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Ch. 5 - Chromosome Mapping in Eukaryotes

Chapter 5, Problem 20

Are mitotic recombinations and sister chromatid exchanges effective in producing genetic variability in an individual? in the offspring of individuals?

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Hi everyone. Welcome back. Let's look at our next problem. It says blank refers to the stained paired sister chroma tides to reveal the sister chromatic exchanges, when we think about a technique that allows um sister chromatic exchanges or recombination to be seen. Um With differential staining, that term that we're looking for is choice B harlequin chromosomes. Harlequin was a character that wore this costume of black and white, which is where that term comes from. And harlequin chromosomes. Or when you have differential staining of sister chroma tides. Um This is done by um bathing the DNA in a solution with a perimeter, an analog that gets taken up instead of timing. Um so that when you replicate the DNA strand, you have you end up with a chrome it'd or a chromosome with sister chroma tides, one of which the original strand can be stained, the other which has taken up this permitting analog doesn't take up the stain in the same way. So you kind of have one dark and one light sister committed. So then you can see when you allow replication to take place where recombination or crossover and recombination has occurred, because you can track each promoted based on the staining. So you can see physically the exchange of pieces of sister Kromah tides by seeing that dark and light bits. Let's just look through our other answer choices to understand why they're not correct. Um We have first choice amy topic recombination. Well that is these events of crossing over and recombination that we're looking for by the use of harlequin chromosomes. So when you have those sister, when you have those chromosomes lined up during meta phase and two of the sister chromosomes next to each other cross over and then exchange those bits of D. N. A. So my topic recombination is what we're looking for using harlequin chromosomes, but it's not the harlequin chromosomes themselves. That's not our correct answer choice C. Is chromosome mapping. And chromosome mapping is that process by which we're looking for the location of every single gene on each chromosome. So actually mapping the physical location of genes on a chromosome. So that's not our answer choice. And then finally we have linkage equilibrium. And linkage equilibrium is a little bit of a tricky concept to explain. And you almost have to explain it by explaining the opposite. So linkage linkage equilibrium is a situation in which linkage disequilibrium is equal to zero. And so linkage disequilibrium is when alleles occur together more often than can be accounted for by chance due to physical proximity on the chromosome. So when the linkage disequilibrium is a way of noticing that these two alleles must be located on the same chromosome close together because you see them um occurring together more often than they would by chance. So they're not independently a sorting. So in linkage equilibrium um the alleles are together no more often than chance they are independently a sorting. So that is not the description of what we're looking for here. So we're going to eliminate that. So again, Choice B. Harlequin chromosomes refers to the stained paired sister chroma tides to reveal the sister chromatic exchanges. See you in the next video.
Related Practice
Textbook Question
Drosophila females homozygous for the third chromosomal genes pink and ebony (the same genes from Problem 16) were crossed with males homozygous for the second chromosomal gene dumpy. Because these genes are recessive, all offspring were wild type (normal). F₁ females were testcrossed to triply recessive males. If we assume that the two linked genes, pink and ebony, are 20 mu apart, predict the results of this cross. If the reciprocal cross were made (F₁ males—where no crossing over occurs—with triply recessive females), how would the results vary, if at all?
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Textbook Question
In Drosophila, two mutations, Stubble (Sb) and curled (cu), are linked on chromosome III. Stubble is a dominant gene that is lethal in a homozygous state, and curled is a recessive gene. If a female of the genotype is to be mated to detect recombinants among her offspring, what male genotype would you choose as a mate?
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Textbook Question
If the cross described in Problem 18 were made, and if Sb and cu are 8.2 map units apart on chromosome III, and if 1000 offspring were recovered, what would be the outcome of the cross, assuming that equal numbers of males and females were observed?
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Textbook Question
What possible conclusions can be drawn from the observations that in male Drosophila, no crossing over occurs, and that during meiosis, synaptonemal complexes are not seen in males but are observed in females where crossing over occurs?
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Textbook Question
An organism of the genotype AaBbCc was testcrossed to a triply recessive organism (aabbcc). The genotypes of the progeny are presented in the following table. 20 AaBbCc 20 AaBbcc 20 aabbCc 20 aabbcc 5 AabbCc 5 Aabbcc 5 aaBbCc 5 aaBbcc If these three genes were all assorting independently, how many genotypic and phenotypic classes would result in the offspring, and in what proportion, assuming simple dominance and recessiveness in each gene pair?
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Textbook Question
An organism of the genotype AaBbCc was testcrossed to a triply recessive organism (aabbcc). The genotypes of the progeny are presented in the following table. 20 AaBbCc 20 AaBbcc 20 aabbCc 20 aabbcc 5 AabbCc 5 Aabbcc 5 aaBbCc 5 aaBbcc Answer part (a) again, assuming the three genes are so tightly linked on a single chromosome that no crossover gametes were recovered in the sample of offspring.

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