Chapter:GeneLinkageRecombinationAnalysis.ppt

Lecture Notes on Genetic Recombination Analysis

1. 1
4
Gene Linkage and
Genetic Mapping

2. 2
If two genes are on
different chromosomes…

3. 3
And half look like they got
a mix of both parents
chromosomes…
Half look like they got a
set of the parents
chromosomes…
If two genes are on
different chromosomes…

4. 4
Now suppose both gene A
and B were next to each
other on the same
chromosome.
What happens to the ratios
in this diagram?

5. 5

6. 6

7. 7
Gene Mapping
•Gene mapping determines the order of genes and the relative
distances between them in map units
•1 map unit = 1 cM (centimorgan)
•Gene mapping methods use recombination
frequencies between alleles in order to determine the relative
distances between them
•Recombination frequencies between genes are inversely
proportional to their distance apart
•Distance measurement: 1 map unit = 1 percent recombination (true
for short distances)

8. 8Fig. 4.6

9. 9
Gene Mapping
•Genes with recombination frequencies less than 50 percent are on
the same chromosome = linked)
•Linkage group = all known genes on a chromosome
•Two genes that undergo independent assortment have
recombination frequency of 50 percent and are located on
nonhomologous chromosomes or far apart on the same
chromosome = unlinked

10. 10
Now cross (AB ab) F1 progeny with (ab ab) tester
to look for recombination on these chromosomes.
Suppose you Get……
AB ab 583 <parental
ab ab 597 <parental
Ab ab 134 <recombinant
aB ab 134 <recombinant
total= 1448
so…. 268 recombinants /1448 progeny =
0.185 recombinants/progeny=
18.5% recombinants=
18.5 mu
Starting with pure breeding lines,
Cross Parent 1(AA BB) with Parent 2(aa bb)
So Parental chromosomes in the F1 have to be AB and ab
Mapping the distance between two genes

11. 11
Cross (ABD abd) F1 progeny with (abd abd) tester
Suppose you Get……
ABD abd 580 <parental
ABd abd 3
abD abd 5 <parental
abd abd 592
AbD abd 45 <recombinant
Abd abd 89
aBD abd 94
aBd abd 40 <recombinant
total= 1448
Mapping (and ordering) three genes
Starting with pure breeding lines, Cross Parent 1(AA BB DD) with Parent 2(aa bb dd)
So you know the Parental chromosomes in the F1 have to be ABD and abc
Ab + aB = (45+89)+(94+40) recom
268 recom/1448 total =0.185
A-B =18.5mu
Bd + bD = (3+40)+(5+45)
93 recom/1448 total= 0.064
B-D =6.4mu
Ad + aD = (3+89)+(5+94)
191 recom/1448 total= 0.132
A-D =13.2mu
so the order must be A-----D---B
-13.2--6.4-
----18.5----

12. 12
Cross (ABD abd) F1 progeny with (abd abd) tester
Suppose you Get……
ABD abd 580 <parental
ABd abd 3
abD abd 5 <parental
abd abd 592
AbD abd 45 <recombinant
Abd abd 89
aBD abd 94
aBd abd 40 <recombinant
total= 1448
Ab + aB = (45+89)+(94+40) recom
268 recom/1448 total =0.185
A-B =18.5mu
A-----D---B
-13.2--6.4-
----18.5----
So How come 13.2 + 6.4 does not equal 18.5?
We missed the double recombinants on the first pass…
longer the distance, more
potential to underestimate
recomb freq.
(45+89)+(94+40)+2(3+5) recom
284recom/1448 total = 0.196
A-B =19.6mu

13. 13
Chromosome interference: crossovers in one region decrease the
probability of a second crossover close by
Coefficient of coincidence = observed number of double recombinants
divided by the expected number
\
Interference = 1-Coefficient of coincidence
If the two crossovers were independent,
we would expect that the probability of seeing two recombination events occur would be
0.132 between A-D AND 0.064 between D-B
0.132 X 0.064 = 0.008
For every 1448 progeny, this would be (1448x0.008)=12.23 double recombinants
We actually observed only (5+3)= 8 double recombinants
So the Coefficient of coincidence = observed / expected = 8/12.23 =0.65
Interference = 1-Coefficient of coincidence
= 1- 0.65
= 0.35

14. 14
Mapping Genes in Human Pedigrees
•Methods of recombinant DNA technology are used to
map human chromosomes and locate genes
•Genes can then be cloned to determine structure and
function
•Human pedigrees and DNA mapping are used to
identify dominant and recessive disease genes
•Polymorphic DNA sequences are used in human
genetic mapping.

15. 15
Genetic Polymorphisms
•The presence in a population of two or more relatively
common forms of a gene or a chromosome is called
polymorphism
•Changes in DNA fragment length produced by presence or
absence of the cleavage sites in DNA molecules are known
as restriction fragment length polymorphism (RFLP)
•A prevalent type of polymorphism is a single base pair
difference, simple-nucleotide polymorphism (SNP)
•A genetic polymorphism resulting from a tandemly repeated
short DNA sequence is called a simple sequence repeat
(SSR)

16. 16
RFLPs
•Restriction endonucleases are used to map genes as
they produce a unique set of fragments for a gene
•There are more than 200 restriction endonucleases in
use, and each recognizes a specific sequence of DNA
bases
•EcoR1 cuts double-stranded DNA at the sequence
5’-GAATTC-3’ wherever it occurs

17. 17Fig. 4.18

18. 18
RFLPs
•Differences in DNA sequence generate different recognition
sequences and DNA cleavage sites for specific restriction enzymes
•Two different genes will produce different fragment patterns when cut
with the same restriction enzyme due to differences in DNA sequence
Fig. 4.19

19. 19
SNPs
•Single-nucleotide polymorphisms (SNPs) are abundant in
the human genome
•Rare mutants of virtually every nucleotide can probably be
found, but rare variants are not generally useful for family
studies of heritable variation in susceptibility to disease
•For this reason, in order for a difference in nucleotide
sequence to be considered as an SNP, the less-frequent
base must have a frequency of greater than about 5% in
the human population.
•By this definition, the density of SNPs in the human
genome averages about one per 1300 bp

20. 20
SSRs
•A third type of DNA polymorphism results from differences
in the number of copies of a short DNA sequence that
may be repeated many times in tandem at a particular site
in a chromosome
•When a DNA molecule is cleaved with a restriction
endonuclease that cleaves at sites flanking the tandem
repeat, the size of the DNA fragment produced is
determined by the number of repeats present in the
molecule
•There is an average of one SSR per 2 kb of human DNA

21. 21
Mapping Genes in Human Pedigrees
•One source of the utility of SNPs and SSRs in human
genetic mapping is their high density across the
genome
•Additional source of utility of SSRs in genetic mapping
is the large number of alleles that can be present in any
human population.

22. 22
Mapping Genes in Human Pedigrees
•Human pedigrees can be analyzed for the inheritance
pattern of different alleles of a gene based on differences in
SSRs and SNPs
•Restriction enzyme cleavage of polymorphic alleles that are
different in RFLP pattern produces different size fragments
by gel electrophoresis

23. 23
A rare recessive disease that affects people late in life
is 90% linked to the RFLP marker on the gel below.
What’s the probability that the grandchildren (row III) are carriers?
90% 10% 90%10% 10% 90% 90%
What’s the probability that II1 is a carrier of the disease?
dd DD
Dd
100%

24. 24
A rare recessive disease that affects people late in life
is 90% linked to the RFLP marker on the gel below.
What’s the probability that the grandchildren (row III) are carriers?
10% 90% 10% 90% 90% 10% 10%
What’s the probability that II1 is a carrier of the disease?
dd DD
Dd
100%

25. 25
Tetrad Analysis
•In some species of fungi, each meiotic tetrad is
contained in a sac-like structure, called an
ascus

•Each product of meiosis is an ascospore, and
all of the ascospores formed from one meiotic
cell remain together in the ascus
•Several features of ascus-producing organisms
are especially useful for genetic analysis:
They are haploid, so the genotype is
expressed directly in the phenotype
They produce very large numbers of
progeny
Their life cycles tend to be short

26. 26
Tetrad Analysis
•In tetrads when two pairs
of alleles are segregating,
three patterns of
segregation are possible
•parental ditype (PD) =
two parental genotypes
•nonparental ditype
(NPD) = only recombinant
combinations
•tetratype (TT) = all four
genotypes observed
B
B
b
b
B
B
b
b
B
B
b
b
B
B
b
b
Parental
ditype
tetrad
Parental
ditype
tetrad

27. 27
Tetrad Analysis
•When genes are unlinked, the parental ditype tetrads
and the nonparental ditype tetrads are expected in
equal frequencies: PD = NPD
•Linkage is indicated when nonparental ditype tetrads
appear with a much lower frequency than parental
ditype tetrads: PD » NPD
•Map distance between two genes that are sufficiently
close that double and higher levels of crossing-over
can be neglected, equals
1/2 x (Number TT / Total number of tetrads) x 100

28. 28
Neurospora: Ordered Tetrads
•Ordered asci also can be
classified as PD, NPD, or
TT with respect to two
pairs of alleles, which
makes it possible to
assess the degree of
linkage between the
genes
•The fact that the
arrangement of meiotic
products is ordered also
makes it possible to
determine the
recombination frequency
between any particular
gene and its centromere

29. 29
Tetrad Analysis: Ordered Tetrads
•Homologous centromeres of parental chromosomes
separate at the first meiotic division
•The centromeres of sister chromatids separate at the
second meiotic division
•When there is no crossover between the gene and
centromere, the alleles segregate in meiosis I
•A crossover between the gene and the centromere
delays segregation alleles until meiosis II

30. 30
•The map distance
between the gene and
its centromere equals
•1/2 x (Number of asci
with second division
segregation/ Total
number of asci) x 100
•This formula is valid
when the gene is close
enough to the
centromere and there
are no multiple
crossovers

31. 31
Gene Conversion
•Most asci from heterozygous Aa diploids demonstrate normal Mendelian
segregation and contain ratios of
2A : 2a in four-spored asci, or 4A : 4a in eight-spored asci
Occasionally, aberrant ratios are also found, such as
3A : 1a or 1A : 3a and 5A : 3a or 3A : 5a. The aberrant asci are said to result from
gene conversion because it appears as if one allele has “converted” the other allele
into a form like itself
•Gene conversion is frequently accompanied by recombination between genetic
markers on either side of the conversion event, even when the flanking markers are
tightly linked
•Gene conversion results from a normal DNA repair process in the cell known as
mismatch repair
•Gene conversion suggests a molecular mechanism of recombination

32. 32
•One of two ways to resolve the resulting structure, known
as a Holliday junction, leads to recombination, the other
does not
•The breakage and rejoining is an enzymatic function
carried out by an enzyme called the Holliday junction-
resolving enzyme

33. 33