Using designations “Group A", "Group B," etc., organize the mutant strains into complementation groups. If the result of any cross does not fit your groupings, indicate which cross(es) you have excluded. 99 Consider the consequence if one of the mutations was dominant. You should see that complementation tests could not be used to provide information about dominant mutations. a) If a mutation were dominant, how would the diploid be affected? What effect would this have on the determination of your complementation groups? b) How would you test if any of these trp¯ mutations were dominant?

Using designations “Group A", "Group B," etc., organize the mutant strains into complementation groups. If the result of any cross does not fit your groupings, indicate which cross(es) you have excluded. 99 Consider the consequence if one of the mutations was dominant. You should see that complementation tests could not be used to provide information about dominant mutations. a) If a mutation were dominant, how would the diploid be affected? What effect would this have on the determination of your complementation groups? b) How would you test if any of these trp¯ mutations were dominant?

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Description: Using designations "Group A", “Group B," etc., organize the mutant strains intocomplementation groups. If the result of any cross does not fit your groupings, indicatewhich cross(es) you have excluded.Consider the consequence if one of the mutations

Human Heredity: Principles and Issues (MindTap Course List)

11th Edition

ISBN:9781305251052

Author:Michael Cummings

Publisher:Michael Cummings

Chapter6: Cytogenetics: Karyotypes And Chromosome Aberrations

Section: Chapter Questions

Problem 2QP: Given the karyotype shown at right, is this a male or a female? Normal or abnormal? What would the...

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Given the karyotype shown at right, is this a male or a female? Normal or abnormal? What would the phenotype of this individual be?
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 form
Using the pedigree chart, explain: a) The number of generations seen. b) If all blue-coloured shapes are affected with disease X- how many males are affected? how many females are affected? c) Does this disease have a dominant or recessive inheritance pattern? Justify your answer.
Neurospora of genotype a + c are crossed withNeurospora of genotype + b +. (Here, + is shorthandfor the wild-type allele.) The following tetrads areobtained (note that the genotype of the four sporepairs in an ascus are listed, rather than listing alleight spores):a + c a b c + + c + b c a b + a + ca + c a b c a + c a b c a b + a b c+ b + + + + + b + + + + + + c + + ++ b + + + + a b + a + + + + c + b +137 141 26 25 2 3a. In how many cells has meiosis occurred to yieldthese data?b. Give the best genetic map to explain these results.Indicate all relevant genetic distances, both betweengenes and between each gene and the centromere.c. Diagram a meiosis that could give rise to oneof the three tetrads in the class at the far right inthe list
. 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 chromosome
In the experiment of Figure shown, Stern followed the inheritancepattern in which females carried two abnormal X chromosomesto correlate genetic recombination with the physical exchange ofchromosome pieces. Is it necessary to use a strain carrying twoabnormal chromosomes, or could he have used a strain in whichfemales carried one normal X chromosome and one abnormal Xchromosome with a deletion at one end and an extra piece of the Ychromosome at the other end?
A single yeast cell placed on a solid agar will dividemitotically to produce a colony of about 107cells. Ahaploid yeast cell that has a mutation in the ade2 genewill produce a red colony; an ade2+ colony will bewhite. Some of the colonies formed from diploidyeast cells with a genotype of ade2+/ ade2− willcontain sectors of red within a white colony.a. How would you explain these sectors?b. Although the white colonies are roughly the samesize, the red sectors within some of the whitecolonies vary markedly in size. Why? Do youexpect the majority of the red sectors to be relativelylarge or relatively small?
Wild-type strains of the haploid fungus Neurospora canmake their own tryptophan. An abnormal allele td renders the fungus incapable of making its own tryptophan.An individual of genotype td grows only when its medium supplies tryptophan. The allele su assorts independently of td; its only known effect is to suppress the tdphenotype. Therefore, strains carrying both td and su donot require tryptophan for growth.a. If a td ; su strain is crossed with a genotypically wildtype strain, what genotypes are expected in the progenyand in what proportions?b. What will be the ratio of tryptophan-dependent totryptophan-independent progeny in the cross of part a?
In a haploid yeast strain, eight recessive mutationswere found that resulted in a requirement for theamino acid lysine. All the mutations were found to revert at a frequency of about 1 × 10−6 except mutations5 and 6, which did not revert. Matings were madebetween a and α cells carrying these mutations. Theability of the resultant diploid strains to grow onminimal medium in the absence of lysine is shown inthe following chart (+ means growth and − means nogrowth.)1 2 3 4 5 6 7 81 − + + + + − + −2 + − + + + + + +3 + + − − − − − +4 + + − − − − − +5 + + − − − − − +6 − + − − − − − −7 + + − − − − − +8 − + + + + − + −a. How many complementation groups were revealedby these data? Which point mutations are foundwithin which complementation groups?The same diploid strains are now induced to undergosporulation. The vast majority of resultant spores areauxotrophic; that is, they cannot form colonies whenplated on minimal medium (without lysine). However,particular diploids can produce rare spores…
12) An individual with genotype AaBbCCDdEE can make many kinds of gametes. Which of the following correctly describes why this situation is possible? A) recurrent mutations form new alleles B) crossing over during prophase I leads to genetic variety C) different possible assortment of chromosomes/alleles into gametes occurs D) there is a tendency for dominant alleles to segregate together 13) Let P = purple
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…
Females heterozygous for the recessive second chromosome mutations px, sp, and cn are mated to a male homozygous for all three mutations. The offspring are as follows: Must show all work 1.) What gene is in the middle? and what are the map distances between the genes in centimorgans? 2.) What is the interference?