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Chapter 6 Genetic Recombination
Two mechanisms of generating variation (recombinants) during meiosis
Independent assortment and crossing over
Independent assortment of genes
on different chromosomes
Parental cells (PURELINES) AA;BB and aa;bb
Parent gametes A;B and a;b
so ALL F1 progeny are
A a; B b
Generate a Dihybrid (double heterozygous)
to see this when
measuring recombination freq
Compare parental gametes to F1 gametes
To determine
Parental
vs
recombinant
Recombination frequency
RF=(recombinants)/(recombinant+parental)
***Always 50% recombinants
when alleles are on different chromosomes
F1 gametes
Selfing a dihybrid (ie the F2):
9/16 DomDom
3/16 DomRec
3/16 RecDom
1/16 RecRec
Can determine by Punnett square (draw) 9:3:3:1
Or Product Rule:
Aa X Aa are ¾ dominant;
Bb X Bb are ¾ dominant;
So….. ¾ x ¾ or 9/16 are DomDom
Product rule works for as many genes as you want
(Also do a Dihybrid X Tester) Aa Bb X aa bb 1:1:1:1
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Crossing over between genes on the same chromosome
Linked genes
Start with pure breeding lines to make F1 with known chromosomes
Cell1(AA BB) X Cell2(aa bb) so chromosomes are Cell 1(AB AB) and Cell 2 (ab ab)
Parental chromosomes have to be AB or ab
so … F1 has two chromosomes that are AB and ab
(draw)
(You need to know what’s on each chromosome)
The utility of the tester line: Cross (AB ab) cell with (ab ab) cell
always see express. of dom or rec. allele because you have to inherit an ab chromosome
(draw as chromosomes)
AB ab
Dom Dom
so F2 phenotypes gives away
ab ab
rec rec
all the gametes inherited from red parent
Ab ab
Dom rec
aB ab
rec Dom
Also get recombinants
(Draw far then close) but….
Recombinants < 50% suggests linked on chromosome
The less the recombination, the closer together they are
1% recombination=1 map unit=1centiMorgan
Linkage maps
Cross (AB ab) cell with (ab ab) cell same as above
Suppose you Get……
AB ab
583 <parental
ab ab
597
<parental
Ab ab
134
<recombinant
aB ab
134 <recombinant
total= 1448
so….268 recomb /1448 total
or 18.5% recombinants=18.5 mu
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Mapping using trihybrid crosses Start with ABD ABD X abd abd (draw as chromosomes)
(just add in 2 poss for each D or d)
Cross ABD abd X abd abd tester
Check pairs
ABD abd
580 still
AB + ab = 583+597 parent
ABd abd
3
still
Ab + aB = (45+89)+(94+40) or 268 recom/1448 total
A-B 18.5%=18.5mu
abD abd
5
abd abd
592
Bd + bD = (3+40)+(5+45) or 93 recom/1448 total
B-D 6.4%=6.4mu
AbD abd
45
so order is either A---------B---D or A-----D---B
Abd abd
89
Ad + aD = (3+89)+(5+94) or 191 recom/1448 total
aBD abd
94
A-D 13.2%=13.2mu
aBd abd
40
so must be A-----D---B
total=1448
-13.2--6.4-mu
----18.5----mu
Interference
(why does 13.2+6.4 not equal 18.5?) underestimates long dist. due to double crossovers.
(Draw)
When we Measured A-B, we didn’t count rarest 3 or 5 because they look parental at ends.
But these had two crossovers (middle is recombinant) so shouldve counted twice.
Single events + single events + 2*double events = recomb events
(49+89)+(94+40)+2(5+3)=284/1448 or 19.6% recombinant =19.6 mu
longer the distance, more potential to underestimate recomb freq.
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Chi square test: is it linked?
If it is, we don’t know how close it is… so we don’t know what number to test for.
But if it isn’t… we know we should get 50% recombination… so we can test for that.
Thus, the hypothesis is that recombinants are equal to 50%.
Cross Parent 1(AB AB) X Parent 2(ab ab)
All F1 are (AB or ab)…..then cross with tester cell (ab ab)
Suppose we get…..
AB ab
142 <parental
ab ab
133 <parental
Ab ab
113 <recombinant
aB ab
112 <recombinant
so….225 recomb /500 total
total= 1448
or 45% recombinants
is it linked? Or is this within the liklihod of happening by chance?
(The Equation)
Calculating Observed and Expected.
Observed: set up your punnet square (but put in the numbers)
Expected: In this case, each allele should be ¼ of total if ideal.
But consider what would happen if only 10% of a’s live?
Wouldn’t see equal ratios… but a could be sorting randomly….
so really all we want to know is does a distribute evenly when it a’s do live?
Use the product rule to calc AB from total frequency of each allele
Then, plug into equation
So…
Then find the probability in the chart.
Degrees freedom=(classes in rows-1)X(classes in column -1) explain why last place fixed
If the probability of generating this much of a deviation by chance is < 0.05 (<5% likely) we
conclude the deviation is significant.
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Mechanism of crossing over
Holliday junctions, branch migration can create heteroduplex DNA
If heteroduplexes are differentially repaired before post meiotic replication, aberrant
ratios can be produced.