Title: Linkage and crossing over
1Chapter 8
- Linkage and crossing over
2Where were going
- Some concepts- that of linkage, which isnt hard
- Calculating recombination, and what it means
crosses involved - The dreaded 3 point cross.
- Thinking required!!
3Linkage Fig 8-1-
- 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
4Individual is AaBb, but the A and B genes are on
the same chromosome!
These will show crossover readily- not very close!
5We find linkage by doing testcrosses
- 8-2 If the gene is on the X chromosome, we use
a male with the recessive traits for the cross-
Fig. 8-3.
6NOTE only 2, NOT 4 gamete types!
7NOTE A 121 ratio, NOT 9331 ratio. bw bw/
hvhv Bw bw/hv hv bw bw/hv hv NO bw bw/
hv hv
Only 2, not 4, progeny types!
8Here we have linkage, but not complete- with
crossover!
9(No Transcript)
10- If you have enough genes, you can locate them
consecutively, through their recombination
frequencies. Problem 14
11- 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 gt 29 - Note these are idealized frequencies tend
towards 50, the further away two genes are from
each other!
12Mapping genes the (dreaded) three-point cross.
- Here, youre trying to locate the arrangement of
three genes on the chromosome, as well as the
distances between them.
13Fig 8-8
- 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 DCOs
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 DCOs get counted twice. - The best way to do this, is to work problems!
14Here, we know the order- on a problem, we dont
15DCOs get counted twice!
Sum 10,000
16More problems
17Interference coefficient of coincidence-
- You can calculate the expected level of DCOs by
multiplying the product of each. For the maize
cross (fig 8-10), it would be 0.2236 0.434
0.097 9.7 - Â NOTE Adding the DCOs to the SCOs is important
in this case! - Â
18- The actual frequency of DCOs 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 DCOs 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! - Somatic cell hybridization save for if we have
time.
19.224 .145 .078!
This tells you that pr is in the
middle. V--.224----pr--.434--------bm
20Things to know
- How to determine the map distance, given a cross
- Three point cross
- Interference (with formulas)