Biology 423 Research Paper:Genetics behind cloning of a human gene:

Outlines due today

Mapping genes by recombination frequency

Test cross to monitor recombination between different genes

Frequency of recombination is directly related to distance between genes (loci) on chromosome

Three point cross

Drosophila, a model organism for genetics

Traits for our three point crossBody color; yellow vs wild typeBristles: forked vs straight (wild type)Crossveins: crossveinless vs wild type

Fig. 5.12

Test cross

vg b pr / vg+ b+ r+ X vg b pr / vg b pr

Punnet square:

Male Femalevg+b+pr+ vg b+pr+ vg+b pr+ vg b pr+ vg+b+ pr vg b+pr vg+b pr vg b pr

Vg b pr vg+b+pr+ vg b+pr+ vg+b pr+ vg b pr+ vg+b+ pr vg b+pr vg+b pr vg b pr

1:1:1:1:1:1:1:1 ratio of phenotypes if genes are not linked

If genes are linked, parental combinations of alleles are overrepresented in progeny

Fig. 5.12

3 genes, which is in the middle?

Fig. 5.13

Calculate distance between pairs: vg to pr: add up all classes with a recombination event between vg+ and pr or vg and pr+

252 + 241 + 13 + 9 = 525

Divide by total number of chromosomes scored:

525/4197 X 100 cM = 12.5

Calculate distance between pairs: pr to b: add up all classes with a recombination event between pr+ and b or pr and b+

131 + 118 + 13 + 9 = 271

Divide by total number of chromosomes scored:

271/4197 X 100 cM = 6.4

The distance between vg and b is the sum of the distance between vg-pr and pr-b

12.3 + 6.4 = 17.7

Fig. 5.12

Fig. 5.15

How do we map genes in humans?

Relative association of markers:Allelic variants will co-segregate if the genes are closely linked on a chromosome.

Map distances depend on frequency of recombination

Markers can be traits, proteins or DNA sequences

Anything that is polymorphic can be mapped

To map a human genetic trait:Look for association between mapped markers and a trait of interest

We can translate map position into DNA sequence by determining the linkage between DNA-based markers and traits.

Pedigree Analysis: symbols

Screen family members for RFLP markers linked to trait

RFLP polymorphisms reveal genetic differences

1. Cut genomic DNA with Restriction enzymes

2. Separate DNA fragments by size on an agarose gel

3. Hybridize to single copy radioactive probe- Southern Blot

Test degree of linkage: odds of linkage

Data looks like M1 is linked to SF. Mother has two M1 alleles.Her chromosome is uninformative, like a test-cross. Father has two different M alleles. Recombination of his alleles can be seen in this pedigree.

In this family, there are 8 informative chromosomes.1 has a recombination event.Therefore, we estimate map distance as 1/8X 100 cM. 12.5%

Odds of Linkage is

(Probability gene and marker are linked at a certain map distance) divided by (Probability they are unlinked).

In our case:

L = 6.1

Log of L or LOD = 0.8

Maximum likelihood odds of linkage; Change estimated linkage Distance p(.1) to get the best LOD score for the data.

Test degree of linkage: odds of linkage

Calculate Lod score

• Odds of linkage– Probability gene and marker are linked at

a certain map distance / Probability they are unlinked.

– Calculate maximum odds for data. Predicted linkage distance gives best odds

– Add up log of odds for many families to get more data

To achieve significant LOD score:

Combine odds of linkage for many families:

p1(L)/p1(NL) x p2(L)/p2(NL) xp3(L)/p3(NL)

In practice we combine the log of odds:

LOD1 + LOD2 + LOD3.

Continue until LOD > 3.0 before linkage is acceptedLinkage distance is based on the linkage distance that gives the maximum value for the data.

If genes and markers are unlinked the p(L)/p(NL) will be <1.0 in some families and the LOD will be Negative.

Therefore, as you add more families the LOD will only increase if the data of the majorityof families supports linkage.

1. Probability SF and M loci are unlinked:Father is SF/sf and M1/M2. Mother is sf/sf and M1/M1. Her gametes are all the same.Her alleles can be ignored here.

Chance of each allele combination in children if SF and M are unlinked is:

1SFM1;1SFM2;1sfM1;1sfM2Probability of any genotype is .25

How to calculate LOD score for the example used in lecture

With 8 children of genotypes:sfM2;SFM1;sfM2;SFM1; sfM2;SFM1;sfM2;SFM2

P(SF and M unlinked) =.25 x .25 x .25 x .25 x .25 x .25 x .25 x .25 = .0000153

Probability SF and M1 are linked at 10 map units: (10 is chosen arbitrarily to start. Other map distances will also be tested).Chance of each allele combination in children is .45 SFM1; .05 SFM2; .05 sfM1; .45 sfM2

With 8 children of genotypes:sfM2;SFM1;sfM2;SFM1;sfM2;SFM1;sfM2;SFM2

p(SF and M1 are linked at 10 cM) =.45 x .45 x .45 x .45 x .45 x .45 x .45 x .05 = .003736

BUT -formally you don’t know the phase of the two alleles of SF and M: If the genes were linked so that SF and M2 were on the same chromosome:

Chance of each allele combination in children is .05 SFM1; .45 SFM2; .45 sfM1; .05 sfM2

With 8 children of genotypes:sfM2;SFM1;sfM2;SFM1; sfM2;SFM1;sfM2;SFM2

p(SF and M2 are linked at 10 cM) =.05 x .05 x .05 x .05 x .05 x .05 x .05 x .45 = .000023

Odds of Linkage is

L = {½ p(SF and M1 are linked at 10 cM) + ½ p(SF and M2 are linked at 10 cM)}Divided by p(SF and M are unlinked)

In our case:

L = [½(.003736) + ½(.000023)]/.0000153L = 6.1

Log of L or LOD = 0.8

Maximum likelihood odds of linkage; Change estimated linkage Distance p(.1) to get the best LOD score for the data.

To achieve significant LOD score:

Combine odds of linkage for many families:

p1(L)/p1(NL) x p2(L)/p2(NL) xp3(L)/p3(NL)

In practice we combine the log of odds:

LOD1 + LOD2 + LOD3.

Continue until LOD > 3.0 before linkage is acceptedLinkage distance is based on the linkage distance that gives the maximum value for the data.

If genes and markers are unlinked the p(L)/p(NL) will be <1.0 in some families and the LOD will be Negative.

Therefore, as you add more families the LOD will only increase if the data of the majorityof families supports linkage.

Summary

1. 100’s of DNA markers mapped onto each chromosome – high density linkage map.

the relative location of 100s of polymorphic DNA markers on chromosomes can be mapped using mapping panels.

2. identify markers linked to trait of interest by recombination analysis.Use LOD score to determine if markers are linked to gene inhuman families. The LOD score allows you to compare familiesin which marker and gene are either in repulsion or in coupling.