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.

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.

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Description: 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 a

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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The pedigree below shows that inheritance of a disease that is caused by a late onset, dominant, autosomal mutation that is rare, but only 50% penetrant. The gene that is mutated in the disease is linked at a distance of 10 cm to a microsatellite marker that has alleles numbered 1, 2, and 3. The marker alleles detected in each individual are indicated below. What is the probability that individual A will develop the disease? Explain using an illustration of this occurs.
Three autosomal recessive mutations in yeast, all producing the same phenotype (m1, m2, and m3), are subjected to complementation analysis. Of the results shown below, which, if any, are alleles of one another? Predict the results of the cross that is not shown—that is, m2 * m3. Cross 1: m1 * m24 F1: all wild-type progeny Cross 2: m1 * m34 F1: all mutant progeny
In an autotetraploid Chinese primrose (Primula sinensis L.), the gene controlling stigma color is very near the centromere of the chromosome carrying it. The allele G for green stigma is dominant to g for red stigmas. A homozygous green autotetraploid strain is crossed with a homozygous red autotetraploid strain. d. If the G locus were 50 or more map units from the centromere, what types and proportions of gametes would the F1 be expected to produce? Derive the expected F2 phenotypic ratio.
In mice, the trait for high cholesterol is specified by a dominant allele designated HC, whereas the wild-type allele for normal cholesterol levels is designated hc. Black fur is specified by a recessive allele designated bl, whereas the wild-type allele which gives brown fur is designated BL. The genes for both of these traits are 30cM apart on the same autosome. A brown female (#1) with high cholesterol is mated to a black male (#2) with normal cholesterol. The progeny from this cross include a brown male (#3) with high cholesterol and a black female (#4) with normal cholesterol. What is the probability that the black mouse in the progeny of the first cross will also have high cholesterol?
Explain why it is possible for the proband in the following pedigree to have children of blood types A, B, and AB. Considering epistatic genes, what are the possible genotypes of II-2?
In mice, the trait for high cholesterol is specified by a dominant allele designatedHC, whereas the wild-type allele for normal cholesterol levels is designated hc.Black fur is specified by a recessive allele designated bl, whereas the wild-typeallele which gives brown fur is designated BL. The genes for both of these traitsare 30cM apart on the same autosome. A brown female (#1) with high cholesterolis mated to a black male (#2) with normal cholesterol. The progeny from this crossinclude a brown male (#3) with high cholesterol and a black female (#4) withnormal cholesterol. Mouse #3 is mated to a brown female (#5) with normalcholesterol. If 4 mice are produced from this cross, what is the probability that all 4will have brown fur?
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?
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?
In the fruit fly, recessive mutations in either of two independently assorting genes, brown and purple, prevent the synthesis of red pigment in the eyes. Thus, homozygotes for either of these mutations have brownish-purple eyes. However, heterozygotes for both of these mutations have dark red, that is, wild-type eyes. If such double heterozygotes are intercrossed, what kinds of progeny will be produced, and in what proportions?
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The mutant genes for vestigial wings and singed bristles are approximately 30 map units apart on chromosome II in Drosophila. Assume that a vestigial-winged female was mated to a singed-bristle male, and that the resulting F1 phenotypically wild type females were mated to vestigial singed males. Of 1000 offspring, which phenotype class would represent the product of crossing over between the genes, and how many would you expect? a) vestigial, 300 flies b) vestigial, 150 flies wild type, 300 flies d) singed, 150 flies e) vestigial, singed double mutants, 150 flies
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