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- . 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 chromosomeDo all of them True/False 31) The process by which an electrical charge is used to introduce DNA into a cell to produce a transgenic organism is called electroporation.Answer: 32) Reproductive cloning is used to produce large amounts of mammalian proteins from transgenic agricultural animals such as cattle.Answer: 33) In gene addition, homologous recombination is used to remove the original gene and replace it with the cloned gene.Answer: 34) All stem cells have the potential to differentiateAnswer: 35) A bone marrow transplant involves the transfer of multipotent stem cellsAnswer: 36) The fact that in mammalian systems multiple genes may compensate for the loss of a gene is called gene redundancy.Answer:● Pedigree for a recessive disease I Signatures of a recessive trait: • Occurs rarely unless consanguineous. What can you infer about genotypes? ● = Cc || ||| IV VI VII What is the probability VII-2 be a carrier? A) 1/4 B) C) D) 1/3 E) 2/3 F) 1 1 сс Copyright © The McGraw-Hill Companies, Inc. Permission requir or 2 9999 3 9999 2 3 2 3 4 CC 5
- If you wanted to make a mouse model for any of the following human genetic conditions (a–d), indicate which of thefollowing types of mice (i–vi) would be useful to your studies. If more than one answer applies, state which type ofmouse would most successfully mimic the human disease:(i) transgenic mouse overexpressing a normal mouse protein; (ii) transgenic mouse expressing normal amounts of amutant human protein; (iii) transgenic mouse expressing adominant negative form of a protein; (iv) a knockout mouse;(v) a conditional knockout mouse; and (vi) a knockin mousein which the normal allele is replaced with a mutant allelethat is at least partially functional. In all cases, the transgeneor the gene that is knocked out or knocked in is a form of thegene responsible for the disease in question.a. Marfan syndrome (a dominant disease caused byhaploinsufficiency for the FBN1 gene);b. A dominantly inherited autoinflammatory diseasecaused by a hypermorphic missense mutation in thegene PLCG2;c.…Null mutations are valuable genetic resources becausethey allow a researcher to determine what happens to anorganism in the complete absence of a particular protein. However, it is often not a trivial matter to determinewhether a mutation represents the null state of the gene.a. Geneticists sometimes use the following test forthe nullness of an allele in a diploid organism: If theabnormal phenotype seen in a homozygote for theallele is identical to that seen in a heterozygote(where one chromosome carries the allele in question and the homologous chromosome is known tobe completely deleted for the gene) then the alleleis null. What is the underlying rationale for thistest? What limitations might there be in interpreting such a result?b. Can you think of other methods to determinewhether an allele represents the null state of a particular gene?dd-ons Help в I U A Calibri 12 三 三1 |:三 6. Consider a guinea pig with a homozygous genotype and a white fur color phenotype. a. What is the probability this parent will produce a gamete with the dominant allele? b. What is the probability this parent will produce a gamete with the recessive allele? C. If 31 sperm cells are collected from this guinea pig, how many would you expect to have the recessive allele (as determined by sequencing the gene)? !!!
- Researchers in search of loci in the human genome that arelikely to contribute to the constellation of factors leading tohypertension have compared candidate loci in humans and rats[Stoll, M., et al. (2000). New Target Regions for Human Hypertensionvia Comparative Genomics. Genome Res. 10:473–482].Through this research, they identified 26 chromosomal regionsthat they consider likely to contain hypertension genes. Howcan comparative genomics aid in the identification of genesresponsible for such a complex human disease? The researchersstate that comparisons of rat and human candidate loci tothose in the mouse may help validate their studies. Why mightthis be so?3. Animation 12.1: This table shows results from allele-specific oligonucleotide hybridization analyses conducted on three different patients. The analyses all used the same probes constructed to test for the presence of a mutant gene. A plus sign (+) indicates hybridization occurred and a negative sign (-) indicates no hybridization occurred. Patient #1 Patient #2 Patient #3 Probe for normal allele Probe for mutant allele Which patient(s) is/are homozygous for the mutation and which patient(s) is/are heterozygous for the mutation? O Patients #2 and #3 are homozygous and Patient #1 is heterozygous for the mutation. O Patient #2 is homozygous and Patient #1 is heterozygous for the mutation. O Patient #1 is homozygous and Patients #2 and #3 are heterozygous for the mutation. O Patient # 1 is homozygous and Patient #2 is heterozygous for the mutation.. 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.?
- 30: 3. The genes for the human blood types MN and Ss are closely linked genes on chromosome 4. A sample of 2000 British people returns the following numbers of each type of chromosome: MS Ms F2 a) Estimate the gametic and allele frequencies. b) Calculate D. English (United States) Accessibility: Unavailable c) Test whether the two loci are in linkage disequilibrium in this population. #3 E d) What is the amount of disequilibrium relative to its maximum (or minimum) value? D e) What are the expected frequencies of chromosomes assuming linkage equilibrium? 80 F3 $ 474 4 Q F4 R F % 5 611 S F5 T G 6 NS 142 F6 MacBook Air Y H & 7 F7 U Ns 773 * 00 8 J DII F8 ( 9 Focus E F9 K 0 F10 L E I PYou are using nitrosoguanidine to “revert” mutant nic-2(nicotinamide-requiring) alleles in Neurospora.You treat cells, plate them on a medium without nicotinamide, and look for prototrophic colonies. You obtainthe following results for two mutant alleles. Explainthese results at the molecular level, and indicate how youwould test your hypotheses.a. With nic-2 allele 1, you obtain no prototrophs at all.b. With nic-2 allele 2, you obtain three prototrophiccolonies A, B, and C, and you cross each separately with awild-type strain. From the cross prototroph A × wildtype, you obtain 100 progeny, all of which are prototrophic.From the cross prototroph B × wild type, you obtain100 progeny, of which 78 are prototrophic and 22 arenicotinamide requiring. From the cross prototrophC × wild type, you obtain 1000 progeny, of which 996 areprototrophic and 4 are nicotinamide requiring.Price et al. [(1999). J. Bacteriol. 181:2358–2362] conducteda genetic study of the toxin transport protein (PA) of Bacillusanthracis, the bacterium that causes anthrax in humans. Withinthe 2294-nucleotide gene in 26 strains they identified five pointmutations—two missense and three synonyms—among differentisolates. Necropsy samples from an anthrax outbreak in 1979revealed a novel missense mutation and five unique nucleotidechanges among ten victims. The authors concluded that thesedata indicate little or no horizontal transfer between differentB. anthracis strains. Question: Which types of nucleotide changes (missense or synonyms)cause amino acid changes?