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
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In Figure 6-2, explain how the mutant polypeptide acts
as a spoiler and what its net effect on
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- Equalizing the Expression of X Chromosome Genes in Males and Females Individuals with an XXY genotype are sterile males. If one X is inactivated early in embryogenesis, the genotype of the individual effectively becomes XY. Why will this individual not develop as a normal male?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?3.27 Complementation tests of the recessive mutant g genes a through f produced the data in the accom- panying matrix. The circles represent missing data. beaim Assuming that all of the missing mutant combina- motions would yield data consistent with the entries ithat are known, complete the table by filling each oneel circle with a + or – as needed. - rt yd bat 21 igi a rıb a O + b c d e fi 612 nobsdi b liw insiqans bosale yns mon las 29 ni YOu C ono e 66 BA AA f AA
- . The human IGF2 gene is autosomal and maternallyimprinted. Copies of the gene received from themother are not expressed, but copies received fromthe father are expressed. You have found two allelesof this gene that encode two different forms of theIGF2 protein distinguishable by gel electrophoresis.One allele encodes a 60K (Kilodalton) blood protein;the other allele encodes a 50K blood protein. In ananalysis of blood proteins from a couple named Billand Joan, you find only the 60K protein in Joan’sblood and only the 50K protein in Bill’s blood. Youthen look at their children: Jill is producing only the50K protein, while Bill Jr. is producing only the 60Kprotein.a. With these data alone, what can you say about theIGF2 genotype of Bill Sr. and Joan?b. Bill Jr. and a woman named Sara have two children, Pat and Tim. Pat produces only the 60K protein and Tim produces only the 50K protein. Withthe accumulated data, what can you now say aboutthe genotypes of Joan and Bill Sr.?. A diploid strain of yeast was made by mating a haploidstrain with a genotype w−, x−, y−, and z− with a haploidstrain of opposite mating type that is wild type for thesefour genes. The diploid strain was phenotypically wildtype. Four different X-ray-induced diploid mutantswith the following phenotypes were produced fromthis diploid yeast strain. Assume a single new mutation is present in each strain.Strain 1 w− x+ y− z+Strain 2 w+ x− y− z−Strain 3 w− x+ y− z−Strain 4 w− x+ y+ z+When these mutant diploid strains of yeast go throughmeiosis, each ascus is found to contain only two viablehaploid spores.a. What kind of mutations were induced by X-rays tomake the listed diploid strains?b. Why did two spores in each ascus die?c. Are any of the genes w, x, y, or z located on thesame chromosome?d. Give the order of the genes that are found on thesame chromosomeA female patient 19 years old, whose symptoms areanemia and internal bleeding due to a massive buildupof leukemic white blood cells, is diagnosed withchronic myelogenous leukemia (CML). Karyotypeanalysis shows that the leukemic cells of this patientare heterozygous for a reciprocal translocation involving chromosomes 9 and 22. However, none of thenormal, nonleukemic cells of this patient contain thetranslocation. Which of the following statements istrue and which is false?a. The translocation results in the inactivation (loss offunction) of a tumor-suppressor gene.b. The translocation results in the inactivation (loss offunction) of an oncogene.c. There is a 50% chance that any child of this patientwill have CML.d. This patient is a somatic mosaic in terms of thekaryotype.e. DNA extracted from leukemic cells of this patient,if taken up by normal mouse tissue culture cells,could potentially transform the mouse cells intocells capable of causing tumors.f. The normal function of the…
- Flies homozygous for recessive null mutations in thesevenless (sev) or bride-of-sevenless (boss) genes have the same mutant phenotype: Every ommatidium(facet) in their eyes lacks photoreceptor cell 7 (R7).The R7 cells enable flies to detect UV light.a. Given that flies normally move toward light, suggest a screening method that would enable you toidentify mutations in additional genes required forR7 determination.b. Would you be able to recover mutations in everygene required for R7 development with yourmethod? Explain.c. How could you tell whether any of the new mutationsyou found in your screen are alleles of sev or boss?d. Suppose you found one recessive mutant allele ofa gene not previously known to be involved in eyedevelopment. How could you use this allele in anew mutagenesis screen to find additional allelesof this gene? Why might you want additional mutant alleles to study the process?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.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.
- 5 f There are two genes that determine the coat colour expression in dogs: eumelanin and merle. These genes are located on two separate chromosomes. For the eumelanin gene, black coat colour (E) is dominant over red coat colour (e). The merle gene controls the degree to which these coat colours are expressed through incomplete dominance. The following table describes the merle gene expression. Genotype MM Mm mm Phenotype White AAY Half colour (Grey or light red) Full colour (Black or Red) 1. If a dog breeder wants to determine the unknown genotype of a dog, what colour of dog must she use as a testcross? 2. A white dog was testcrossed, and all of the puppies had light red coats. What is the genotype of the white dog? Use a Punnett square to show your work. Hint - When a trait is controlled by two genes, an individual's genotype consists of 4 letters. = = = B I 123 ||| 14 E E I ос C GO * ●● ● ●0 00 00 00 ✓ C U up X₂ x²In 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?When the His− Salmonella strain used in the Ames testis exposed to substance X, no His+ revertants are seen.If, however, rat liver supernatant is added to the cellsalong with substance X, revertants do occur. Is substanceX a potential carcinogen for human cells? Explain