Finding “the gene” for cystic fibrosis

Finding “the gene” for cystic fibrosis

Finding “the gene” for cystic fibrosis

Why is this in quotes?

A. CF is not caused by a gene, it’s caused by multiple genes.

B. CF is not caused by genetic factors.

C. CF is not caused by a gene, it’s caused by a mutation.

How to find genetic determinants of naturally

varying traits?

Genetic markers

Fig. 10.3

(microsatellite)

Genetic markers

Fig. 10.3

(microsatellite)

Table 11.1

Lots of benign variation between us.

How do you find polymorphisms?

Fig. 11.6


Fig. 11.6

Introduced in lecture 9/15.


Fig. 11.6


Fig. 11.6


Fig. 11.6


Fig. 11.6


Fig. 11.6

Hybrid mapping: location of probe

QuickTime™ and aTIFF (Uncompressed) decompressor

are needed to see this picture.



www3.mdanderson.org/depts/cellab/fish1.htm

mouse human/mouse hybrid

Hybrid mapping: location of probe



Back then, no technique to see 6kb

at cytological resolution.

Who cares about benign polymorphisms?

Remember Sturtevant?

Fig. 5.10

Who cares about benign polymorphisms?

We are going to do a two-point cross.

One of our genetic loci is represented by phenotype; the other is a DNA marker.

Mapping a disease locus

Fig. 11.A

(Autosomal dom)

A1

A2


Fig. 11.A

(Autosomal dom)phenotype (variation in locus 1)

A1

A2


Fig. 11.A


marker genotype (variation in locus 2)

A1

A2


Fig. 11.A


marker genotype (variation in locus 2)

How close are they in genetic distance?

A1

A2


Fig. 11.A

A1 D

A2 d

(Autosomal dom)

A1

A2


Fig. 11.A

A1 D

A2 d

(Autosomal dom)

A1

A2

(assume phase)


Fig. 11.A

A1 D

A2 d

A1 d

A1 d

A1

A2


Fig. 11.A

A1 D

A2 d

A1 d

A1 d

A2 d

A1

A2


Fig. 11.A

A1 D

A2 d

A1 d

A1 d

A1 D

A1

A2


Fig. 11.A

A1 D

A2 d

A1 d

A1 d

A2 d

A1

A2


Fig. 11.A

A1 D

A2 d

A1 d

A1 d

?

A1

A2


Fig. 11.A

A1 d

A1 d

?

A1 D

A2 d(sperm)

A1

A2


Fig. 11.A

A1 d

A1 d

A2 D

A1 D

A2 d(sperm)

A1

A2


Fig. 11.A

A1 d

A1 d

A1 D

A2 d

A1

A2

In total, 7 of the kids are non-recombinants and 1

is a recombinant.


Fig. 11.A

What is the apparent RF between the DNA marker and the disease mutation?

A. 1/10B. 1/8C. 1/20

A1

A2


is a recombinant.


Fig. 11.A

What is the apparent RF between the DNA marker and the disease mutation?

1/8 = 12.5 m.u.

A1

A2

A. 1/10B. 1/8C. 1/20


is a recombinant.

Why do I say “apparent RF?”

What if…

observed recombination fraction = 1/8 = 12.5 cM

Disease-causing mutation

Restriction fragment length polymorphism

True distance 30 cM

What if…




True distance 30 cM

You could say this will never happen. But…

What if…




True distance 30 cM

this is our observation

What if…




True distance 30 cM

The observed number of recombinants is just a point estimate, with some error associated.


12 cM, 18 cM…who cares?

Further experiments need to find the causal variant, not just a marker. If distances are wrong, could be

hunting for years.


Fig. 11.A

We now know the mutation is near (linked to) the marker.

1/8 = 12.5 m.u.

A1

A2



marker (known)

1/8 = 12.5 m.u.

A1

A2



window containing causative mutation

1/8 = 12.5 m.u.marker (known)

A1

A2


1/8 = 12.5 m.u.

How significant?

A1

A2


1/8 = 12.5 m.u.

How significant?If RF = 0.5 (unlinked), would be like flipping a coin 8 times.

How likely would you be to get 7 heads and 1 tail?

A1

A2

If RF = 0.5 (unlinked), would be like flipping a coin 8 times.How likely would you be to get 7 heads and 1 tail?

How much MORE likely is a model of RF < 0.5?

If RF = 0.5 (unlinked), would be like flipping a coin 8 times.How likely would you be to get 7 heads and 1 tail?

How much MORE likely is a model of RF < 0.5?

For large cross between known parents, would use 2 to evaluate significance.

Here we can’t.

LOD scores

1 recomb, 7 non-recomb.

r = genetic distance between marker and disease locus

A1

A2

LOD scores


Odds = P(pedigree | r)

P(pedigree | r = 0.5)


A1

A2

LOD scores





“How likely are the data given our model?”

A1

A2

LOD scores

k = 1 recomb, n = 7 non-recomb.




Odds = (1-r)n • rk

0.5n • 0.5k

A1

A2

LOD scores





0.5(total # meioses)

A1

A2


LOD scores






We have an idea of true r, but it is imprecise.


A1

A2

Remember?




True distance 30 cM

The observed number of recombinants is just a point estimate, with some error associated.


LOD scores







A1

A2

This formalism allows any r value. Let’s guess r = 0.3.

LOD scores






Odds = 0.77 • 0.31

0.58


A1

A2


LOD scores






Odds = 0.77 • 0.31

0.58

= 6.325


A1

A2


LOD scores






Odds = 0.77 • 0.31

0.58

= 6.325

Data >6 times more likely under LINKED hypothesis than under UNLINKED hypothesis.


A1

A2

LOD scoresr odds

0.1 12.244

0.2 10.737

0.3 6.325

0.4 2.867

0.5 ??






LOD scores





Odds at r=0.5?A. 2.5B. 0C. 1D. 10

r odds

0.1 12.244

0.2 10.737

0.3 6.325

0.4 2.867

0.5 ??

LOD scores

What’s the best (most likely) value of r?

A. 0.1B. 0.2C. 0.3D. 0.4E. 0.5

r odds

0.1 12.244

0.2 10.737

0.3 6.325

0.4 2.867

0.5 1

What problems will look like

1,2 1,2 1,1 1,2 1,1 1,2 1,1 1,2 1,2 1,1

A1

A2


1,2 1,2 1,1 1,2 1,1 1,2 1,1 1,2 1,2 1,1


1,2 1,2 1,1 1,2 1,1 1,2 1,1 1,2 1,2 1,1

Count number of recombinants, calculate odds.

Reading and chapter problems on web site.

Documents

Finding “the gene” for cystic fibrosis