. Miniature wings (X") in Drosophila result from an X-linked allele that is recessive to the allele for long wings (X *). Give the genotypes of the parents in each of the following crosses. Male Female Male offspring Female offspring parent parent 231 long, 250 a. long long 560 long miniature 250 miniature b. miniature long 610 long 632 long 410 long, 417 miniature 412 long, 415 c. miniature long miniature 417 miniature 415 miniature d. long 761 long 630 long miniature 753 miniature e. long long 625 long
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- . The normal eye color of Drosophila is red, but strains inwhich all flies have brown eyes are available. Similarly,wings are normally long, but there are strains with shortwings. A female from a pure line with brown eyes andshort wings is crossed with a male from a normal pureline. The F1 consists of normal females and short-wingedmales. An F2 is then produced by intercrossing the F1.Both sexes of F2 flies show phenotypes as follows:38 red eyes, long wings38 red eyes, short wings18 brown eyes, long wings18 brown eyes, short wingsDeduce the inheritance of these phenotypes; useclearly defined genetic symbols of your own invention.State the genotypes of all three generations and thegenotypic proportions of the F1 and F2.Unpacking Problem 31Before attempting a solution to this problem, try answering thefollowing questions:1. What does the word normal mean in this problem?2. The words line and strain are used in this problem.What do they mean, and are they interchangeable?3. Draw a simple…The maternal-effect mutation bicoid (bcd) is recessive. Inthe absence of the bicoid protein product, embryogenesis isnot completed. Consider a cross between a female heterozygousfor the bicoid mutation (bcd+/ bcd-) and a homozygousmale(bcd-/ bcd-). Predict the outcome (normal vs. failed embryogenesis) inthe F1 and F2 generations of the cross described.The XG locus on the human X chromosome has twoalleles, XG+ and XG. The XG+ allele causes the presence of the Xg surface antigen on red blood cells,while the recessive XG allele does not allow antigento appear. The XG locus is 10 m.u. from the STSlocus. The STS+ allele produces normal activity ofthe enzyme steroid sulfatase, while the recessive STSallele results in the lack of steroid sulfatase activityand the disease ichthyosis (scaly skin). A man withichthyosis and no Xg antigen has a normal daughterwith Xg antigen. This daughter is expecting a child.a. If the child is a son, what is the probability he willlack Xg antigen and have ichthyosis?b. What is the probability that a son would have boththe antigen and ichthyosis?c. If the child is a son with ichthyosis, what is theprobability he will have Xg antigen?
- 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?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 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.
- a. In Drosophila, crosses between F1 heterozygotes ofthe form A b / a B always yield the same ratio ofphenotypes in the F2 progeny regardless of the distance between the two genes (assuming completedominance for both autosomal genes). What is thisratio? Would this also be the case if the F1 heterozygotes were A B / a b? (Hint: Remember that inDrosophila, recombination does not take placeduring spermatogenesis.)b. If you intercrossed F1 heterozygotes of the formA b / a B in mice, the phenotypic ratio among the F2progeny would vary with the map distance betweenthe two genes. Is there a simple way to estimate themap distance based on the frequencies of the F2phenotypes, assuming rates of recombination areequal in males and females? Could you estimatemap distances in the same way if the mouse F1heterozygotes were A B / a b?Female fruit flies homozygous for the X-linked white-eye alleleare crossed to males with red eyes. On very rare occasions, an offspringof such a cross is a male with red eyes. Assuming these rareoffspring are not due to a new mutation in one of the mother’s Xchromosomes that converted the white-eye allele into a red-eyeallele, explain how a red-eyed male arises.In Drosophila, the gene loci for curved wings and purple eyes are 20 centimorgans (map units) apart from each other on an autosomal chromosome. Wild-type flies were mated with double mutant flies with cv cv p p genotypes and curved wing/purple eye phenotypes. AlIl of the F1 flies had wild type phenotypes. The F, flies were crossed with double mutant flies with cv cv p'p genotypes and curved wing/purple eye phenotypes. What are the predicted percentages of flies in the F2 generation with wild type wings and eyes, curved wings and wild type eyes, wild type wings and purple eyes, and curved wings and purple eyes?
- The recessive, X-linked z1mutation of the Drosophilagene zeste (z) can produce a yellow (zeste) eye coloronly in flies that have two or more copies of the wildtype white (w) gene. Using this property, tandem duplications of the w+ gene called w+Rwere identified.Males with the genotype y+ z1w+R spl+ / Y thus havezeste eyes. These males were crossed to females withthe genotype y z1 w+R spl / y+ z1 w+R spl+. (These fourgenes are closely linked on the X chromosome, in theorder given in the genotype, with the centromere tothe right of all these genes: y = yellow bodies; y+ =tan bodies; spl = split bristles; spl+ = normal bristles.) Out of 81,540 male progeny of these females,the following exceptions were found:Class A 2430 yellow bodies, zeste eyes, wild-type bristlesClass B 2394 tan bodies, zeste eyes, split bristlesClass C 23 yellow bodies, wild-type eyes, wild-type bristlesClass D 22 tan bodies, wild-type eyes, split bristlesa. What were the phenotypes of the remainder of the81,540 males…In Drosophila, a female fly is heterozygous for three mutations, Bareyes (B), miniature wings (m), and ebony body (e). Note that Bar isa dominant mutation. The fly is crossed to a male with normal eyes,miniature wings, and ebony body. The results of the cross are asfollows.111 miniature29 wild type117 Bar26 Bar, mimatue101 Bar, ebony31 Bar, miniature, ebony35 ebony115 miniature, ebonyInterpret the results of this cross. If you conclude that linkage isinvolved between any of the genes, determine that map distance(sbetween them.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.