Two models for dominance of a mutation a Model 1: Model 2: Phenotype t Haploinsufficiency Dominant negative p +/+ a 2 "doses" of product Dimer Wild type be MIM Mutant a 0 "dose" ti ti +/M Mutant t 1 "dose" (inadequate) tl sufficiency and the a« mutation is illustrated FIGURE 6-2 Amutation may be dominant because (left) a single wild-type gene does not produce enough protein product for proper function or (right) the mutant allele acts as a dominant negative that produces a "spoiler" protein product. KEY CONCEPT expression (such gen recessive. Harmful mu Mutations in genes the dominant negatives, a

Two models for dominance of a mutation a Model 1: Model 2: Phenotype t Haploinsufficiency Dominant negative p +/+ a 2 "doses" of product Dimer Wild type be MIM Mutant a 0 "dose" ti ti +/M Mutant t 1 "dose" (inadequate) tl sufficiency and the a« mutation is illustrated FIGURE 6-2 Amutation may be dominant because (left) a single wild-type gene does not produce enough protein product for proper function or (right) the mutant allele acts as a dominant negative that produces a "spoiler" protein product. KEY CONCEPT expression (such gen recessive. Harmful mu Mutations in genes the dominant negatives, 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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In an in situ hybridization experiment, a certain clonebound to only the X chromosome in a boy with no diseasesymptoms. However, in a boy with Duchenne musculardystrophy (X-linked recessive disease), it bound to theX chromosome and to an autosome. Explain. Could thisclone be useful in isolating the gene for Duchenne muscular dystrophy?
. 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?
2. Uniparental disomy is a rare phenomenon in whichonly one of the parents of a child with a recessivedisorder is a carrier for that trait; the other parent ishomozygous normal. By analyzing DNA polymorphisms, it is clear that the child received both mutantalleles from the carrier parent but did not receive anycopy of the gene from the other parent.a. Diagram at least two ways in which uniparentaldisomy could arise. (Hint: These mechanismsall require more than one error in cell division,explaining why uniparental disomy is so rare.)Is there any way to distinguish between thesemechanisms to explain any particular case ofuniparental disomy?
2. Uniparental disomy is a rare phenomenon in whichonly one of the parents of a child with a recessivedisorder is a carrier for that trait; the other parent ishomozygous normal. By analyzing DNA polymorphisms, it is clear that the child received both mutantalleles from the carrier parent but did not receive anycopy of the gene from the other parent.a. Diagram at least two ways in which uniparentaldisomy could arise. (Hint: These mechanismsall require more than one error in cell division,explaining why uniparental disomy is so rare.)Is there any way to distinguish between thesemechanisms to explain any particular case ofuniparental disomy?b. How might the phenomenon of uniparental disomyexplain rare cases in which girls are affected withrare X-linked recessive disorders but have unaffectedfathers, or other cases in which an X-linked recessive disorder is transmitted from father to son?c. If you were a human geneticist and believed oneof your patients had a disease syndrome caused…
The drug ivacaftor has recently been developed totreat cystic fibrosis in children with the rare G551Dmutant allele of CFTR.a. Do you think that ivacaftor would be effective onlyin patients homozygous for the G551D mutation,or might it work as well in compound heterozygotes in which one copy of chromosome 7 hadG551D and the other copy a different allele ofCFTR, such as the more prevalent allele ΔF508?(The protein encoded by G551D folds up properlyand inserts into the cell membrane, but is inefficient in chloride ion transport. Ivacaftor increasesthe efficiency of G551D’s ion transport. TheΔF508 protein does not fold up properly and therefore does not get inserted into the cell membrane.)b. Why do you think ivacaftor would be more effectivein children than in older cystic fibrosis patients?c. The scientists who developed ivacaftor had a modelfor cystic fibrosis: a line of cells that grow in culture and that are homozygous for G551D. Thesecells accumulate mucus at their surfaces that…
Drosophila females heterozygous for each of three recessive autosomal mutations with independent phenotypic effects [thread antennae (th), hairy body (h), andscarlet eyes (st)] were testcrossed to males showingall three mutant phenotypes. The 1000 progeny of thistestcross werethread, hairy, scarlet 432wild type 429thread, hairy 37thread, scarlet 35hairy 34scarlet 33a. Show the arrangement of alleles on the relevantchromosomes in the triply heterozygous females.b. Draw the best genetic map that explains these data.c. Calculate any relevant interference values
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.
F1 hybrids between two species of cotton, Gossypium barbadenseand Gossypium hirsutum, are very vigorous plants. However, F1crosses produce many seeds that do not germinate and a high percentageof very weak F2 offspring. Suggest two reasons for theseobservations.
Tay–Sachs disease is caused by loss-of-function mutations ina gene on chromosome 15 that encodes a lysosomal enzyme.Tay–Sachs is inherited as an autosomal recessive condition.Among Ashkenazi Jews of Central European ancestry, about1 in 3600 children is born with the disease. What fraction ofthe individuals in this population are carriers?
Figure 19-18a shows a plot of P values (represented bythe dots) along the chromosomes of the dog genome.Each P value is the result of a statistical test of association between a SNP and body size. Other than the clusterof small P values near IGF1, do you see any chromosomalregions with evidence for a significant association between a SNP and body size? Explain
EXTRANUCLEAR INHERITANCE In drosophila, a recessive mutation (m-) of a maternal effect gene (MEG) results in an abnormal phenotype wherein homozygous (m-m-) females produce eggs that cannot support embryonic development. Homozygous (m-m-) males, however, can still produce viable sperm. Using m+ to denote a normal gene, determine the genotypes and phenotypes of the F1s produced by a cross between a heterozygous female and a recessive male. From the offspring, backcross the recessive female with the paternal strain. What are the genotypes and phenotypes of the F2s? with COMPLETE cross for both cases.