from part (e), short-kinked-tail females and short-kinked-tail males, and youcross them. If each model is correct, what phenotypic classes do you expect toobserve, and in what ratios?

b pr vg

6 cM 13 cM

Mapping in flies by Two- and Three- factor crosses

1. Consider a portion of an autosome in Drosophila, which carries the following threemutations, each of which cause the corresponding recessive phenotype: b– (blackbody), pr– (purple eyes), and vg– (vestigial wings). Wild-type flies have brown bodies,red eyes, and large wings. The corresponding wild-type alleles of each gene aredesignated b+, pr+, and vg+. A genetic map of this portion of the chromosome is shownbelow:

The measured two-factor distance between the b and pr loci is 6 cM, and the distancebetween the pr and vg loci is 13 cM.

(a) Imagine that you want to set up a cross to verify these map distances and you havein the lab a true-breeding strain with a black body and vestigial wings. What type oftrue-breeding fly would you want to mate this fly to in order to carry out a three-factorcross to map the b, pr, and vg loci?

(b) For the cross described in part (a), what would the F1 generation look like?

(c) For the cross described in part (a),what type of strain would be the best to mate theF1 generation to in order to score the gamete genotypes passed to the next generation?

(d) What would be the rarest phenotypic class(es) produced from the cross in part (c)?

(e) Write out the eight possible genotypes produced in the cross described in part (c).

(f) If 1,000 progeny were produced in the cross from part (c), how many of each of theeight genotypes would you expect in this F2 generation?

(g) What would you expect the measured distance between b and vg to be in a two-factor cross between these markers?

(h) Explain why the distance you gave in part (f) is different than the sum of the b–prand pr–vg distances from the map above (i.e. 19 cM)?

2. Consider the three autosomal hypothetical recessive traits in Drosophilia: bentwings (caused by the bt allele), yellow body (caused by the yl allele), and long bristles(caused by the lg allele). You are given a fly that looks like wild-type, but is in factheterozygous for all three traits. You set up a test cross of your fly to a fly that ishomozygous recessive for all three traits. From 500 progeny, the following types arefound. (+ denotes a wild-type phenotype.)

Phenotype numberbent wings yellow body long bristles 34+ + + 46bent wings + + 13+ yellow body long bristles 7bent wings yellow body + 202+ + long bristles 193bent wings + long bristles 3+ yellow body + 2

(a) If the fly that you were given came from a cross of two true-breeding lines, whatwould be the phenotypes of the two true-breeding lines?

(b) What is the distance between the bt and lg loci in cM?

(c) Produce a genetic map showing the order and the relative positions of the genesthat control these three traits. Include all pairwise genetic distances in your diagram.

3. Two different true-breeding Drosoplila lines are crossed, and F1 females from thiscross are then crossed to males from a line that is homozygous for alleles that causefour different recessive traits. A total of 1000 progeny from these crosses are thenevaluated for each of the four autosomal traits. Each of these four traits in question aredetermined by a single gene. For simplicity, the recessive traits are designated a, b, c,and d, whereas the corresponding dominant traits are designated with a “+”. Thephenotypes and number of flies falling into each of the sixteen possible phenotypicclasses are given below:

Phenotype Number Phenotype Number

a b c d 10 + b c d 14+ + + + 13 a + + + 11

a b c + 217 a b + + 25+ + + d 198 + + c d 29

a b + d 5 a + + d 203+ + c + 2 + b c + 214

a + c d 28 + b + d 3+ b + + 26 a + c + 2

(a) Determine the two-factor cross distance between markers c and d in cM.

(b) What can you conclude about the genotypes of the two true-breeding parentallines? (If it is not possible to specify their exact genotype, note the nature of theambiguities that remain.)

(c) Draw a genetic map showing the relative positions and genetic distances betweenthe genes that influence the four traits in question. Linked genes should be groupedtogether on the same chromosomal segment and any unlinked genes should be placedon a different chromosomal segment.

4. You have been studying eye color mutations in Drosophila, which normally havered eyes. White eyes is a recessive mutant trait that is caused by w, a mutant allelefound on the X chromosome. You have isolated a new mutation, ap, that causes therecessive phenotype of apricot colored eyes. The location of the ap locus in the flygenome is unknown.

(a) A female from a true-breeding ap strain is crossed to a male from a true-breeding wstrain. The females in the F1 progeny have very pale peach colored eyes. Explain whatthis result tells you about the relationship between the w and ap mutations and why.

(b) What would you expect the phenotype of the male F1 progeny to be and why?

(c) F1 females are crossed to wild-type males, and 10,000 male progeny are examined.Most of these males have either white eyes or apricot eyes, however, one of the maleshave red eyes. What is the origin of this red-eyed male?

(d) What is the distance between the w and ap loci in cM?

(e) Crossveinless is a recessive phenotype caused by the cv mutation. The cv locusmaps about 10 cM away from the w locus. A female from a true-breeding cv, w strainis crossed to a male from a true-breeding ap strain. The females from this cross arethen crossed to wild-type males, and a very large number of the resulting male progenyare examined. Eight males with red eyes are found; seven of these have normal wingsand one has crossveinless wings. Draw a map showing the relative positions of the cv,w, and ap loci.

5. The hypothetical Drosophila traits big-head, curly-wings, and weak-knees areautosomal recessive, and determined by the alleles bh, cw, and wk respectively. True-breeding wild-type Drosophila have the alleles bh+, cw+, and wk+, and are small-headed, straight-winged, and strong-kneed. The three genes in question are autosomaland are linked to one another. The gene order and two of the distances between genesare shown in the following genetic map.

A fly from a big-headed strain is crossed to a fly from a weak-kneed curly-winged strainand all of the F1 progeny look normal. Females from the F1 are collected and crossedto males from a big-headed, weak-kneed, curly-winged strain.

(a) List all eight of the theoretically possible phenotypes that could result from this cross.Indicate the two phenotypes that should be the most abundant and the two phenotypesthat should be the least abundant.

(b) If 80 progeny flies were examined from this cross, how many flies of each of theeight phenotypic classes would you expect?

(c) The three factor crosses we discussed in class typically give 8 phenotypic classes.Why did this particular cross you did not give eight classes?

20 cM 5 cM

bh wkcw

Mapping in yeast by Tetrad analysis

1. You have isolated two different yeast mutants that will not grow on medium thatlacks the amino acid arginine. You call these mutants arg1– and arg2–.

(a) Mating of either arg1– or arg2– haploid mutant yeast to wild-type yeast producesdiploids that can grow without arginine. Mating of arg1– haploid mutant yeast to arg2–haploid mutant yeast produces a diploid that also can grow on medium without arginine.What do these results tell you about the arg1– and arg2– mutations?

(b) You induce sporulation of the diploids produced by the mating of arg1– haploidmutant yeast to arg2– haploid mutant yeast. This yields tetrads of the following types:

type 1 tetrad type 2 tetrad

1 Arg+ spore and 3 Arg– spores 4 Arg– spores

Out of the 20 tetrads you analyze, two are type 1, and eighteen are type 2. Categorizeeach of the tetrad types as parental ditype (PD), tetratype (TT), or nonparental ditype(NPD).

(c) Are the arg1 and arg2 loci linked? If so, give the distance between them in cM.

(d) Now you isolate a third mutant called arg3–. When this haploid mutant is mated towild-type haploid yeast, the resulting diploids cannot grow without arginine. Whenarg3– haploid mutant yeast is mated to arg1– haploid mutant yeast, the resulting diploidcannot grow without arginine. What do these results tell you about arg3– and itsrelationship to arg1–?

(e) When the diploid produced by mating arg3– haploid mutant yeast to arg1– haploidmutant yeast is induced to sporulate, tetrads of three types are produced.

type 1 type 2 type 3

1 Arg+, 3 Arg– 4 Arg– 2 Arg+, 2 Arg–

Out of 20 tetrads, twelve are type 1, five are type 2, and three are type 3. Categorizeeach of the tetrad types as parental ditype (PD), tetratype (TT), or nonparental ditype(NPD).

(f) Are the arg1 and arg3 loci linked? If so, give the distance between them in cM.

(g) Does the result from part (e) tell you anything new about the gene(s) mutated inarg1– and arg3– mutants? If so, what does it tell you that is new?

(h) Based on these results, deduce the relationship between arg3– and arg2–. In youranswer, describe the phenotype you would see of a diploid that results from the matingof arg3– haploid yeast to arg2– haploid yeast.

(i) If the resulting diploid from part (h) were induced to sporulate, and 60 resultingtetrads were examined, how many tetrads of each type (PD, NPD, TT) would youexpect?

(j) If the resulting diploid from part (h) were induced to sporulate, what would thephenotypic spore composition of each type of tetrad (PD, NPD, TT) be (i.e. # of Arg+spores and # of Arg– spores)? What would the genotypic spore composition of eachtype of tetrad be (i.e. # of each of 2+3+ and 2–3– and 2–3+ and 2+3– spores)?

2. You want to construct a haploid yeast strain with two different mutations in theLeu2 gene. To do this, you mate a Leu2-a haploid mutant yeast to a Leu-2b haploidmutant yeast. (Each of these single mutant strains are phenotypically Leu–; that is, theywill not grow unless the amino acid leucine is provided in the growth medium.) Youinduce sporulation of the resulting diploid. You find that the resulting tetrads are of twotypes:

Type One Type Two 4 Leu – spores 3 Leu – spores: 1 Leu+ spores

49 tetrads were of Type One, whereas only 1 tetrad was of Type Two.

(a) Explain why this cross produced only two of the three possible tetrad types.

(b) The desired Leu2-a Leu2-b double mutant haploid yeast was one of the three Leu–spores in the Type Two tetrad. Explain why complementation tests couldn’t be used tofind the one double mutant out of the three Leu– spores.

(c) Describe a procedure that you would use to identify the desired double mutant. Beas specific as possible about the crosses that you would perform, how you wouldanalyze the resulting tetrads, and how many tetrads you would analyze for each cross.Also describe the results you expect to see.

(d) Say that you have isolated a new Leu– mutant that you call LeuX. You do a crossbetween LeuX mutant haploid yeast and the Leu2-a Leu2-b double mutant haploidyeast. You induce sporulation of the resulting diploid, and find that the following tetradtypes are obtained:

Type Three Type Four Type Five 4 Leu – 3 Leu – : 1 Leu + 2 Leu – : 2 Leu +

From 50 tetrads you obtain, 35 of Type Three, 13 of Type Four, and 2 of Type Five.What is the relationship between the LeuX mutation and the Leu2 gene?

3. You have isolated three new His– yeast mutants in mating type α. Each of thesethree mutant strains cannot grow without the amino acid histidine being supplementedin the growth medium. You cross each of the haploid mutants to a wild-type (His+)haploid strain of mating type a, and then induce sporulation of the resulting diploid. Youthen evaluate 50 of the resulting tetrads for the His+ or His– phenotype. In principle,there are five different possible types of tetrads that could be obtained.

Type One Type Two Type Three Type Four Type Five4 His+ 3 His+ : 1 His– 2 His+ : 2 His– 1 His+ : 3 His– 4 His–

(a) In the cross of mutant 1 to wild-type, all of the resulting tetrads are of Type Three.What does this tell you about the His– mutation(s) in mutant 1?

(b) In the cross of mutant 2 to wild-type, 40 tetrads are Type Three; 8 tetrads are TypeFour; and 2 tetrads are Type Five. Classify each relevant tetrad type as PD, NPD, orTT.

(c) Propose a genetic mechanism that would explain the behavior of mutant 2 by givingas much information as possible about the His– mutation(s) in mutant 2. (For example:How many mutations are contained within this mutant, and if there are multiplemutations, how far apart are the loci at which those mutations lie?)

(d) In the cross of mutant 3 to wild-type, 45 tetrads are Type Three and 5 tetrads areType Two. Classify each relevant tetrad type as PD, NPD, or TT.

(e) Propose a genetic mechanism that would explain the behavior of mutant 3 by givingas much information as possible about the His– mutation(s) in mutant 3. (For example:How many mutations are contained within this mutant, and if there are multiplemutations, how far apart are the loci at which those mutations lie?)

4. You are studying biosynthesis of the amino acid serine in yeast and you know thatthree different genes that are required (Ser1, Ser2, and Ser3). A mutation in any one ofthese genes will cause the yeast to be unable to grown unless serine is provided in thegrowth medium. You have isolated a collection of new Ser– mutants, and all but onecan be placed in one of the three Ser genes by complementation tests. This lastmutation, designated SerX, gives a dominant Ser– phenotype, and thus can’t beanalyzed by complementation testing. Therefore you decide to cross SerX haploidyeast to a haploid yeast containing one loss-of-function mutation in one of the three Sergenes. The results you get are described below, depending on whether the haploidyeast you mate SerX yeast to was Ser1–, Ser2–, or Ser3–. In all three experiments,when you induce sporulation of the resulting diploid, three potential types of tetradscould be produced.

Type One Type Two Type Three2 Ser– spores: 2 Ser+ spores 3 Ser – spores: 1 Ser + spores 4 Ser – spores

(a) In the cross of SerX haploid mutant yeast to Ser1– haploid mutant yeast, 4 tetradsare of Type One, 15 are of Type Two, and 6 are of Type Three. What does this resulttell you about the relationship between Ser1 and SerX?

(b) In the cross of SerX haploid mutant yeast to Ser2– haploid mutant yeast, 25 tetradsare examined and all are of Type Three. What is the upper limit of the distancebetween the SerX and Ser2 loci? Given that the average yeast gene is about 2 kbp inlength and the recombination rate in yeast is about 2 kbp/cM -- is SerX likely to be anallele of Ser2? Why or why not?

(c) In the cross of SerX haploid mutant yeast to Ser3– haploid mutant yeast, 20 tetradsare of Type Three and 5 tetrads are of Type Two. What is the distance between theSerX and Ser3 loci?

Mapping in phage

1. You have isolated two temperature-sensitive mutations in an essential gene ofphage λ. These phage mutants are called ts-1 and ts-2. Each mutant will formplaques in a bacterial lawn at 30˚C but not at 42˚C. You cross ts-1 to ts-2 phage bycoinfecting E. coli at the permissive temperature of 30˚C with both of these phagestrains. When the resulting phage lysate is plated at 30˚C, you count 105 plaques perml of phage lysate. But when the same resulting phage lysate is plated at 42˚C, thereare only 300 plaques per ml.

(a) What is the distance between the ts-1 and ts-2 loci in m.u.?

You next cross a ts-1 phage strain to a ts-2 phage strain that also carries a r– mutation,which gives plaques that are rough. When the resulting lysate is plated out at 42˚C and100 plaques are examined, 85 are rough and 15 have normal smooth edges.

(b) If the phage produced from this cross were plated at 30˚C, what fraction of theplaques would you expect to be rough?

(c) Draw a map showing the relative order of the r, ts-1 and ts-2 loci.

2. In a phage cross that you perform, bacteria are simultaneously infected with wild-type phage and a phage that is mutant at three different loci: a, b, and c. Each bacterialcell is infected with more than one phage of each genotype, thereby allowingrecombination between the phage to occur. Millions of progeny phage can be producedfrom such a cross, allowing recombination frequencies to be determined veryaccurately. The genetic map for the phage you are using is shown below. The distancebetween a and b is 1 m.u. and the distance between b and c is 10 m.u. What are theeight possible genotypes that would result from this cross and what fraction of the totalprogeny would each represent? You can designate the mutant alleles with lower caseletters and wild-type alleles with the same letter followed by a + sign.

a b c

1 m.u.

10 m.u.

3. Wild type phage λ produces turbid plaques because about 10% of the infectedcells become lysogens (that is, the bacterial host chromosomes harbor a dormant copyof the phage genome) and are therefore immune to further infection and lysis by otherphage. In order to study the genes responsible for allowing phage λ to make lysogens,you isolate a collection of phage λ mutants that cannot form lysogens and thereforemake clear (rather than turbid) plaques. The mutants are numbered 1 – 5. In order toassign your mutants to complementation groups, you carry out coinfections of E. colicells with two different clear plaque mutants. You then assay the E. coli cells harboringtwo different phage for the presence of lysogens in these doubly-infected cells. In thetable below, “+” indicates the presence of lysogens in the doubly-infected bacterial cells,and “–“ indicates no lysogens.

(a) Organize the five phage mutations into complementation groups.

Next, you carry out coinfections of E. coli cells with pairs of mutant phage. You thenexamine the progeny phage that are produced for the presence of recombinant progenyphage that can form turbid plaques when a single recombinant phage infects a new cell.In the table below, the frequency of progeny phage that give turbid plaques resultingfrom each pairwise infection is shown.

(b) Draw a genetic map showing the relative positions of the five loci that are mutatedin your strains. Give distances in map units, indicate the gene boundaries, and noteany ambiguities in gene order that might remain.

From the crosses that are performed above, you are able to isolate a double mutantphage that contains mutations 1 and 5, as well as a double mutant phage that containsmutations 3 and 5. As a way to determine the relative order of your mutations, youperform the following three factor crosses. First, you coinfect E. coli with the 1, 5 doublemutant phage and mutant 3 phage. You find that, among the resulting progeny phage,the frequency of phage that form turbid plaques is 2 x 10-4.Second, when E. coli are coinfected with 3, 5 double mutant phage and mutant 1phage, the frequency of phage that form turbid plaques is 2 x 10-6.

(c) Draw out the possible relative orders of the 1, 3, and 5 loci.

(d) For each possible order from part (c), show the types of recombination events thatwould need to occur between the two mutant phage in a bacterial cell to produce wild-type phage for the FIRST three factor cross.

(e) For each possible order from part (c), show the types of recombination events thatwould need to occur between the two mutant phage in a bacterial cell to produce wild-type phage for the SECOND three factor cross.

(f) Choose the correct gene order among the possibilities from part (c).

4. Imagine that you are studying a new type of phage. The genome of this phagehas been measured to be 8x104 base pairs in length and, by addition of many differentmap distances, the total genetic length of the chromosome is estimated to be 200 m.u..You have isolated two different phage mutants that appear to be in the same gene.This gene controls plaque size, and each of the two mutants causes the formation ofsmall plaques. You call these mutants sm-1 and sm-2.

(a) You cross an sm-1 mutant phage with an sm-2 mutant phage by coinfecting E. coliwith these two different types of phage. You plate out the resulting phage lysate andfind that 11 out of 1000 plaques have normal large plaques while the rest are small.What is the distance between the sm-1 and sm-2 loci in m.u.?

(b) You have identified the protein product of the phage sm gene – it is a protein of 50kDa (one kDa = 1000 daltons). On examination of the sm protein in phage lysates frommutant phage, you find that the sm-1 mutant phage produces an sm protein of 45 kDawhile the sm-2 mutant phage produces an sm protein of 50 kDa. What type(s) of singlenucleotide substitutions would explain the behavior of the sm-1 mutant? What type(s) ofmutations would explain the behavior of the sm-2 mutant? In thinking about youranswer, you may find it useful to consider that the average molecular weight of anamino acid is approximately 110 Da.

(c) Based on everything that you know about the sm-1 and sm-2 mutants, draw anapproximate map of the sm gene, showing the coding sequence, the direction oftranscription of the mRNA from this gene, and the approximate positions of the sm-1and sm-2 loci. Mark distances on the map by using what you know about the effects ofthe mutations on the sm protein product, the map distance between sm-1 amd sm-2,and the relationship between the genetic and physical lengths of the phage genome.

(d) You have isolated a new phage mutant that gives clear plaques (the wild-type phagenormally produces turbid plaques). You call this mutant cl-1. You construct a cl-1 sm-1 double mutant phage (which makes small clear plaques) and then you cross thisdouble mutant to an sm-2 single mutant by coinfecting E. coli with these two differenttypes of mutant phage. You plate the resulting phage lysate, and note that theresulting plaques are all small, except for 9 clear large plaques and 1 turbid largeplaque. Is the cI gene closer to the sm-1 locus, or the sm-2 locus?

5. You have isolated two temperature-sensitive mutations in phage λ that lie in thesame essential gene. These phage mutants are called ts-3 and ts-4. Each mutant willform plaques on bacterial lawns when incubated at 35˚C but not at 42˚C. You cross ts-3 phage to ts-4 phage by coinfecting E. coli at the permissive temperature of 35˚C withboth of these types of phage. When the resulting phage lysate is plated at 35˚C, youdetermine that there are 10,000 plaques per ml in the phage lysate, but when the samephage lysate is plated at 42˚C, there are only 80 plaques per ml.

(a) What is the distance between the ts-3 and ts-4 loci in m.u.?

(b) If the total size of the phage genome is 5 x 104 basepairs, and the total genetic maplength of the phage genome is 100 m.u., about how far apart are the ts-3 and ts-4 lociin base pairs?

You next cross a ts-3 mutant phage to a phage strain that carries ts-4 in addition to amutation called sm, which gives small plaques. You coinfect E. coli with these twotypes of phage, and harvest the resulting phage lysate. When the resulting lysate isplated out at 42˚C and 100 plaques are examined, 25 are small and 75 are of normalsize.

(c) Draw a map showing the relative order of the sm, ts-3 and ts-4 loci.

6. You have two useful strains of phage λ with mutations in the cI gene. The cI-1mutation maps very close to the beginning of the cI gene coding sequence while the cI-2 mutation maps very close to the end of the coding sequence. Both mutations causethe phage to form clear plaques rather than the normal turbid plaques. Phage with cI-1are crossed to phage with cI-2 by coinfecting E. coli with both types of phage. Whenthe resulting phage lysate is examined, four plaques out of 1000 are turbid.

(a) What is the distance between cI-1 and cI-2 in map units?

(b) Given your answer for part (a) and that the cI repressor protein is 240 amino acidslong, estimate how many kb correspond to one map unit for this phage ( one kb = 103

base pairs).

(c) You isolate a new mutation in the cI gene and find that, in crosses between the newmutant phage and cI-1 phage, turbid plaques are produced at twice the frequency as incrosses between the new mutant phage and cI-2 phage. You discover that your newmutation introduces a stop codon into the coding sequence of the cI gene. Given thatthe average molecular weight of an amino acid is 110 Daltons, what is the expectedmolecular weight of the product of your new mutant version of the cI gene?

(d) A number of different mutagens cause what are known as transition mutations, inwhich a T•A base pair is converted to C•G (or a C•G base pair is converted to an T•A).By examining the table for the genetic code, determine the sense codons (and theamino acids which they code for) that can be converted into a stop codon by a singletransition mutation.