Transcript PPT file - University of Evansville Faculty Web sites

Chapter 6
Genetic Recombination in
Eukaryotes
Linkage and genetic diversity
31 January, 2 February, 2005
Overview
• In meiosis, recombinant products with new combinations of
parental alleles are generated by:
– independent assortment (segregation) of alleles on nonhomologous
chromosomes.
– crossing-over in premeiotic S between nonsister homologs.
• In dihybrid meiosis, 50% recombinants indicates either that
genes are on different chromosomes or that they are far apart
on the same chromosome.
• Recombination frequencies can be used to map gene loci to
relative positions; such maps are linear.
• Crossing-over involves formation of DNA heteroduplex.
Overview
Recombination
Detailed
Independent
Assortment
Independent assortment (2)
• For genes on different (nonhomologous) pairs of chromosomes,
recombinant frequency is always 50%
A/A ; B/B  a/a ; b/b
A/A ; b/b  a/a ; B/B
A/a ; B/b
A/a ; B/b
¼A;BP
¼A;BR
¼A;b
R
50%
¼A;b
P
¼a;B
R
recombinants
¼a;B
P
¼a;b
P
¼a;b
R
Testcross of a Dihybrid
Self Cross of a Dihybrid
Independent assortment: multiple loci
• Calculations can be made for any gene
combination using predicted outcomes at
single loci and the product rule
P1 A/a ; B/b ; C/c ; D/d  P2 a/a ; B/b ; C/c ; D/D
# gametes P1
# gametes P2
# genotypes in F1
2 x 2 x 2 x 2 = 16
1x2x2x1=4
2 x 3 x 3 x 2 = 36
# phenotypes in F1 2 x 2 x 2 x 1 = 8
Frequency of
½ x ¾ x ¾ x 1 = 9/32
A/– ; B/– ; C/– ; D/–
Crossing-over
• Breakage and rejoining of homologous
DNA double helices
• Occurs only between nonsister chromatids
at the same precise place
• Visible in diplotene as chiasmata
• Occurs between linked loci on same
chromosome
– cis: recessive alleles on same homolog (AB/ab)
– trans: recessive alleles on different homologs
(Ab/aB)
Linkage maps
# observed
140
50
60
150
• RF is (60+50)/400=27.5%, clearly less than 50%
• Map is given by:
A
B
27.5 m.u.
Trihybrid testcross
• Sometimes called three-point testcross
• Determines gene order as well as relative
gene distances
• 8 categories of offspring
– for linked genes, significant departure from
1:1:1:1:1:1:1:1
• Works best with large numbers of offspring,
as in fungi, Drosophila
Trihybrid testcross example
v (vestigial) v+ (long) wings
b (black) b+ (gray) body
p (purple) p+ (red) eyes
v+ v+ . b b . pp x vv. b+ b+ . p+ p+
v/ v+. b/ b+. p/ p+
v/ v+. b/ b+. p/ p+
x
v/v.b/b.p/p
Progeny:
v
b+
p+
580
Parental
v+
b
p
592
Parental
v
b
p+
45
v+
b+
p
40
v
b
p
89
v+
b+
p+
94
v
b+
p
3
v+
b
p+
5
v.b and v+.b+
45 + 40 + 89 + 94 = 268 / 1448 = .185
v.p and v+.p+
89 + 94 + 3 + 5 = 191 / 1448 = .132
b.p+ and b+.p
45 + 40 + 3 + 5 = 93 / 1448 = .064
v
13.2
p
6.4
Why do 13.2 + 6.4 = 19.6 instead of 18.5?
b
The failure to count double crossover gametes as v.b recombinants
results in an underestimation of the v.b distance. A better v.b
number would be 268 + 3 + 3 + 5 + 5 = 284 / 1448 = 19.6. Aaah!
In general, to minimize the effect of double crossovers, it is
necessary to measure a number of small RF distances and sum to
larger distances.
Interference
• Crossing-over in one region of chromosome
sometimes influences crossing-over in an
adjacent region
• Interference = 1 – (coefficient of coincidence)
# observed double recombinan ts
c.o.c. 
# expected double recombinan ts
• Usually, I varies from 0 to 1, but sometimes it
is negative, meaning double crossing-over is
enhanced
In the previous example, expect (.132 x .064 = .0084)(1448) =
12 double crossovers, but saw 8, so I = 1-(8/12) = 4/12 = 1/3.
In regions where double crossovers are forbidden, observed = 0, so
I = 1.
What do you think negative interference means?
Genetic maps
•Useful in understanding and experimenting
with the genome of organisms
•Available for many organisms in the
literature and at Web sites
•Maps based on RF are supplemented with
maps based on molecular markers, segments
of chromosomes with different nucleotide
sequences
Chi-square test
• Statistical analysis of goodness of fit
between observed data and expected
outcome (null hypothesis)
• Calculates the probability of chance
deviations from expectation if hypothesis is
true
• 5% cutoff for rejecting hypothesis
– may therefore reject true hypothesis
– statistical tests never provide certainty, merely
probability
Chi-square application to linkage
• Null hypothesis for linkage analysis
– based on independent assortment, i.e., no
linkage
– no precise prediction for linked genes in
absence of map
(O  E )
 
E
2
2
for all classes
• Calculated from actual observed (O) and
expected (E) numbers, not percentages
What is E?
For the AbBb testcross example from the text, E is not just 125
for each of the genotypes, as there may be allele effects on
viability. Instead, get expected values from the data:
B
A
142
a
112
254
b
113
133
246
255
245
So, E (AB) would be (255/500) x (254/500) x 500 = 129.54
Other E are found similarly, and Chi-square is 4.97
Mechanism of meiotic crossing-over
•Exact mechanism with no gain or loss of
genetic material
•Current model: heteroduplex DNA
–hybrid DNA molecule of single strand from
each of two nonsister chromatids
–heteroduplex resolved by DNA repair
mechanisms
•May result in aberrant ratios in systems that
allow their detection
Recombination within a gene
• Recombination between alleles at a single
locus
• In diploid heterozygous for mutant alleles of
the same gene, recombination can generate
wild-type and double mutant alleles
a1/a2  a+ and a1,2
• Rare event, 10-3 to 10-6, but in systems with
large number of offspring, recombination can
be used to map mutations within a gene
Assignment: Concept map, solved
problems 1 and 2, all other basic and
challenging problems
Continue with PubMed section of the
Web tutorial.