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Lecture Outline 9/13

• Finish talking about chromosome structureand packaging

• Linkage of genes on chromosomes– Estimating recombination fraction

• Chi-square tests for independence– Two-point mapping– Predicting offspring phenotypes when genes are

linked– Three-point mapping (if we have time)

I have posted a new homework assignment on the web page

What’s wrong with this picture?

Metaphase, Meiosis 1

A a

b B

What’s wrong with this picture?

Metaphase, Meiosis 1

A Aa a

b bB B

How are chromosome packedinto cells?

Fig 3.20

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Chromosome packaging

Mitochonrdia and Chloroplastshave their own chromosomes

Fig 3.44

Fig. 7.17(TE Art)

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

All whitechloroplasts

All greenchloroplasts

White chloroplastGreen chloroplast

Nucleus

Variegatedleaf

or

Pedigree for Mitochondrialtrait

Strictly maternal inheritance

Fig 3.45

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Genes are linked onchromosomes Morgan’s data

• Purple-eye vestigial wing x red-eye normalwing– What do you expect from this cross if genes are

independent?» Morgan’s data don’t fit very well . . . X2=1747.6

• F2 Phenotype Obs.• pr+ vg+ 1339• pr+ vg 151• pr vg+ 154• pr vg 1195

Test crossHeterozygous F1 x double recessive

pr+/pr vg+/vg x pr/pr vg/vg

Expect: ???

Morgan’s data• Purple-eye vestigial wing x red-eye normal

wing– Morgan guessed that the two genes must be

linked on the same chromosome

• F2 Phenotype Obs.• pr+ vg+ 1339• pr+ vg 151• pr vg+ 154• pr vg 1195

CO 5

Recombination

• Recombination is a result of crossing overduring meiosis

• When does it happen??

• Crossing over is random• occurs on any part of the chromosome with equal

probability

• Recombination is more likely when genesare far apart on a chromosome

-> Therefore, use the recombination frequency to estimatedistances between genes

Creighton and McClintockproved that crossing over resulted in

genetic recombinationWx C

wx c

wx C

wx cwww.dnalc.org

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See animation of meiosis

• The book web site has severalexample problems and animations

http://bcs.whfreeman.com/iga8e• Linkage animation

Morgan’s data

• Purple-eye vestigal wing x red-eye normalwing

• Phenotype number expected• pr+ vg+ 1339 709.75• pr+ vg 151 709.75• pr vg+ 154 709.75• pr vg 1195 709.75

2839 2839

Which are the recombinants?How do you know?

• What fraction of the flies are recombinants?305/2839 = 0.107

Mapping

• Define 1 “map unit” (m.u.) == 1%recombination

• Also known as one “centimorgan” (cM)

pr vg

10.7 cM

Recombination frequency = 0.107

Mapping

• Will any cross work?– How about

• PPVV x ppvv?• PpVv x PPVV?• PpVv x PpVv?

• Unlinked genes can be on the samechromosome– The maximum recombination fraction is 50%, no matter how

far apart they are.– Recombination frequency will always underestimate the

actual map distance

Try this one on your own if you want a challenge!

Genetic Maps

• In 1911, Alfred Sturtevant produced thefirst genetic map, of X chromosomegenes in Drosophila

y w v r m0 1.0 30.7 33.7 57.6

Genetic Maps

Compare his map with his data

%Recombination between:• y and w: 1%• y and v: 32%• y and m: 37.5 <-- what went wrong here?

y w v r m0 1.0 30.7 33.7 57.6

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Correcting map distances

If a pair of genes is far apart, there will bedouble crossovers that you can’t detect.

Maximum recombinationfrequency is 50%

• For well-separated genes (far enough that at leasttwo crossovers occur) there will be on average 50%recombination.

• Types of double crossover:– 2 chromatid: 0 apparent recombinants– 3 chromatid: 50% apparent recombinants– 4 chromatid: 100% apparent recombinants

– Average is 50%

We now have linkage mapsfor many organisms

Partial linkage map for Drosophila melanogaster

Partial map of Tomato genes

• Test Cross AaBb x aabb• Results:

• A-B- 70• A-bb 20• aaB- 25• aabb 85

Total = 200

Are these loci independent?If not, How far apart are they?

Chi-squareTest for Independence

• Test if genes assort independently into GAMETES• Results of test cross:

• A-B- 70• A-bb 20• aaB- 25• aabb 85

Total = 200

Chi-squareTest for Independence

What did the parental gametes have to be?

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Chi-squareTest for Independence• 2 way table of gamete frequencies for each

locus

20011090total

1058520b

952570B

totalaA

Number of gametes that had big A and little b

Chi-squareTest for IndependenceExpected value =Row total * Column total / Grand Total

20011090total

10585E=57.75

20E=47.25

b

9525E=95*110/200E=52.25

70E=95*90/200E=42.75

B

totalaA

Chi-squareTest for IndependenceDegrees of freedom=(Rows-1) * (Cols-1)

20011090total

10585E=57.75

20E=47.25

b

9525E=52.25

70E=42.75

B

totalaA

Chi-squareTest for IndependenceX2=Sum((O-E)2/E) <-- This formula is the same as before

20011090total

10585E=57.75

20E=47.25

b

9525E=52.25

70E=42.75

B

totalaA

X2= 17.37 +14.21+15.71+12.85 = 60.14 Compare this to thetable of X2 for 1 df

Testcross AaBb x aabb

• Results:• A-B- 70• A-bb 20• aaB- 25• aabb 85• Total = 200

• Are these loci independent?– No

• How far apart are they?– what fraction of the gametes

are recombinant?

45/200 = 0.225= 22.5 cM

For next time:

• Finish reading chapter 4

• Work on the new homework problems,due next week