3.13 In Drosophila, the mutant allele bwdts causing brown eyes (normal eyes are red) is temperature sensitive. In flies reared at 29°C the mutant allele is dominant, but in flies reared at 22°C the mutant allele is recessive. In a cross of bwdts/+ × bwdts/+, where the + sign denotes the wild-type allele of bwdts, what is the expected ratio of brown-eyed flies to red-eyed flies if the progeny are reared at 29°C? At 22°C?
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- 10 cM separates two hypothetical autosomal human genes. The dominant alleles are have complete penetrance and will result to Crossed eyes (e+) and short thumbs (th+). Four children are born to a normal guy and cross-eyed, small-thumbed woman. Two of the children have short thumbs and the other two have crossed eyes. She is carrying her fifth child. What is the probability that this fifth child will be cross-eyed and have short thumbs?A rare blinding disease that has a relation to dengenerative factors is partially penetrant. In the following pedigrees for two families, the affected symptomatic individuals (black circles and squares) have been diagnosed with this disease due to the mutation in mitochondrial DNA m.14484T>C. If III.4 is homoplasmic for m.14484T>C in hair, blood, urine and other tissues examined. What will occur with IV.7 then?Male Drosophila expressing the autosomal recessivemutations sc (scute), ec (echinus), cv (crossveinless),and b (black) were crossed to phenotypically wildtype females, and the 3288 progeny listed wereobtained. (Only mutant traits are noted.)653 black, scute, echinus, crossveinless670 scute, echinus, crossveinless675 wild type655 black71 black, scute73 scute73 black, echinus, crossveinless74 echinus, crossveinless87 black, scute, echinus84 scute, echinus86 black, crossveinless83 crossveinless1 black, scute, crossveinless1 scute, crossveinless1 black, echinus1 echinusa. Diagram the genotype of the female parent.b. Map these loci.c. Do the data provide evidence of interference?Justify your answer with numbers.
- In a cross in Drosophila, a female heterozygous for the autosomallylinked genes a, b, c, d, and e (abcde/ + + + + +) was testcrossedwith a male homozygous for all recessive alleles (abcde/abcde).Even though the distance between each of the loci was at least3 map units, only four phenotypes were recovered, yielding thefollowing data: Phenotype No. of Flies+ + + + + 440a b c d e 460+ + + + e 48a b c d + 52 Total = 1000 Why are many expected crossover phenotypes missing? Can anyof these loci be mapped from the data given here? If so, determinemap distances.2. a. A Drosophila male from a true-breeding stockwith scabrous eyes was mated with a female from atrue-breeding stock with javelin bristles. Both scabrous eyes and javelin bristles are autosomal recessive mutant traits. The F1 progeny all had normaleyes and bristles. F1 females from this cross weremated with males with both scabrous eyes andjavelin bristles. Write all the possible phenotypicclasses of the progeny that could be produced from the cross of the F1 females with the scabrous, javelin males, and indicate for each class whether it is arecombinant or parental type.b. The cross in part (a) yielded the following progeny:77 scabrous eyes and normal bristles; 76 wild type(normal eyes and bristles); 74 normal eyes andjavelin bristles; and 73 scabrous eyes and javelinbristles. Are the genes governing these traits likelyto be linked, or do they instead assort independently? Why?c. Suppose you mated the F1 females from the crossin part (a) to wild-type males. Why would thiscross fail…7-5. In corn a pair of genes determines leaf shape and another pair determines pollen shape. A ragged-leafed plant with round pollen was crossed to a ragged-leafed plant with angular-pollen, and the resultant progeny were classified as follows: Class Phenotypes 186 ragged-leaf round-pollen 11 IV 174 ragged-leaf angular-pollen 57 smooth-leaf round-pollen Total 63/480 smooth-leaf angular-pollen (a) Using alphabetical letters of your choice, designate the genes for the different leaf and pollen characters. (b) On the basis of the symbols given in (a) 1 provide genotypes two parents. (c) (forth)/(nbers) According to your hypothesis what would you have expected for each of the four classes of progeny? (d) After re viewing Chapter 8, use the chi-square method (p. 132) to test your hypothesis and indicate whether you accept or reject it.
- Cinnabar eyes (cn) and reduced bristles (rd) are linked autosomal recesive characters in Drosophila fruit flies. A homozygous wild-type female was crossed to a reduced, cinnabar male, and the F1 females were then crossed with cinnabar reduced males to obtain the F2. Of the 200 F2 offspring obtained, 86 were wild type, 13 were cinnabar, 17 were reduced and 84 were reduced and cinnabar.. What is the map distance between the cn and rd alleles? 8,4 mu 15 mu 17 mu 30 mu 13 mua. 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?. The production of pigment in the outer layer of seedsof corn requires each of the three independently assorting genes A, C, and R to be represented by at leastone dominant allele, as specified in Problem 64. Thedominant allele Pr of a fourth independently assortinggene is required to convert the biochemical precursorinto a purple pigment, and its recessive allele pr makesthe pigment red. Plants that do not produce pigmenthave yellow seeds. Consider a cross of a strain of genotype A/A ; C/C ; R/R ; pr/pr with a strain of genotypea/a ; c/c ; r/r ; Pr/Pr.a. What are the phenotypes of the parents?b. What will be the phenotype of the F1?c. What phenotypes, and in what proportions, willappear in the progeny of a selfed F1?d. What progeny proportions do you predict from thetestcross of an F1?
- . A white-eyed female Drosophila is mated with a redeyed (wild-type) male, the reciprocal cross of the oneshown in Figure 15.4. What phenotypes and genotypesdo you predict for the offspring from this cross?34 a,b Note that the starting heterozygous plant does not have its parents genotypes listed. However, we can determine that the starting heterozygous plant has S,U,tu alleles on one chromosome and s,u,Tu alleles on the other because they are most common combinations in the offspring.In Drosophila, males from a true-breeding stock withraspberry-colored eyes were mated to females from atrue-breeding stock with sable-colored bodies. In theF1 generation, all the females had wild-type eye andbody color, while all the males had wild-type eyecolor but sable-colored bodies. When F1 males andfemales were mated, the F2 generation was composedof 216 females with wild-type eyes and bodies, 223females with wild-type eyes and sable bodies, 191males with wild-type eyes and sable bodies, 188 maleswith raspberry eyes and wild-type bodies, 23 maleswith wild-type eyes and bodies, and 27 males withraspberry eyes and sable bodies. Explain these resultsby diagramming the crosses and calculating any relevant map distances