A GENETICIST DISCOVERS A NEW MUTATION IN PEAS THAT CAUSES MOTTLED FLOWER COLOR. SHE CALLS THIS MUTATION SPOTTY AND DETERMINES THAT IT IS DUE TO AN AUTOSOMAL RECESSIVE ALLELE (S). SHE WANTS TO DETERMINE WHETHER THE GENE ENCODING SPOTTY IS LINKED TO THE RECESSIVE GENE FOR FLOWER COLOR (WHITE, W, IS RECESSIVE TO PURPLE). SHE CROSSES A PLANT HETEROZYGOUS FOR BOTH TRAITS WITH A PLANT HOMOZYGOUS RECESSIVE FOR BOTH TRAITS AND OBTAINED THE FOLLOWING PROGENY: NORMAL, PURPLE: 86 NORMAL, WHITE: 14 SPOTTY, PURPLE: 10 SPOTTY, WHITE: 90 • WHAT ARE THE GENOTYPES FOR EACH OF THESE FOUR PROGENY ● . TYPES? WHICH ALLELES ARE FOUND ON EACH CHROMOSOME OF THE DOUBLE HETEROZYGOTE PARENT? (USE SLASH NOTATION.) WHAT IS THE DISTANCE IN CM BETWEEN THESE TWO GENES?
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- A corn geneticist wants to obtain a corn plant that hasthe three dominant phenotypes: anthocyanin (A), longtassels (L), and dwarf plant (D). In her collection ofpure lines, the only lines that bear these alleles are AALL dd and aa ll DD. She also has the fully recessive lineaa ll dd. She decides to intercross the first two and testcross the resulting hybrid to obtain in the progeny aplant of the desired phenotype (which would have to beAa Ll Dd in this case). She knows that the three genesare linked in the order written, that the distance between the A/a and the L/l loci is 16 m.u., and that thedistance between the L/l and the D/d loci is 24 m.u.a. Draw a diagram of the chromosomes of the parents,the hybrid, and the tester.b. Draw a diagram of the crossover(s) necessary toproduce the desired genotype.c. What percentage of the testcross progeny will be ofthe phenotype that she needs?d. What assumptions did you make (if any)?A GENETICIST DISCOVERS A NEW MUTATION IN PEAS THAT CAUSES MOTTLED FLOWER COLOR. SHE CALLS THIS MUTATION SPOTTY AND DETERMINES THAT IT IS DUE TO AN AUTOSOMAL RECESSIVE ALLELE (S). SHE WANTS TO DETERMINE WHETHER THE GENE ENCODING SPOTTY IS LINKED TO THE RECESSIVE GENE FOR FLOWER COLOR (WHITE, W, IS RECESSIVE TO PURPLE). SHE CROSSES A PLANT HETEROZYGOUS FOR BOTH TRAITS WITH A PLANT HOMOZYGOUS RECESSIVE FOR BOTH TRAITS AND OBTAINED THE FOLLOWING PROGENY:NORMAL, PURPLE: 86 NORMAL, WHITE: 14 SPOTTY, PURPLE: 10 SPOTTY, WHITE: 90• WHAT ARE THE GENOTYPES FOR EACH OF THESE FOUR PROGENY TYPES?• WHICH ALLELES ARE FOUND ON EACH CHROMOSOME OF THE DOUBLE HETEROZYGOTE PARENT? (USE SLASH NOTATION.)• WHAT IS THE DISTANCE IN CM BETWEEN THESE TWO GENES?. In 1919, Calvin Bridges began studying an X-linkedrecessive mutation causing eosin-colored eyes inDrosophila. Within an otherwise true-breedingculture of eosin-eyed flies, he noticed rare variantsthat had much lighter cream-colored eyes. By intercrossing these variants, he was able to make a truebreeding cream-eyed stock. Bridges now crossedmales from this cream-eyed stock with true-breedingwild-type females. All the F1 progeny had red (wildtype) eyes. When F1 flies were intercrossed, the F2progeny were 104 females with red eyes, 52 maleswith red eyes, 44 males with eosin eyes, and14 males with cream eyes. Assume that thesenumbers represent an 8:4:3:1 ratio.a. Formulate a hypothesis to explain the F1 and F2results, assigning phenotypes to all possiblegenotypes.b. What do you predict in the F1 and F2 generations if the parental cross is between truebreeding eosin-eyed males and true-breedingcream-eyed females?c. What do you predict in the F1 and F2 generationsif the parental cross is…
- In individuals affected by cystic fibrosis, salt crystals may appear afterperspiration dries up. In addition, the disease causes respiratory disorderswhich can be both debilitating and lethal. It occurs in individuals homozygousfor the recessive gene. Two normal parents had a daughter with thesymptoms of this disease, and a normal son who marries a normal womanwith an afflicted A test (salt concentration in perspiration of heterozygotes ishigher than normal) disclosed that both are indeed carriers of the gene. If thefirst child born to the mating in (b) was defective, what is the probability thatthe 2nd child would also be defective?Express answer in fraction formIn 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.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-). How is it possible for a male homozygous for the mutationto exist?
- A geneticist discovers a new mutation in Drosophila melanogaster that causes the flies to shake and quiver. She calls this mutation quiver (qu)and determines that it is due to an autosomal recessive gene. She wants to determine whether the gene encoding the quiver phenotype is linked to the recessive gene encoding vestigial (reduced) wings (vg). She crosses a fly homozygous for quiver and vestigial traits with a fly homozygous for the wild-type traits and then uses the resulting F1 females in a testcross. She obtains the following flies from this testcross: vg+ qu+ 230 vg qu 224 vg qu+ 97 vg+ qu 99 Total 650Are the genes that cause vestigial wings and the quiver phenotype linked? Do a chi-square test of independence to determine whether the genes have assorted independentlyAn autosomal recessive form of deafness is caused by a mutation in the ESPN gene (E = dominant allele and e = recessive allele). An autosomal recessive disease called Usher syndrome is caused by a mutation in the USH2A gene (H = dominant allele and h = recessive allele). The ESPN and USH2A genes are far apart on chromosome 1. Iris's genotype= eeHH Zacharias' genotype= EEhh Zacharias and Iris have a daughter named Gianna (EeHh) who produces an eh gamete. Here are some pictures of metaphase I cells before recombination has been resolved. Which one of the pictures correctly depicts Gianna's metaphase I cell that could divide to produce her eh gamete? (Hint: it is useful to consider the gametes that Iris and Zacharias each contributed to Gianna)You have a Drosophila line that is homozygous for autosomal recessive alleles a, b, and c, linked in that order. Youcross females of this line with males homozygous for thecorresponding wild-type alleles. You then cross the F1 heterozygous males with their heterozygous sisters. You obtain the following F2 phenotypes (where letters denoterecessive phenotypes and pluses denote wild-type phenotypes): 1364 + + +, 365 a b c, 87 a b +, 84 + + c,47 a + +, 44 + b c, 5 a + c, and 4 + b +.a. What is the recombinant frequency between a andb? Between b and c? (Remember, there is no crossingover in Drosophila males.)b. What is the coefficient of coincidence?
- Bloom syndrome is an autosomal recessive disease that exhibitshaploinsufficiency. A recent survey showed that people heterozygousfor mutations at the BLM locus are at increased risk of colon cancer.Suppose you are a genetic counselor. A young woman is referred to youwhose mother has Bloom syndrome; the young woman’s father has nofamily history of Bloom syndrome. The young woman asks whether sheis likely to experience any other health problems associated with herfamily history of Bloom syndrome. What advice would you give her?In 1995, doctors reported a Chinese family in whichretinitis pigmentosa (progressive degeneration of theretina leading to blindness) affected only males. Allsix sons of affected males were affected, but all of thefive daughters of affected males (and all of thechildren of these daughters) were unaffected.a. What is the likelihood that this form of retinitispigmentosa is due to an autosomal mutationshowing complete dominance?b. What other possibilities could explain the inheritance of retinitis pigmentosa in this family? Whichof these possibilities do you think is most likely?Familial retinoblastoma, a rare autosomal dominant defect, arose in a large family that had no prior history of the disease. Consider the following pedigree (the darkly colored symbols represent affected individuals): a. Circle the individual(s) in which the mutation most likely occurred. b. Is the person who is the source of the mutation affected by retinoblastoma? Justify your answer. c. Assuming that the mutant allele is fully penetrant, what is the chance that an affected individual will have an affected child?