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Chapter 7

Chapter 7. Linkage and crossing over. Where we’re going. Some concepts- that of linkage, which isn’t hard Calculating % recombination, and what it means; crosses involved The dreaded 3 point cross. Thinking required!!. Linkage: Fig 7-1-.

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Chapter 7

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  1. Chapter 7 Linkage and crossing over

  2. Where we’re going • Some concepts- that of linkage, which isn’t hard • Calculating % recombination, and what it means; crosses involved • The dreaded 3 point cross. • Thinking required!!

  3. Linkage: Fig 7-1- • if two genes are close on the same chromosome, they will be linked. Often, the linkage is incomplete- the closer two genes are to each other, the more complete the linkage

  4. Individual is AaBb, but the A and B genes are on the same chromosome! These will show crossover readily- not very close!

  5. We find linkage by doing testcrosses • 7-2; If the gene is on the X chromosome, we use a male with the recessive traits for the cross- Fig. 7-3.

  6. NOTE: only 2, NOT 4 gamete types!

  7. NOTE: A 1:2:1 ratio, NOT 9:3:3:1 ratio. bw bw/ hv+hv+ Bw+ bw/hv+ hv bw+ bw+/hv hv ; NO bw bw/ hv hv Only 2, not 4, progeny types!

  8. Here we have linkage, but not complete- with crossover!

  9. If you have enough genes, you can locate them consecutively, through their recombination frequencies. Problem 13:

  10. 8 8 13 6 17 • adp--------c--------vg-------------pr------b-----------------d • Here, we use the information that adp is at one end, then arrange the results consistent with the data. e.g.: vg has to be between adp and pr, since the adp-vg frequency is 16%, and the pr-vg is 13%; if vg was to the right of pr, then the adp-vg frequency would be > 29% • Note: these are idealized; frequencies tend towards 50%, the further away two genes are from each other!

  11. Mapping genes: the (dreaded) three-point cross. Here, you’re trying to locate the arrangement of three genes on the chromosome, as well as the distances between them.

  12. Fig 7-8: • Two classes are obvious: the non-crossovers, and the double crossovers. If all the wild-type alleles are on one parental chromosome, and all the mutant alleles on the other, then the DCO’s will have 2 + and one mutant allele, or two mutant and one + allele (y, w+ ec, y+ w ec+). This determines the gene in the middle. To get the recombination frequency between two genes, divide the total # recombinants by the total # of offspring. The DCO’s get counted twice. • The best way to do this, is to work problems!

  13. Here, we know the order- on a problem, we don’t

  14. DCO’s get counted twice! Sum= 10,000

  15. More problems;sample 1&2, 15&17

  16. Fig 7-10 Here, the one that is different from the parental chromosomes in the DCO is pr .224= .145+ .078! .434= .356+ .078! This tells you that pr is in the middle. V--.224----pr--.434--------bm

  17. coefficient of coincidence& Interference • Coefficient of coincidence = C= actual frequency of DCO/expected frequency of DCO’s • Interference is 1-C Fig. 7-10: • The expected frequency of DCO’s is 9.7%- (.224X.434) • The actual frequency of DCO’s is 7.8%- This produces a coefficient of coincidence (C) of 7.8/9.7= 0.804. Interference is 1-C. If you have few DCO’s interference is high, and the value is close to 1. If you have as many as expected, then C is close to1 and interference is close to 0. There are cases where you have recombination “hot spots”, and the interference can be negative!

  18. Things to know • How to determine the map distance, given a cross • Three point cross • Interference (with formulas)

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