EVOLUTION CONNECTION Crossing over is thought to beevolutionarily advantageous because it continually shufflesgenetic alleles into novel combinations. Until recently, it wasthought that the genes on the Y chromosome might degenerate because they lack homologous genes on the X chromosomewith which to pair up prior to crossing over. However, when theY chromosome was sequenced, eight large regions were foundto be internally homologous to each other, and quite a few ofthe 78 genes represent duplicates. (Y chromosome researcherDavid Page has called it a “hall of mirrors.”) Explain what mightbe a benefit of these regions.

EVOLUTION CONNECTION Crossing over is thought to beevolutionarily advantageous because it continually shufflesgenetic alleles into novel combinations. Until recently, it wasthought that the genes on the Y chromosome might degenerate because they lack homologous genes on the X chromosomewith which to pair up prior to crossing over. However, when theY chromosome was sequenced, eight large regions were foundto be internally homologous to each other, and quite a few ofthe 78 genes represent duplicates. (Y chromosome researcherDavid Page has called it a “hall of mirrors.”) Explain what mightbe a benefit of these regions.

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 3QP: A colleague e-mails you saying that she has identified an interesting chromosome variation at 21q13....

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Researchers have discovered that some regions of chromosomesare much more likely than others to cross over. We might callsuch a region a “hot spot” for crossing over. Let’s suppose thattwo genes, gene A and gene B, are 5,000,000 bp apart on thesame chromosome. Genes A and B are in a hot spot for crossingover. Two other genes, let’s call them gene C and gene D, are also5,000,000 bp apart but are not in a hot spot for crossing over. Ifwe conducted two-factor crosses to compute the map distancebetween genes A and B and other two-factor crosses to compute themap distance between genes C and D, would the calculated mapdistance between A and B be the same as that between C and D? Explain.
DRAW IT Suppose, as in the questionat the bottom of Figure 15.9, the parental(P generation) flies were true-breedingfor gray body with vestigial wings andblack body with normal wings. Draw thechromosomes in each of the four possiblekinds of eggs from an F1 female, andlabel each chromosome as “parental”or “recombinant.”
Female fruit flies homozygous for the X-linked white-eye alleleare crossed to males with red eyes. On very rare occasions, an offspringof such a cross is a male with red eyes. Assuming these rareoffspring are not due to a new mutation in one of the mother’s Xchromosomes that converted the white-eye allele into a red-eyeallele, explain how a red-eyed male arises.
Mitotic recombination can occasionally produce a twin spot.Let’s suppose an animal species is heterozygous for two genesthat govern fur color and length: one gene affects pigmentation,with dark pigmentation (A) dominant to albino (a); the other geneaffects hair length, with long hair (L) dominant to short hair (l).The two genes are linked on the same chromosome. Let’s assume ananimal of this species is AaLl; A is linked to l, and a is linked to L.Draw the chromosomes labeled with these alleles, and explainhow mitotic recombination could produce a twin spot with onespot having albino pigmentation and long fur and the other havingdark pigmentation and short fur.
The karyotype of a young girl who is affected with familialDown syndrome revealed a total of 46 chromosomes. Her olderbrother, however, who is phenotypically unaffected, actually had45 chromosomes. Explain how this could happen. What wouldyou expect to be the numbers of chromosomes in the parents ofthese two children?
The human genome consists of 23 pairs of chro-mosomes (22 pairs of autosomes and one pair of sex chro-mosomes). During meiosis, the maternal and paternal setsof homologs pair, and then are separated into gametes, sothat each contains 23 chromosomes. If you assume thatthe chromosomes in the paired homologs are randomlyassorted to daughter cells, how many potential combina-tions of paternal and maternal homologs can be gener-ated during meiosis? (For the purposes of this calculation,assume that no recombination occurs.)
The Drosophila chromosome 4 is extremely small;virtually no recombination occurs between genes onthis chromosome. You have available three differentlymarked chromosome 4s: one has a recessive allele ofthe gene eyeless (ey), causing very small eyes; one hasa recessive allele of the cubitus interruptus (ci) gene,which causes disruptions in the veins on the wings;and the third carries recessive alleles of both genes.Drosophila adults can survive with two or three, butnot with one or four, copies of chromosome 4.a. How could you use these three chromosomes tofind Drosophila mutants with defective meiosescausing an elevated rate of nondisjunction?b. Would your technique allow you to discriminatenondisjunction occurring during the first meioticdivision from nondisjunction occurring during thesecond meiotic division?
The Drosophila chromosome 4 is extremely small;virtually no recombination occurs between genes onthis chromosome. You have available three differentlymarked chromosome 4s: one has a recessive allele ofthe gene eyeless (ey), causing very small eyes; one hasa recessive allele of the cubitus interruptus (ci) gene,which causes disruptions in the veins on the wings;and the third carries recessive alleles of both genes.Drosophila adults can survive with two or three, butnot with one or four, copies of chromosome 4.a. How could you use these three chromosomes tofind Drosophila mutants with defective meiosescausing an elevated rate of nondisjunction?b. Would your technique allow you to discriminatenondisjunction occurring during the first meioticdivision from nondisjunction occurring during thesecond meiotic division?c. What progeny types would you expect if a flyrecognizably formed from a gamete produced bynondisjunction were testcrossed to a fly homozygous for a chromosome 4 carrying both ey…
A cytogeneticist has collected tissue samples from members of acertain butterfly species. Some of the butterflies were located inCanada, and others were found in Mexico. Through karyotyping,the cytogeneticist discovered that chromosome 5 of the Canadianbutterflies had a large inversion compared with chromosome 5 ofthe Mexican butterflies. The Canadian butterflies were inversionhomozygotes, whereas the Mexican butterflies had two normalcopies of chromosome 5.Explain whether a mating between Canadian and Mexicanbutterflies would produce phenotypically normal offspring?
in the experiment of following chromosomal dna mvement througgh meiosis, why. do you use non-sister chromatids to demobstrate crossing over? what combinatiobns of alleles could result from a crossover between BD and bd chromosomes? Identify two ways that meiosis contributes to genetic recombination. Why is it necessary to reduce the number of chromosomes in gametes, but not in other cell?
. 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?
You are an allele on a single uncondensed chromosome in a Yellowfever-carrying mosquito (Aedes aegypti) diploid cell (2n = 6). Demonstrate in as much detail as possible (1) how you will get to each genetically identical diploid cell of the multicellular offspring; (2) how you will get to the next generation of sexually reproduced progeny. Help, I no understand. Perhaps you will?