Problem set questions from Exam 1 Unit Basic Genetic Tests, Setting up and Analyzing Crosses, 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 of threonine, an amino acid. These Thr- mutants do not grow on minimal medium, but they do grow on minimal medium supplemented with threonine. Ten of your Thrmutants (numbered 1 through 10) were isolated in a strain of mating type alpha (MAT α). The other 10 Thr- mutants (numbered 11 through 20) were isolated in a strain of mating type a (MAT a). You cross each of the MAT a haploid strains to each of the MAT α haploid strains, and you include crosses to the appropriate wild-type haploid strains. Your experimental observations are shown in the table below, where (-) indicates diploids that did not grow on minimal medium and (+) indicates diploids that did grow on minimal medium (that had NOT been supplemented with threonine). (a) Unfortunately, when all the data were collected and the plates discarded, you lost some of your data. From the data that remains, see if you can reconstruct the full table. strains of strains of mating type α mating type a: wild-type 1 2 3 4 5 6 7 8 9 10 wild-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 for threonine 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 of the usual white colonies of wild-type yeast. You cross each of the haploid mutants to a wild-type haploid strain and obtain the resulting diploids shown below. Mutant 1 X wild-type diploid Mutant 2 X wild-type diploid Mutant 3 X wild-type diploid Mutant 4 X wild-type diploid Mutant 5 X wild-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 the opposite mating type. From the results shown below deduce as much as you can about how many different colony color genes you have isolated mutations in, and which specific mutations lie in the same gene. Mutant 1 X Mutant 2 diploid Mutant 2 X Mutant 4 diploid Mutant 1 X Mutant 3 diploid Mutant 2 X Mutant 5 diploid Mutant 1 X Mutant 4 diploid Mutant 3 X Mutant 4 diploid Mutant 1 X Mutant 5 diploid Mutant 3 X Mutant 5 diploid Mutant 2 X Mutant 3 diploid Mutant 4 X Mutant 5 diploid (c) Clearly state any remaining ambiguities.
3. You have isolated five new Arg mutants in yeast. Each mutant cannot grown unless the amino acid arginine is provided in the growth medium. After obtaining versions of each mutant in mating type a and in mating type α, you perform all of the possible pairwise matings shown in the table below. A "+" at the intersection of the two parental strains indicates that the diploid can grow without arginine added to the medium, whereas a " " indicates that the diploid can t grow without arginine. mutants of mating type α 1 2 3 4 5 mutants of mating type a 1 2 3 4 5 + + + + + (a) Give as much information as you can about your new Arg mutants. Indicate which mutants have dominant Arg phenotypes, and which have recessive Arg phenotypes. Also state how many genes are represented by your collection of five mutants, and which 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 transport canavanine, which is a toxic analog of arginine. Mutations that interfere with this permease s transporter activity can block canavanine uptake and will therefore allow cells to grow in the presence of an amount of canavanine that would kill a wild-type yeast cell. You have isolated a set of five canavanine resistant mutants. As before, you obtain versions of each mutant in mating type a and mating type α, and you perform all of the possible pairwise matings shown in the table below. In this table a "+" at the intersection of the two parental strains indicates that the diploid can grow in the presence of high levels of canavanine, whereas a " " indicates that the diploid is as sensitive to canavanine as wild-type.
mutants of mating type α 1 2 3 4 5 mutants of mating type a 1 2 3 4 5 + + + + + + + + + + Give as much information as you can about your new Can R mutants. Indicate which mutants have dominant Can R phenotypes, and which have recessive Can R phenotypes. Also state how many genes are represented by your collection of five mutants, and which mutations lie the same gene. Assume each strain carries only a single Can R mutation. 4. You have isolated twenty His yeast mutants. Each single mutant cannot grow unless the amino acid histidine is supplemented in the growth medium. Mutants 1 10 are of mating type α and mutants 11 20 are of mating type a. You performed complementation tests by mating each α strain to each a strain. The table below shows the results, where a "+" at the intersection of the two parental strains indicates that the diploid can grow without histidine added to the medium, and a " " indicates that the diploid 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.
(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 same complementation group. Any remaining ambiguities in the assignment of mutations to complementation 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. = female showing trait 1 only = male showing trait 2 only = male showing both traits? (a) Assuming that the two traits are due to unlinked autosomal genes, calculate the probability 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 that child 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 that child is born female.
2. Each of the families below exhibits a different very rare genetic disorder where individuals expressing the disorder are shown by solid symbols. Assume complete penetrance and also assume that no new mutations have arisen in these families. Give all possible modes of inheritance that are consistent with each pedigree (your choices are: autosomal recessive, X-linked recessive, or autosomal dominant). Also indicate the predicted genotypes of each individual in the pedigree using: A for autosomal alleles giving dominant phenotypes, a for autosomal alleles giving recessive phenotypes, X A for X-linked alleles giving dominant phenotypes, and X a for X-linked alleles giving recessive phenotypes. Do this for each possible mode of inheritance. In ambiguous cases, give all possible genotypes. (a) (b)? (c)??
3. Each of the following pedigrees contains individuals carrying two different rare recessive 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. For each pedigree, calculate the probabilities that the individual indicated by? will have only trait 1. (a) (b) (c)
4. You have just been hired as a genetic counselor for a royal family that still engages in a significant amount of inbreeding. As your first assignment, you are presented with the following pedigree where the filled symbol represents a male in the royal family who has 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 no unrelated individual is a carrier (because this is a very rare disease). Also assume complete penetrance. (a) If the disease is autosomal recessive, what is the probability that the child indicated by the? will have the disease? (b) If the disease is autosomal recessive, and the first child born has the disease, what is 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 the disease if that child is born male? (d) If the disease is X-linked, what is the probability that a child will be born with the disease if that child is born female?
5. The producers of a soap opera have hired you as a consultant. The story line includes two families, each containing individuals that have a rare trait. The families are diagramed below individuals are numbered, and those expressing the trait are represented by the filled symbols. The scriptwriters are contemplating a number of different couplings between individuals in the two families. Because they are concerned with the genetic accuracy of the story, they want you to figure out what the offspring from each possible mating might be like. Assume no new mutations and complete penetrance. 1 2 5 6 3 4 7 8 (a) Assume that the rare trait is autosomal recessive. Consider the possible matings described below. For each, calculate the probability that the child will have the rare trait. 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 the possible matings given below, calculate the probability that the child will have the rare trait. 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 possible matings 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 where the 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 where individuals expressing the disorder are shown by solid symbols. Assume complete penetrance and also assume that no new mutations have arisen in these families. Give all possible modes of inheritance that are consistent with each pedigree (your choices are: autosomal recessive, X-linked recessive, or autosomal dominant). Also calculate the probability that the next child indicated by a (?) will be affected given each mode of inheritance. In the case of X-linked recessive inheritance, calculate separate 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 two mutations, each of causes the loss-of-function in an enzyme that normally leads to synthesis of a red eye pigment and thus yield mutants with white eyes. You establish that the white-eyed phenotype is recessive, which makes sense, because loss-offunction mutations normally yield recessive phenotypes. It is possible that both mutations you have isolated are in the same gene. But it is also possible that you have isolated mutations in two different genes, and that there is thus a pathway in which two different enzymes each operate to produce red pigment. If there are two genes that encode enzymes that produce red pigment, two different pathways for pigment production can be drawn. The two genes might act in series such that a mutation in either gene would block the formation of red pigment. Alternatively, the two genes could act in parallel such that mutations in both genes would be required to block the formation of red pigment. Series Gene 1 Gene 2 red pigment Gene 1 Parallel Gene 2 red pigment If there are two genes, assume that they are unlinked. Further complexity arises from the possibility that mutations in either gene that lead to a block in enzymatic activity could 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.
For each of the six possible cases outlined above, determine the expected phenotypic ratios 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-eyed male 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 both genes if it is a two-gene model), which you cross to a wild-type female. All of the F1 progeny have normal red eyes. Crosses among these F1 insects yield 24 progeny -- 5 with white eyes and 19 with red eyes. Determine whether this data is consistent with each of the six possibilities outlined in part (a). Use the table below of chi-squared probabilities for your statistical tests. For each chi square test you do, give the observed and expected phenotypic ratios, the degrees 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 null hypothesis. 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 2. Consider the following mouse breeding experiment involving two different rare traits. Assume that each rare trait is caused by a specific allele of a single gene. A male mouse with both traits is crossed to a normal female, and all of the offspring appear normal. A female offspring from this cross is mated multiple times to a normal male to produce several litters of offspring. A total of 32 offspring are scored as having the 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 null hypothesis (that is, that the two traits are unlinked) differs significantly from the observed data. There are a number of different ways to set up this test, but there is one best way to test for linkage. Show your work and use the table below which gives p values as a function of chi-square values and degrees of freedom. For your answer, give the observed and expected phenotypic ratios, the degrees of freedom, your calculated value for χ2, and a rough estimate of the p value. Also, state whether or not you can reject your null hypothesis. p value:.995.975 0.9 0.5 0.1 0.05 0.025 0.01 0.005 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 (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 female from a true-breeding tan strain, all of the F 1 progeny are gray. (a) Based on the information that you have at this stage, is it possible that a single gene determines the differences in coat color among the two parental strains and progeny? If so, what coat colors should appear in the F 2 generation and at what frequencies? (b) In fact, when F 1 mice are crossed among themselves, the following F 2 progeny are produced: 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 F 2 mice. For your answer give the genotypes of the parental mice, the F 1 mice and each class of F 2 mice.
(c) Use the Chi-square test to show that the observed frequencies fit with the expected frequencies based on your model. For your answer, give the observed and expected phenotypic ratios, the degrees 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 null hypothesis. 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 (d) Returning to the original true-breeding parental strains, you do a different cross, and find that things are even more complicated. A male from the tan strain is crossed to a female from the black strain. As expected all of the female F 1 mice are gray, but to your surprise, all of the male F 1 mice are black. Propose a genetic model to account for this new data. Give the genotypes of the male and female parental mice and the male and female F 1 mice. Finally, predict the types of mice that will appear in the F 2 generation. Specify the coat colors, sex, and expected frequency of each class. (e) Given the genetic model that you have developed in part (d), return to the crosses described in parts (a) and (b). For these crosses, give the genotypes of the male and female parental mice and the male and female F 1 mice. Finally, predict the types of mice that will appear in the F 2 generation. Specify the coat colors, sex, and expected frequency of each class. 4. Being a mouse geneticist, you maintain a large colony of mice. One day you spot a mouse in your colony with a novel and interesting phenotype: a kinked tail. You breed the kinked-tail mouse (a male) with several wild-type females and observe that about half of the offspring (both males and females) have kinked tails and half have normal tails. (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 the resulting mice would you expect to have kinked tails? (c) When you cross kinked-tail offspring from part (a), you find that one-third of the resulting kinked-tail males produce no sperm and thus are sterile. The other two-thirds of 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 in which males produce no sperm but have normal tails, and in which females are phenotypically normal (fertile; normal tails). You explain to your colleague that this is impossible. Why? (e) Your colleague discovers a mutant that displays a dominant phenotype in both sexes short tail. Through extensive breeding, your colleague identifies a series of short-tail females that, when crossed with wild-type males, produce exclusively short-tail progeny. You cross these short-tail females with fertile, kinked-tail males and observe the following offspring: 35 short-kinked-tail females, 32 short-non-kinked-tail females, 28 short-kinked-tail males, 36 short-non-kinked-tail males. (All of these offspring are fertile.) 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 the offspring from part (e), short-kinked-tail females and short-kinked-tail males, and you cross them. If each model is correct, what phenotypic classes do you expect to observe, and in what ratios?
Mapping in flies by Two- and Three- factor crosses 1. Consider a portion of an autosome in Drosophila, which carries the following three mutations, each of which cause the corresponding recessive phenotype: b (black body), 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 are designated b +, pr +, and vg +. A genetic map of this portion of the chromosome is shown below: b pr vg 6 cm 13 cm The measured two-factor distance between the b and pr loci is 6 cm, and the distance between 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 have in the lab a true-breeding strain with a black body and vestigial wings. What type of true-breeding fly would you want to mate this fly to in order to carry out a three-factor cross 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 the F1 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 the eight genotypes would you expect in this F2 generation? (g) What would you expect the measured distance between b and vg to be in a twofactor cross between these markers? (h) Explain why the distance you gave in part (f) is different than the sum of the b pr and pr vg distances from the map above (i.e. 19 cm)?
2. Consider the three autosomal hypothetical recessive traits in Drosophilia: bent wings (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 fact heterozygous for all three traits. You set up a test cross of your fly to a fly that is homozygous recessive for all three traits. From 500 progeny, the following types are found. (+ denotes a wild-type phenotype.) Phenotype number bent wings yellow body long bristles 34 + + + 46 bent wings + + 13 + yellow body long bristles 7 bent wings yellow body + 202 + + long bristles 193 bent wings + long bristles 3 + yellow body + 2 (a) If the fly that you were given came from a cross of two true-breeding lines, what would 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 genes that control these three traits. Include all pairwise genetic distances in your diagram.
3. Two different true-breeding Drosoplila lines are crossed, and F 1 females from this cross are then crossed to males from a line that is homozygous for alleles that cause four different recessive traits. A total of 1000 progeny from these crosses are then evaluated for each of the four autosomal traits. Each of these four traits in question are determined 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 +. The phenotypes and number of flies falling into each of the sixteen possible phenotypic classes 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 parental lines? (If it is not possible to specify their exact genotype, note the nature of the ambiguities that remain.) (c) Draw a genetic map showing the relative positions and genetic distances between the genes that influence the four traits in question. Linked genes should be grouped together on the same chromosomal segment and any unlinked genes should be placed on a different chromosomal segment.
4. You have been studying eye color mutations in Drosophila, which normally have red eyes. White eyes is a recessive mutant trait that is caused by w, a mutant allele found on the X chromosome. You have isolated a new mutation, ap, that causes the recessive phenotype of apricot colored eyes. The location of the ap locus in the fly genome is unknown. (a) A female from a true-breeding ap strain is crossed to a male from a true-breeding w strain. The females in the F 1 progeny have very pale peach colored eyes. Explain what this result tells you about the relationship between the w and ap mutations and why. (b) What would you expect the phenotype of the male F 1 progeny to be and why? (c) F 1 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 males have 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 locus maps about 10 cm away from the w locus. A female from a true-breeding cv, w strain is crossed to a male from a true-breeding ap strain. The females from this cross are then crossed to wild-type males, and a very large number of the resulting male progeny are examined. Eight males with red eyes are found; seven of these have normal wings and 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 are autosomal recessive, and determined by the alleles bh, cw, and wk respectively. Truebreeding wild-type Drosophila have the alleles bh+, cw+, and wk+, and are smallheaded, straight-winged, and strong-kneed. The three genes in question are autosomal and are linked to one another. The gene order and two of the distances between genes are shown in the following genetic map. 20 cm 5 cm bh cw wk A fly from a big-headed strain is crossed to a fly from a weak-kneed curly-winged strain and all of the F 1 progeny look normal. Females from the F 1 are collected and crossed to 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 phenotypes that should be the least abundant. (b) If 80 progeny flies were examined from this cross, how many flies of each of the eight 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?
Mapping in yeast by Tetrad analysis 1. You have isolated two different yeast mutants that will not grow on medium that lacks 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 produces diploids 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 haploid mutant 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. Categorize each 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 to wild-type haploid yeast, the resulting diploids cannot grow without arginine. When arg3 haploid mutant yeast is mated to arg1 haploid mutant yeast, the resulting diploid cannot grow without arginine. What do these results tell you about arg3 and its relationship to arg1? (e) When the diploid produced by mating arg3 haploid mutant yeast to arg1 haploid mutant 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. Categorize each 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 in arg1 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 your answer, describe the phenotype you would see of a diploid that results from the mating of arg3 haploid yeast to arg2 haploid yeast. (i) If the resulting diploid from part (h) were induced to sporulate, and 60 resulting tetrads were examined, how many tetrads of each type (PD, NPD, TT) would you expect? (j) If the resulting diploid from part (h) were induced to sporulate, what would the phenotypic 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 each type 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 the Leu2 gene. To do this, you mate a Leu2-a haploid mutant yeast to a Leu-2b haploid mutant yeast. (Each of these single mutant strains are phenotypically Leu ; that is, they will not grow unless the amino acid leucine is provided in the growth medium.) You induce sporulation of the resulting diploid. You find that the resulting tetrads are of two types: 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 to find the one double mutant out of the three Leu spores.
(c) Describe a procedure that you would use to identify the desired double mutant. Be as specific as possible about the crosses that you would perform, how you would analyze 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 cross between LeuX mutant haploid yeast and the Leu2-a Leu2-b double mutant haploid yeast. You induce sporulation of the resulting diploid, and find that the following tetrad types 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 these three mutant strains cannot grow without the amino acid histidine being supplemented in 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. You then 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 Five 4 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 Type Four; and 2 tetrads are Type Five. Classify each relevant tetrad type as PD, NPD, or TT. (c) Propose a genetic mechanism that would explain the behavior of mutant 2 by giving as 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 multiple mutations, 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 are Type 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 giving as 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 multiple mutations, 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 that three different genes that are required (Ser1, Ser2, and Ser3). A mutation in any one of these genes will cause the yeast to be unable to grown unless serine is provided in the growth medium. You have isolated a collection of new Ser mutants, and all but one can be placed in one of the three Ser genes by complementation tests. This last mutation, designated SerX, gives a dominant Ser phenotype, and thus can t be analyzed by complementation testing. Therefore you decide to cross SerX haploid yeast to a haploid yeast containing one loss-of-function mutation in one of the three Ser genes. The results you get are described below, depending on whether the haploid yeast 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 tetrads could be produced. Type One Type Two Type Three 2 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 tetrads are of Type One, 15 are of Type Two, and 6 are of Type Three. What does this result tell you about the relationship between Ser1 and SerX? (b) In the cross of SerX haploid mutant yeast to Ser2 haploid mutant yeast, 25 tetrads are examined and all are of Type Three. What is the upper limit of the distance between the SerX and Ser2 loci? Given that the average yeast gene is about 2 kbp in length and the recombination rate in yeast is about 2 kbp/cm -- is SerX likely to be an allele of Ser2? Why or why not? (c) In the cross of SerX haploid mutant yeast to Ser3 haploid mutant yeast, 20 tetrads are of Type Three and 5 tetrads are of Type Two. What is the distance between the SerX and Ser3 loci?
Mapping in phage 1. You have isolated two temperature-sensitive mutations in an essential gene of phage λ. These phage mutants are called ts-1 and ts-2. Each mutant will form plaques in a bacterial lawn at 30 C but not at 42 C. You cross ts-1 to ts-2 phage by coinfecting E. coli at the permissive temperature of 30 C with both of these phage strains. When the resulting phage lysate is plated at 30 C, you count 10 5 plaques per ml of phage lysate. But when the same resulting phage lysate is plated at 42 C, there are 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 and 100 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 the plaques 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 wildtype phage and a phage that is mutant at three different loci: a, b, and c. Each bacterial cell is infected with more than one phage of each genotype, thereby allowing recombination between the phage to occur. Millions of progeny phage can be produced from such a cross, allowing recombination frequencies to be determined very accurately. The genetic map for the phage you are using is shown below. The distance between a and b is 1 m.u. and the distance between b and c is 10 m.u. What are the eight possible genotypes that would result from this cross and what fraction of the total progeny would each represent? You can designate the mutant alleles with lower case letters and wild-type alleles with the same letter followed by a + sign. a b c 10 m.u. 1 m.u.
3. Wild type phage λ produces turbid plaques because about 10% of the infected cells become lysogens (that is, the bacterial host chromosomes harbor a dormant copy of the phage genome) and are therefore immune to further infection and lysis by other phage. 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 therefore make clear (rather than turbid) plaques. The mutants are numbered 1 5. In order to assign your mutants to complementation groups, you carry out coinfections of E. coli cells with two different clear plaque mutants. You then assay the E. coli cells harboring two different phage for the presence of lysogens in these doubly-infected cells. In the table 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 then examine the progeny phage that are produced for the presence of recombinant progeny phage 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 resulting from each pairwise infection is shown.
(b) Draw a genetic map showing the relative positions of the five loci that are mutated in your strains. Give distances in map units, indicate the gene boundaries, and note any ambiguities in gene order that might remain. From the crosses that are performed above, you are able to isolate a double mutant phage that contains mutations 1 and 5, as well as a double mutant phage that contains mutations 3 and 5. As a way to determine the relative order of your mutations, you perform the following three factor crosses. First, you coinfect E. coli with the 1, 5 double mutant 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 1 phage, 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 that would need to occur between the two mutant phage in a bacterial cell to produce wildtype phage for the FIRST three factor cross. (e) For each possible order from part (c), show the types of recombination events that would need to occur between the two mutant phage in a bacterial cell to produce wildtype 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 phage has been measured to be 8x10 4 base pairs in length and, by addition of many different map 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 of small 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. coli with these two different types of phage. You plate out the resulting phage lysate and find 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 50 kda (one kda = 1000 daltons). On examination of the sm protein in phage lysates from mutant phage, you find that the sm-1 mutant phage produces an sm protein of 45 kda while the sm-2 mutant phage produces an sm protein of 50 kda. What type(s) of single nucleotide substitutions would explain the behavior of the sm-1 mutant? What type(s) of mutations would explain the behavior of the sm-2 mutant? In thinking about your answer, you may find it useful to consider that the average molecular weight of an amino acid is approximately 110 Da. (c) Based on everything that you know about the sm-1 and sm-2 mutants, draw an approximate map of the sm gene, showing the coding sequence, the direction of transcription of the mrna from this gene, and the approximate positions of the sm-1 and sm-2 loci. Mark distances on the map by using what you know about the effects of the 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 phage normally 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 this double mutant to an sm-2 single mutant by coinfecting E. coli with these two different types of mutant phage. You plate the resulting phage lysate, and note that the resulting plaques are all small, except for 9 clear large plaques and 1 turbid large plaque. 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 the same essential gene. These phage mutants are called ts-3 and ts-4. Each mutant will form 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 with both of these types of phage. When the resulting phage lysate is plated at 35 C, you determine that there are 10,000 plaques per ml in the phage lysate, but when the same phage 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 10 4 basepairs, and the total genetic map length of the phage genome is 100 m.u., about how far apart are the ts-3 and ts-4 loci in base pairs? You next cross a ts-3 mutant phage to a phage strain that carries ts-4 in addition to a mutation called sm, which gives small plaques. You coinfect E. coli with these two types of phage, and harvest the resulting phage lysate. When the resulting lysate is plated out at 42 C and 100 plaques are examined, 25 are small and 75 are of normal size. (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-1 mutation 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 cause the phage to form clear plaques rather than the normal turbid plaques. Phage with ci-1 are crossed to phage with ci-2 by coinfecting E. coli with both types of phage. When the 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 acids long, estimate how many kb correspond to one map unit for this phage ( one kb = 10 3 base pairs).
(c) You isolate a new mutation in the ci gene and find that, in crosses between the new mutant phage and ci-1 phage, turbid plaques are produced at twice the frequency as in crosses between the new mutant phage and ci-2 phage. You discover that your new mutation introduces a stop codon into the coding sequence of the ci gene. Given that the average molecular weight of an amino acid is 110 Daltons, what is the expected molecular 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, in which 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 the amino acids which they code for) that can be converted into a stop codon by a single transition mutation.