Biology 163 Research Paper:
Genetics behind cloning of a human gene:
Topics due today
Outline due Oct. 31.
Writing a scientific paper:
Choose a topic
Collect papers
Define the problem you want to present
Make a title – active statements work well
eg. Grant’s disease is caused by a defective potassium pump
Make an outline:
Define a title for each section – even each paragraph.
Use active statements again.
Eg. Grant’s disease is a genetic disease that affects breathing
The Grant’s disease mutation is on Chromosome 7
A mutation in a potassium pump is linked to Grant’s disease
Expression of wild type potassium pump reverts Grant’s disease
effects in cultured cells
Grant’s potassium pump message is expressed in lung cells
Treatment of patients with potassium pump blockers has
no effect on progress of disease.
Microarray analysis suggests potential treatment.
These can be used as section titles or as topic sentences (see below).
The sentences of the outline could be strung together to make an abstract.
Choose figures to illustrate your points.
You can copy figures from some of the papers you read.
If so, cite them properly eg (from Smith et al., 1987).
The figure with its figure legend should be comprehensible
without reading the text.
Write the paragraphs.
First sentence is a topic sentence.
Last sentence is a concluding sentence.
Active voice
don’t try to sound academic but use proper grammar and no slang
Define your terms.
If referring to a figure, define the figure but do not duplicate the figure legend.
Maintain the same tense, either past or present.
Citations: When you discuss published work, cite the paper.
Do the citation in the first sentence in which the study is mentioned.
Eg. Seven large families with a high incidence of cystic fibrosis
were surveyed for DNA markers linked to the disease (Smith et al., 1987).
References: at the end of the section: in alphabetical order.
Eg. Smith J, Jones, P.A. and White, K. 1987 Family studies map
cystic fibrosis to Chromosome 7 Genetics 130: 147-156.
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 cross
Body color; yellow vs wild type
Bristles: 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
Female
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 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
Fig. 5.13
Fig. 5.12
Fig. 5.15
Map Based Cloning
• 1. 100’s of DNA markers mapped onto each
chromosome – high density linkage map.
• 2. identify markers linked to trait of interest by
recombination analysis
• 3. Narrow region down to a managable length of DNA
– for cloning and sequence comparison
• 4. Compare mutant and wild type sequences to find
differences that could cause mutant phenotype
• 5. Prove that mutation is responsible for phenotype.
1. 100’s of DNA markers mapped onto each
chromosome – high density linkage map.
Loci distinguished by
polymorphism
Can be measured by
DNA analysis or
by phenotypic traits
Mouse mapping Panels:
How to score many loci at one time?
•
•
•
•
Backcross panel
BXS heterozygote cross to either B or S
Progeny are recombinant BS/parental S
– like a test cross.
Each mouse represents one BS recombinant
chromosome.
If markers are linked, they with be the same, either B
or S, in most progeny.
Mouse mapping panel
Loci 1 and 3 have a high chance of linkage
because when a mouse is B at Locus 1
it is also B at locus 3.
Individual mice
1 2 3 4 5 6 7 8 9 101112131415161718192021222324252627282930313233
Locus 1 B S B B S S S B B S B B S S S B S B B S S B S B B B S S S B B S B
2 B S S B B S B S B S S B B S B S S S B B S B B S S B B S B S B S B
3 B S B B S S S B S S B S B S S B B B S S S B B B B B B S S B B S B
Loci 1 and 3 appear to be linked:
1
B S B B S S S B B S B B S S S B S B B S S B S B B B S S S B B S B
3
B S B B S S S B S S B S B S S B B B S S S B B B B B B S S B B S B
Mouse Mapping Panel
For any locus (1) 50% of mice are 1B and 50% are 1S.
Are Locus 1 and 2 linked?
12/50 1B2B: 13/50 1B2S; 12/50 1B2S; 13/50 1S2S
Unlinked:
50% parental : 50% recombinant
Locus 3 linked to locus 1:
38/50 parental : 12/50 recombinant
The frequency of recombinant types indicates
distance between loci.
12 recomb. /50 potential recombs = 24% recombination.
In situ hybridization to locate genes on chromosomes
Calculate relationship
between many loci to
get a high density
linkage map
How do we map genes in humans?
We cannot set up a test cross.
Mating is always random and uncontrolled.
Test degree of linkage: odds of linkage
Data looks like M1 is linked to SF.
Calculate odds:
Probability linkage exists vs
probability linkage does not exist.
Best guess at recombination distance gives highest odds
but number of potential recombination events is limited.
We use log of odds (LOD) to integrate information from
many families.
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
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;1sfM2
Probability of any genotype is .25
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 M are linked at 10 map units:
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 M 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 P and M: If the genes were linked so
that P 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 M 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 = p(.1)/p(.5) = [.5p(.1 in coupling) +.5p(.1 in repulsion)]/p(.5)
p(.5 in coupling) = p(.5 in repulsion)
In our case:
L = [½(.003736) + ½(.000023)]/.0000153
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.
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 accepted
Linkage 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 majority
of 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 in
human families. The LOD score allows you to compare families
in which marker and gene are either in repulsion or in coupling.
Gusella et al 1984. Science 225, 1320-1326
HindIII polymorphism is closely linked to disease
Allele C is always found in affected
individuals with one exception
Some unaffected individuals also have allele C
But in these cases it is from an unaffected parent
Marker G8 is linked to Huntingtons disease at a distance of 2 cM
With a LOD score of 12.1
Lod score
• Odds of Linkage:
– Probability gene and
marker are linked at a
certain map distance /
Probability they are
unlinked.
P(A will be inherited by healthy child if G8 and
Huntingtons linked at 2 cM) = .5 X .98 = .49
P(C will be inherited by affected child) = .49
P (gene and marker are linked) for family is .24
P(A or C will be inherited if not linked to Huntingtins) =.25
P(A or C unlinked) for family is 0.0625
Odds is 3.84 for this family. Log of odds is 0.5
Multiple families to Odds of 1000 or LOD of 3.0
Physical map of region that contains Huntingtons gene
IT15 had more CAG repeats at 5’ end in Huntington’s patients
than in healthy relatives
HDCRG 1994. Cell 72, 971-983
Exon trap used to identify expressed genes
in Huntingtons interval
Bucker et al. 1991 PNAS 88, 4005-4009
Structure of Huntingtin protein
PolyGlutamine repeats (CAG codon) are
expanded in diseased individuals
Several caspase cleavage sites are found
Sawa 2001 J. Mol. Med. 79, 375-381.
Huntingtin defect associated with
loss of mitochondrial function
Huntingtin may function in the nucleus
affecting gene expression
Possible functions of Huntingtin leading to disease