The balancer chromosome includes GFP and a mutation in gene X that causes flies to be sterile in the homozygous state. The recessive mutation that you are studying is in gene Y, and causes flies to develop two tail ends in the homozygous state. This mutation is lethal in the homozygous state. For each of the following phenotypes in the balanced mutant line, what alleles are present and in how many copies? a.) Glows green, fertile (Select) b.) Glows green, sterile ISelect) c.) does not glow green, two tails ( Select] If you crossed a fly from your balanced line with the following flies, what proportion of the offspring would glow green and be fertile? d.) wild-type ISelect ) e.) fly with balancer chromosome from parental generation [ Select ]

The balancer chromosome includes GFP and a mutation in gene X that causes flies to be sterile in the homozygous state. The recessive mutation that you are studying is in gene Y, and causes flies to develop two tail ends in the homozygous state. This mutation is lethal in the homozygous state. For each of the following phenotypes in the balanced mutant line, what alleles are present and in how many copies? a.) Glows green, fertile (Select) b.) Glows green, sterile ISelect) c.) does not glow green, two tails ( Select] If you crossed a fly from your balanced line with the following flies, what proportion of the offspring would glow green and be fertile? d.) wild-type ISelect ) e.) fly with balancer chromosome from parental generation [ Select ]

Name: You ave established a mutant line of flies with a balancer chromosome
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Description: You ave established a mutant line of flies with a balancer chromosome.The balancer chromosome includes GFP and a mutation in gene X thatcauses flies to be sterile in the homozygous state. The recessive mutationthat you are studying is in gene Y, and

Human Heredity: Principles and Issues (MindTap Course List)

11th Edition

ISBN:9781305251052

Author:Michael Cummings

Publisher:Michael Cummings

Chapter11: Genome Alterations: Mutation And Epigenetics

Section: Chapter Questions

Problem 16QP: Familial retinoblastoma, a rare autosomal dominant defect, arose in a large family that had no prior...

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. A diploid strain of yeast was made by mating a haploidstrain with a genotype w−, x−, y−, and z− with a haploidstrain of opposite mating type that is wild type for thesefour genes. The diploid strain was phenotypically wildtype. Four different X-ray-induced diploid mutantswith the following phenotypes were produced fromthis diploid yeast strain. Assume a single new mutation is present in each strain.Strain 1 w− x+ y− z+Strain 2 w+ x− y− z−Strain 3 w− x+ y− z−Strain 4 w− x+ y+ z+When these mutant diploid strains of yeast go throughmeiosis, each ascus is found to contain only two viablehaploid spores.a. What kind of mutations were induced by X-rays tomake the listed diploid strains?b. Why did two spores in each ascus die?c. Are any of the genes w, x, y, or z located on thesame chromosome?d. Give the order of the genes that are found on thesame chromosome
A Drosophila male is heterozygous for a reciprocaltranslocation between an autosome and the Y chromosome. The part of the autosome now present onthe Y chromosome contains the dominant mutationLyra (shortened wings); the other (normal) copyof the same autosome is Lyra+. This male is nowmated with a true-breeding, wild-type female. Whatkinds of progeny would be obtained, and in whatproportions?
In a haploid yeast strain, eight recessive mutationswere found that resulted in a requirement for theamino acid lysine. All the mutations were found to revert at a frequency of about 1 × 10−6 except mutations5 and 6, which did not revert. Matings were madebetween a and α cells carrying these mutations. Theability of the resultant diploid strains to grow onminimal medium in the absence of lysine is shown inthe following chart (+ means growth and − means nogrowth.)1 2 3 4 5 6 7 81 − + + + + − + −2 + − + + + + + +3 + + − − − − − +4 + + − − − − − +5 + + − − − − − +6 − + − − − − − −7 + + − − − − − +8 − + + + + − + −a. How many complementation groups were revealedby these data? Which point mutations are foundwithin which complementation groups?The same diploid strains are now induced to undergosporulation. The vast majority of resultant spores areauxotrophic; that is, they cannot form colonies whenplated on minimal medium (without lysine). However,particular diploids can produce rare spores…
In Drosophila, a cross (cross 1) was made between twomutant flies, one homozygous for the recessive mutationbent wing (b) and the other homozygous for the recessivemutation eyeless (e). The mutations e and b are alleles oftwo different genes that are known to be very closelylinked on the tiny autosomal chromosome 4. All the progeny had a wild-type phenotype. One of the female progeny was crossed with a male of genotype b e/b e ; we willcall this cross 2. Most of the progeny of cross 2 were of theexpected types, but there was also one rare female ofwild-type phenotype.a. Explain what the common progeny are expected tobe from cross 2.b. Could the rare wild-type female have arisen by (1)crossing over or (2) nondisjunction? Explain.
In Drosophila, a heterozygous female for the X-linkedrecessive traits a, b, and c was crossed to a male that phenotypically expressed a, b, and c. The offspring occurred inthe following phenotypic ratios.+ b c 460a + + 450a b c 32+ + + 38a + c 11+ b + 9 No other phenotypes were observed.(a) What progeny phenotypes are missing? Why?
The a, b, and c loci are all on different chromosomesin yeast. When a b+ yeast were crossed to a+ b yeastand the resultant tetrads analyzed, it was found thatthe number of nonparental ditype tetrads was equal tothe number of parental ditypes, but there were no tetratype asci at all. On the other hand, many tetratypeasci were seen in the tetrads formed after a c+ wascrossed with a+ c, and after b c+ was crossed withb+ c. Explain these results.
The Drosophila chromosome 4 is extremely small;virtually no recombination occurs between genes onthis chromosome. You have available three differentlymarked chromosome 4s: one has a recessive allele ofthe gene eyeless (ey), causing very small eyes; one hasa recessive allele of the cubitus interruptus (ci) gene,which causes disruptions in the veins on the wings;and the third carries recessive alleles of both genes.Drosophila adults can survive with two or three, butnot with one or four, copies of chromosome 4.a. How could you use these three chromosomes tofind Drosophila mutants with defective meiosescausing an elevated rate of nondisjunction?b. Would your technique allow you to discriminatenondisjunction occurring during the first meioticdivision from nondisjunction occurring during thesecond meiotic division?
The Drosophila chromosome 4 is extremely small;virtually no recombination occurs between genes onthis chromosome. You have available three differentlymarked chromosome 4s: one has a recessive allele ofthe gene eyeless (ey), causing very small eyes; one hasa recessive allele of the cubitus interruptus (ci) gene,which causes disruptions in the veins on the wings;and the third carries recessive alleles of both genes.Drosophila adults can survive with two or three, butnot with one or four, copies of chromosome 4.a. How could you use these three chromosomes tofind Drosophila mutants with defective meiosescausing an elevated rate of nondisjunction?b. Would your technique allow you to discriminatenondisjunction occurring during the first meioticdivision from nondisjunction occurring during thesecond meiotic division?c. What progeny types would you expect if a flyrecognizably formed from a gamete produced bynondisjunction were testcrossed to a fly homozygous for a chromosome 4 carrying both ey…
In Drosophila, a heterozygous female for the X-linkedrecessive traits a, b, and c was crossed to a male that phenotypically expressed a, b, and c. The offspring occurred inthe following phenotypic ratios.+ b c 460a + + 450a b c 32+ + + 38a + c 11+ b + 9 No other phenotypes were observed.(a) What is the genotypic arrangement of the alleles ofthese genes on the X chromosome of the female?
Flies homozygous for recessive null mutations in thesevenless (sev) or bride-of-sevenless (boss) genes have the same mutant phenotype: Every ommatidium(facet) in their eyes lacks photoreceptor cell 7 (R7).The R7 cells enable flies to detect UV light.a. Given that flies normally move toward light, suggest a screening method that would enable you toidentify mutations in additional genes required forR7 determination.b. Would you be able to recover mutations in everygene required for R7 development with yourmethod? Explain.c. How could you tell whether any of the new mutationsyou found in your screen are alleles of sev or boss?d. Suppose you found one recessive mutant allele ofa gene not previously known to be involved in eyedevelopment. How could you use this allele in anew mutagenesis screen to find additional allelesof this gene? Why might you want additional mutant alleles to study the process?
In Drosophila, a heterozygous female for the X-linkedrecessive traits a, b, and c was crossed to a male that phenotypically expressed a, b, and c. The offspring occurred inthe following phenotypic ratios.+ b c 460a + + 450a b c 32+ + + 38a + c 11+ b + 9 No other phenotypes were observed.(a) Determine the correct sequence and construct amap of these genes on the X chromosome ?
The following list of four Drosophila mutations indicates the symbol for the mutation, the name of thegene, and the mutant phenotype:Allele symbol Gene name Mutant phenotypedwp dwarp small body, warped wingsrmp rumpled deranged bristlespld pallid pale wingsrv raven dark eyes and bodiesYou perform the following crosses with the indicatedresults:Cross #1: dwarp, rumpled females × pallid, raven males→ dwarp, rumpled males and wild-type femalesCross #2: pallid, raven females × dwarp, rumpled males→ pallid, raven males and wild-type femalesF1 females from cross #1 were crossed to males froma true-breeding dwarp rumpled pallid raven stock.The 1000 progeny obtained were as follows:pallid 3pallid, raven 428pallid, raven, rumpled 48pallid, rumpled 23dwarp, raven 22dwarp, raven, rumpled 2dwarp, rumpled 427dwarp 47Indicate the best map for these four genes, includingall relevant data. Calculate interference values whereappropriate.