




























































































Study with the several resources on Docsity
Earn points by helping other students or get them with a premium plan
Prepare for your exams
Study with the several resources on Docsity
Earn points to download
Earn points by helping other students or get them with a premium plan
Community
Ask the community for help and clear up your study doubts
Discover the best universities in your country according to Docsity users
Free resources
Download our free guides on studying techniques, anxiety management strategies, and thesis advice from Docsity tutors
that results is then crossed to a true-breeding black-eyed male. The following phenotypes are observed in 100 progeny: eyeless. 51 black-eyed 39.
Typology: Study notes
1 / 126
This page cannot be seen from the preview
Don't miss anything!
Exam Questions from Exam 1 – Basic Genetic Tests, Setting up and Analyzing Crosses, and Genetic Mapping
melanogaster. Flies that are homozygous for the hb– mutation are “humpbacked” (wild-
type flies are straight-backed). Flies that are homozygous for the bl– mutation are “blistery-winged” (wild-type flies are smooth-winged). Flies that are homozygous for the st– mutation are “stubby-legged” (wild-type flies are long-legged).
You mate flies from two true-breeding strains, and the resulting F1 flies are all are straight-backed, smooth-winged, and long-legged. F1 females are then mated to males that are humpbacked, blistery-winged, and stubby-legged. In the F2 generation, among 1000 progeny resulting from this cross, you observe the following phenotypes:
Phenotype Number
humpbacked, blistery-winged, and stubby-legged (26 flies) humpbacked, blistery-winged, and long-legged (455 flies) humpbacked, smooth-winged, and long-legged (24 flies) straight-backed, blistery-winged, and stubby-legged (27 flies) straight-backed, blistery-winged, and long-legged (4 flies) straight-backed, smooth-winged, and stubby-legged (442 flies) straight-backed, smooth-winged, and long-legged (22 flies)
(a) The male flies that were bred to the F1 generation in order to produce the F generation were humpbacked, blistery-winged, and stubby-legged. On each of their chromosomes, they have the alleles hb– bl– st–. Using this notation, state the genotype of each of the two true-breeding parental strains (i.e. the two strains in the P generation ).
Genotype of one parental strain:
Genotype of the other parental strain:
Mouse
Mouse
Mouse
Mouse
(b) How many flies are found in the class that is the reciprocal class of the humpbacked, blistery-winged, and stubby-legged flies?
(c) What is the genetic distance between the hb and bl loci? (Label your answer with the proper units.)
(d) What is the genetic distance between the bl and st loci? (Label your answer with the proper units.)
(e) Draw a genetic map showing the correct order of the hb , bl , and st loci.
traits. The mutant trait indicated by the dots is dominant, whereas the mutant trait indicated by the stripes is recessive. Assume 100% penetrance and no new mutations. (Squares = males, circles = females.)
= expressing recessive mutant trait,
= expressing dominant mutant trait,
= expressing both mutant traits
(d) Assuming that the recessive mutant trait is caused by a gene on an autosome and the dominant mutant trait is caused by a gene on the X chromosome, fill in the following chart using the allele notation indicated by the key above. Blocks in the chart that cannot be filled in conclusively should be indicated as “inconclusive.”
Number of “A* ” alleles
Number of “a” alleles
Number of “B” alleles
Number of “b” alleles Mouse #
Mouse #
Mouse #
Mouse #
(e) Assuming that the recessive mutant trait is caused by a gene on an autosome and the dominant mutant trait is caused by a gene on the X chromosome, what is the probability that the mouse indicated by a question mark will show only the recessive mutant trait assuming that the mouse is born female?
white). The “dark tan” phenotype of the haploid cells you are working with is caused by two different mutations in the same strain. The two mutations are designated drk1– and
drk2–.
(a) Mating of the drk1– drk2–^ double mutant to wild-type yeast produces diploids that are white. Sporulation of these diploids yields 50 tetrads. 4 of these tetrads (called “Type One”) contain four light tan spores. 37 of these tetrads (called “Type Two”) contain two dark tan spores and two white spores. 9 of these tetrads (called “Type Three”) contain one dark tan spore, two light tan spores, and one white spore. Categorize each of the tetrad types as parental ditype (PD), tetratype (TT), or nonparental ditype (NPD).
(b) Are the drk1– and drk2– mutations linked? If so, give the distance between them. (Label your answer with the proper units.)
(c) In yeast, 1 cM of genetic distance corresponds to 3,500 base pairs of physical distance. An average yeast gene is about 1,400 base pairs long, and the longest yeast gene is 14,700 base pairs. Keeping this information in mind, you select a “Type Three” tetrad from part (a) and mate the two light tan spores from that tetrad to each other. Can you deduce the color of the resulting diploids? If so , what color would the diploids be?
Next you isolate a mutant strain of yeast that cannot grow on medium lacking leucine. This strain contains a single mutation you call leu1–. The leu1– mutation is near to drk1– on the same chromosome. When the leu1– mutant is mated to wild-type yeast, the resulting diploids cannot grow on medium lacking leucine.
(d) You mate leu1–^ yeast to drk1–^ yeast and sporulate the resulting diploid. You grow the resulting spores on medium containing leucine. You then test for growth on medium lacking leucine. It is apparent that you have isolated only two types of tetrads, 10 tetrads of Type A and 10 tetrads of Type B. On medium lacking leucine, only two spores from each Type A tetrad can grow; both are light tan in color. Complete the chart below so as to indicate: How many spores from each Type B tetrad can grow on medium lacking leucine, and what color is each spore that can grow?
grow on medium lacking leucine
color of each spore that can grow on medium lacking leucine
Type B tetrad
A mutation called eyeless ( ey ) is identified, which gives the autosomal dominant phenotype of having no eyes. You want to map ey relative to bl , but your colleague claims this can't be done since you obviously can't score the presence of black or brown eyes in an eyeless bug. You don't agree that it can’t be done, and you cross a true- breeding single mutant eyeless bug to a true-breeding black-eyed bug. An F1 female that results is then crossed to a true-breeding black-eyed male. The following phenotypes are observed in 100 progeny:
eyeless 51 black-eyed 39 brown-eyed 10
(c) What is the map distance between bl and ey?
wild-type mice, according to the biosynthetic pathway below.
antigen 1
antigen 2
intermediate
enzyme A enzyme C
enzyme B
Mice homozygous for alleles that block the production of enzyme A (genotype a/a) do not make either antigen 1 or antigen 2. Mice homozygous for defects in the gene encoding enzyme B (genotype b/b) do not make antigen 1. Mice homozygous for defects in the gene encoding enzyme C (genotype c/c) do not make antigen 2. All three of these phenotypes of absences of antigen are autosomal recessive phenotypes.
(a) Two different true-breeding strains of mice have been isolated that do not make either antigen 1 or antigen 2. When an individual from one strain is crossed with an individual from the other strain, all of the F1 mice produce both antigens. Write out the genotypes for both strains. (Use “A,” “B,” and “C” to designate the wild-type alleles and “a,” “b,” and “c” to designate the defective alleles of the three genes that encode these enzymes.) Assume that these three genes are unlinked.
(b) Two of the F1 mice are crossed to one another. The possible phenotypes for the F2 progeny are shown below. What proportion of the F2 will be represented by each phenotype on average? Fraction of F
antigen 1+, antigen 2+
antigen 1+, antigen 2–
antigen 1–, antigen 2–
antigen 1–, antigen 2+
(c) Among the F2 progeny, there will be mice of several different genotypes that are
phenotypically antigen 1– and antigen 2–. Suppose you wanted to use a test cross in order to test whether a given F2 mouse that does not express either antigen is defective in production of enzyme A. What genotype would you choose for a mouse to be used for such a test cross of the F2 mouse? Describe the possible outcomes of this cross and how you would interpret them.
biosynthesis form red colonies because of the accumulation of an intermediate in the pathway, which is a red pigment.
(a) You have isolated two different red-colored mutants in haploid yeast strains of
different mating types, which you callade1– andade2–. When either theade1– or ade2– haploid mutant is mated to wild-type haploid yeast, the resulting diploid forms white colonies like those of wild-type yeast. When theade1– haploid mutant is mated to
theade2– haploid mutant, the resulting diploid makes red colonies. From these observations, describe as much as you can about theade1– andade2– mutations and the relationship between them.
(e) What do the results from this tetrad analysis tell you about the relationship between
theade3– andade1– mutations? Mention in your answer whether these mutations can be in the same gene.
(f) Suppose that you wanted to do some experiments with anade3–ade1– double mutant haploid yeast strain. Below is the same image from above of the tetrads formed
from inducing sporulation of the diploid formed by matingade3– andade1– haploid mutants. On this image, circle each spore clone that you can be sure is double mutant haploid strain, without any further testing.
(g) You mate anade3– haploid mutant to anade2– haploid mutant. You then induce sporulation of the resulting diploid, and dissect 18 tetrads. How many Tetratype (TT) tetrads would you expect to see?
different autosomal recessive traits -- apricot eyes, black body, and curly wings. These phenotypes are caused by alleles in three different genes -- a , b , and c respectively. Wild-type flies have red eyes, brown body, and straight wings, and are genotypically a+ , b+ , and c+. Two different true-breeding lines are crossed and the F 1 progeny all appear as wild-type. These F 1 progeny are then crossed to individuals from a true-
breeding line that has all three recessive traits, and 100 progeny from this cross are analyzed. The phenotypes and numbers are as follows:
Phenotype Number wild-type (red eyes, brown body, straight wings) 3 apricot eyes, black body, curly wings 7 apricot eyes, brown body, curly wings 34 red eyes, black body, straight wings 36 red eyes, black body, curly wings 8 apricot eyes, brown body, straight wings 12
(a) What are the genotypes of the two parental true-breeding lines? Use the notation in the introduction to this question.
(b) Why are there only six phenotypic classes, rather than eight?
(c) Give as much information as you can about the chromosomal positions of the three loci a , b , and c. Include in your answer any relevant map distances in cM.
(d) Given the map distances in part (c) , if F 1 insects are crossed to one another, what frequency of the resulting F 2 progeny would have all three recessive traits?
recessive trait. (Assume all phenotypes are completely penetrant and that no new mutations arise).
(a) What is the genotype of mouse 1 if the two traits are X-linked? For your answer use XD^ to designate the allele for the dominant trait (with Xd^ representing the corresponding wild-type allele) and Xr^ to designate the allele for the recessive trait (with XR representing the corresponding wild-type allele).
(b) If the genes for both traits are 30 cM apart on the X chromosome, what is the probability that a progeny mouse indicated by the? will show both traits if she is born female?
mutation (and thus has cinnabar colored eyes) and is homozygous for the shi-1 – mutation (and thus becomes paralyzed at high temperature). You mate this strain to a true-breeding wild-type fly and obtain F1 flies, all of which have the wild type phenotype (red eyes, not paralyzed). F1 females are then mated to males of the starting strain (homozygous cn– and shi-1 –^ ). Among 100 progeny from this cross you observe the following phenotypes: Phenotype Number wild-type (not paralyzed, red eyes) 44 paralyzed, cinnabar eyes 41 not paralyzed, cinnabar eyes 7 paralyzed, red eyes 8
(a) From this data, what is the distance between the cn and shi genes?
(b) You isolate a second allele of theshibire gene designated shi-2–^ , which also causes the recessive phenotype of paralysis at high temperature. Flies from a true-breeding
shi-2–^ strain are crossed to flies from the true-breeding cn–, shi-1–^ strain described above. What is the phenotype of the resulting F1 female flies?
F1 females are then mated to males from the true-breeding cn–, shi-1–^ strain. You collect 10,000 progeny from this cross and note that, although almost all the flies are paralyzed at high temperature, there are 10 that are not paralyzed.
(c) What is the distance between the shi-1 and shi-2 loci?
(d) Among the 10 progeny flies that are not paralyzed that result from the cross described in part (b) , 8 have cinnabar eyes and 2 have normal red eyes. On the basis of this information as well as the results from parts (a) and (b) , draw a genetic map showing the order of the cn, shi-1, and shi-2 loci.
(a) When you mate this haploid mutant to a wild-type haploid yeast strain (that is His +^ ), you find that the resulting diploids are His+^. What does this tell you about the mutant that you isolated?
(b) When you induce sporulation in the His +^ diploid from part ( a ), you find that tetrads of three types are produced. From a total of 100 tetrads, the following tetrad types are seen:
Type: 2 His –^ : 2 His +^ spores 3 His –^ : 1 His +^ spores 4 His –^ spores Number: 65 30 5
What does this result tell you about the original His –^ strain? Give any relevant genetic distances (in cM) that you can calculate.
(c) There are a total of 240 His –^ spore clones in the 100 tetrads from part ( b ). If you picked two of these His–^ clones (of opposite mating type) at random and mated them, what is the probability that the resulting diploid would be His +^? (You may find it helpful to consider the genotypes of the His –^ spores in each tetrad type).
allergy to milk are shown by shaded-in symbols, and unaffected individuals are shown by unshaded symbols.
= female
= male
1
1
1
1
2
2 3 4
(^2 3 4 )
2 3 4
6
5 6 7
(b) When the diploids from part (a) are induced to sporulate, three types of tetrads are found. Type I have 4 red spores. Type II have 1 white spore and 3 red spores. Type III have 2 white spores and 2 red spores
Classify each tetrad type as PD, NPD or TT.
(c) When the number of each tetrad type is tallied, you find that the cross produces 30 Type I tetrads, 16 Type II tetrads, and 4 Type III tetrads. Are the red3 and red4 loci linked? If so, how far apart are they in cM?
(d) One of the Type II tetrads from above is selected for further analysis and you designate the four spore clones a , b , c , and d. Clone a is white, whereas clones b , c , and d are red. Each clone is mated to either a red3 haploid mutant or a red4 haploid mutant, and the color of the resulting diploid is noted.
x red3 haploid → white diploid Clone a (white) x red4 haploid → white diploid
x red3 haploid → red diploid Clone b (red) x red4 haploid → white diploid
x red3 haploid → red diploid Clone c (red) x red4 haploid → red diploid
x red3 haploid → white diploid Clone d (red) x red4 haploid → red diploid
Give the genotypes of each of the four spore clones with respect to red3 and to red.
Each individual is numbered for reference in this problem. PLEASE NOTE that, in this pedigree, the phenotypes of the family members are NOT denoted. They will be described in the text of the question instead. In this problem, assume complete penetrance and no new mutations.
(a) Say that female #1 exhibits a rare recessive X-linked trait, and that male #2 does not exhibit the trait. Because the trait is rare, assume that the individuals #3 and #6 neither have nor are carriers of the trait.
What is the probability that male #4 will have the trait?
What is the probability that female #5 will have the trait?
What is the probability that female #7 will have the trait?
What is the probability that male #8 will have the trait?
What is the probability that male #9 will have the trait?
Next you isolate a second mutation in a different gene ( hb-2 ) that also causes the recessive phenotype of humpback. A female from a true-breeding hb-2 strain is crossed to a male from a true breeding cr , hb-1 strain. An F 1 female from this cross is
then crossed to a true-breeding cr hb-1 hb-2 male (who has curly wings and a humpback) and 500 progeny are examined.
Phenotype Number of flies straight-wings, straight back 5 straight wings, humpback 240 curly-wings, humpback 255
(b) What is the phenotype of the F1 females in this cross?
(c) What is the distance between the hb-1 and hb-2 loci in cM?
(d) Draw a genetic map showing the relative order of the cr , hb-1 , and hb-2 loci.
this true-breeding strain is crossed to a wild-type female, all of the female F 1 progeny
exhibit both traits, whereas all of the male F 1 progeny look wild-type. Assume
complete penetrance and no new mutations.
(a) What is the mode of inheritance of the two traits?
(b) The male and female F 1 mice described above are crossed to one another to
produce F 2 progeny. Of the male F 2 progeny, 40% have both traits (the rest of the F 2
males either appear wild-type or have only one trait or the other). What fraction of the female F 2 progeny would you expect to have both traits?
(c) What is the map distance (in cM) between the genes for the two traits?
your haploid mutant to a haploid wild-type strain, the resulting diploids look like wild- type.
(a) What does this observation tell you about your mutant?
(b) When the diploids from part (a) are induced to sporulate, all of the tetrads appear to be PDs. What does this observation tell us about your mutant?
(c) What is the phenotype of each of the four spores from a PD tetrad described in part (b)?
(d) You isolate a second haploid mutant that also makes small colonies. When a haploid of one small mutant is mated to a haploid of the other small mutant, the resulting diploids appear normal. What is the relationship between the two “small” mutations?
When the diploids from part (d) are induced to sporulate, three types of tetrads are found.
Type I have 4 small spores
Type II have 1 normal and 3 small spores
Type III have 2 normal and 2 small spores
The cross produces 24 type I tetrads, 24 type II tetrads, and 2 type III tetrads.
(e) What is the map distance between the two “small” loci?
(f) Give your best estimate for the number of tetrads (out of 50 total) described in part (d) that resulted from two crossovers in the interval between the two “small” loci.