A bacterial culture is heated up, resulting in protein denaturation and loss of protein function, including proteins necessary for metabolism of certain nutrients. The cellular DNA remains intact and unchanged. O Genotypic Change Only O Phenotypic Change Only O Both Genotypic and Phenotypic Changes O Neither Genotypic nor Phenotypic Change
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- In individuals affected by cystic fibrosis, salt crystals may appear afterperspiration dries up. In addition, the disease causes respiratory disorderswhich can be both debilitating and lethal. It occurs in individuals homozygousfor the recessive gene. Two normal parents had a daughter with thesymptoms of this disease, and a normal son who marries a normal womanwith an afflicted A test (salt concentration in perspiration of heterozygotes ishigher than normal) disclosed that both are indeed carriers of the gene. If thefirst child born to the mating in (b) was defective, what is the probability thatthe 2nd child would also be defective?Express answer in fraction formEhler-Danlos syndrome is a rare disorder caused by a mutation ina gene that encodes a protein called collagen (type 3 A1). Collagenis found in the extracellular matrix that plays an important role inthe formation of skin, joints, and other connective tissues. Peoplewith Ehler-Danlos syndrome have extraordinarily flexible skin andvery loose joints. The following pedigree contains several individualsaffected with this syndrome, shown with black symbols. Basedon this pedigree, does the syndrome follow autosomal recessive,autosomal dominant, X-linked recessive, or X-linked dominantinheritance? Explain your reasoning.Ehler-Danlos syndrome is a rare disorder caused by a mutation ina gene that encodes a protein called collagen (type 3 A1). Collagenis found in the extracellular matrix that plays an important role inthe formation of skin, joints, and other connective tissues. Peoplewith Ehler-Danlos syndrome have extraordinarily flexible skin and very loose joints. The pedigree below contains several individualsaffected with this syndrome, shown with black symbols. Based onthis pedigree, does the syndrome appear to follow autosomalrecessive, autosomal dominant, X-linked recessive, or X-linkeddominant inheritance? Explain your reasoning.
- tl ban o a rne dononant allo le dD), ifs dccessrveallele (h) produces aght bour Another gene has wo alclcs Cm) dork boir n dominant over (b) bionde, A woman ith wooly, blonde lour roarmesan with stranght, dark hai heir daughteT has straight, blonde t What phenorypes and in what proportions can they expcct among fture children? o H X h & 7. The color in Labrador retrievers is an example of epistasis. The two alleles for the pigment gene are: (B) for black coat and its recessive counterpart (b) for chocolate coat. A different autosomal gene affects the expression of color in the coat and has two alleles, (E), which allows the expression of color, and (e), which is epistatic and blocks the expression of the B/b gene. The epistatic allele is recessive and is only expressed as a yellow color in the homozygous condition (ee), regardless of the B and b genotypes. A completely homozygous black lab was mated to a yellow lab, homozygous for the chocolate allele. Their offspring will be the F1…The XG locus on the human X chromosome has twoalleles, XG+ and XG. The XG+ allele causes the presence of the Xg surface antigen on red blood cells,while the recessive XG allele does not allow antigento appear. The XG locus is 10 m.u. from the STSlocus. The STS+ allele produces normal activity ofthe enzyme steroid sulfatase, while the recessive STSallele results in the lack of steroid sulfatase activityand the disease ichthyosis (scaly skin). A man withichthyosis and no Xg antigen has a normal daughterwith Xg antigen. This daughter is expecting a child.a. If the child is a son, what is the probability he willlack Xg antigen and have ichthyosis?b. What is the probability that a son would have boththe antigen and ichthyosis?c. If the child is a son with ichthyosis, what is theprobability he will have Xg antigen?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?
- . Among adults with Turner syndrome, it has beenfound that a very high proportion are genetic mosaics.These are of two types: In some individuals, themajority of cells are XO, but a minority of cellsare XX. In other Turner individuals, the majorityof cells are XO, but a minority of cells are XY.Explain how these two patterns of somatic mosaicscould arise.On rare occasions, an organism may have three copies of achromosome and therefore has three copies of the genes on thatchromosome (instead of the usual number of two copies). Forsuch a rare organism, the alleles for each gene usually segregateso that a gamete will contain one or two copies of the gene. Let’ssuppose that a rare pea plant has three copies of the chromosomethat carries the height gene. Its genotype is TTt. The plant is alsoheterozygous for the seed color gene, Yy, which is found on adifferent chromosome. With regard to both genes, how manytypes of gametes can this plant make, and in what proportions?(Assume that it is equally likely that a gamete will contain oneor two copies of the height gene.)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.
- . An allotetraploid species has a genome composed oftwo ancestral genomes, A and B, each of which havea basic chromosome number (x) of seven. In thisspecies, the two copies of each chromosome of eachancestral genome pair only with each other duringmeiosis. Resistance to a pathogen that attacks the foliage of the plant is controlled by a dominant allele atthe F locus. The recessive alleles Faand Fbconfersensitivity to the pathogen, but the dominant resistancealleles present in the two genomes have slightly different effects. Plants with at least one FAallele areresistant to races 1 and 2 of the pathogen regardlessof the genotype in the B genome, and plants with atleast one FBallele are resistant to races 1 and 3 of thepathogen regardless of the genotype in the A genome.What proportion of the self-progeny of an FA Fa FB Fbplant will be resistant to all three races of the pathogen?The life cycle of the haploid fungus Ascobolus is similar tothat of Neurospora. A mutational treatment producedtwo mutant strains, 1 and 2, both of which when crossedwith wild type gave unordered tetrads, all of the followingtype (fawn is a light brown color; normally, crosses produce all black ascospores):spore pair 1 black spore pair 3 fawnspore pair 2 black spore pair 4 fawna. What does this result show? Explain.The two mutant strains were crossed. Most of the unordered tetrads were of the following type:spore pair 1 fawn spore pair 3 fawnspore pair 2 fawn spore pair 4 fawnb. What does this result suggest? Explain.When large numbers of unordered tetrads were screenedunder the microscope, some rare ones that containedblack spores were found. Four cases are shown here:Case A Case B Case C Case Dspore pair 1 black black black blackspore pair 2 black fawn black abortspore pair 3 fawn fawn abort fawnspore pair 4 fawn fawn abort fawn(Note: Ascospores with extra genetic material…Spherocytosis is an inherited blood disease in whichthe erythrocytes (red blood cells) are spherical insteadof biconcave. This condition can be inherited in adominant fashion, with ANK1 (the nonfunctional mutant allele) dominant to ANK1+. In people with spherocytosis, the spleen recognizes the spherical redblood cells as defective and removes them from thebloodstream, leading to anemia. The spleen in different people removes the spherical erythrocytes withdifferent efficiencies. Some people with sphericalerythrocytes suffer severe anemia and some mild anemia, yet others have spleens that function so poorly nosymptoms of anemia exist at all. When 2400 peoplewith the genotype ANK1 ANK1+ were examined, itwas found that all of them had spherical erythrocytes,2250 had anemia of varying severity, and 150 had nosymptoms. (Assume that ANK1 ANK1 homozygotesdo not exist.)a. Does this description of people with spherocytosisrepresent incomplete penetrance, variable expressivity, or both? Explain…