Gene linkage
seminar
No 405 Heredity
Key words:
complete and incomplete gene linkage, linkage
group, Morgan´s laws, crossing-over,
recombination, cis and trans linkage phases,
strenght of linkage, Morgan´s number - p,
centimorgan, gene map unit, Bateson´s number - c,
gene mapping.
3. Mendel´s principle of combination
(independent assortment):
Members of two and more allelic pairs segregate
independently - there are as many types of gametes
as possible combinations among maternal and
paternal chromosomes (alleles).
It is applied only in the case of different localization
of genes on chromosomes (pairs of homologous
chromosomes).
Gene linkage
3. Mendel‘s law is not valid, if the crossed genes (A,B) are
on the same chromosome (a pair of homologous
chromosome)
= They are in gene linkage
Consequence: the alleles with non-recombinant (parental)
combinations are more frequent, because they are passed
together more frequently into gametes. The alleles with new
recombinant combinations are less frequent – incomplete
linkage
Incomplete linkage – dihybrid cross:
Parental generation – genes on the same chromosome (a
pair) not like in a situation of independent assortment
F1 generation – AaBb =
A
a
=
B
b
AB
ab
incomplete linkage (cis)
A
a
B
b
independent assortment
In the case of incomplete linkage, alleles non-recombinant (AB,
ab) are more frequently than the alleles new – recombinant
(Ab, aB).
In the F2 generation are the same types of genotypes,
but genotype and phenotype ratio is different –
Non-recombinant (parental) alleles are formed more
frequent than new recombinant alleles.
In B1 generation are the same types of
genotypes,
phenotypes, but not in ratio 1:1:1:1 like in the case of independent
assortment. Non-recombinant (parental) alleles are formed
more frequent than new recombinant alleles e.g. 2:2:1:1
Incomplete linkage:
Individual AaBb in F1 forms for F2 and for B1
Alleles
AB
Ab
ab
aB
in cis phase
in trans phase
Non-recombinant
AB ab
Ab
aB
Recombinant
Ab aB
AB
ab
Proof of recombination: always in B1 (testcross)
B1
cis
AB>Ab aB < ab
AB and ab
non-recombinant
(parental) are more
frequent
trans
AB < Ab aB > ab
Ab and aB non-recombinant
(parental) are more more frequent
Mechanism of recombination – crossing over
Probability of c.o. depends on the distance between
alleles
Complete linkage - dihybrid cross
Genes A an B are next to each other, there in no place
for crossing over = recombination. Dihybrid forms
only two types (non-recombinant/parental) of alleles
with ratio 1:1
cis phase F1
AB
A
B
a
b
ab
alleles
trans phase F1
A
b
=
AB
ab
AaBb
a
B
=
Ab
aB
Ab
aB
The whole dihybrid cross with complete gene linkage
cis phase
P
AB
trans phase
ab
Ab
ab
Ab
x
aB
x
AB
F1
AB
ab
F2
aB
Ab
aB
cis
trans
AB
ab
AB
AB
AB
ab
AB
ab
AB
ab
ab
ab
Ab
aB
Ab
Ab
Ab
aB
Ab
aB
Ab
aB
aB
aB
AB
ab
AB
AB
AB
ab
AB
ab
AB
ab
ab
ab
GP
1:2:1
FP
ab
1
Ab
Ab
Ab
aB
Ab
aB
Ab
aB
aB
aB
1:2:1
AB
ab
AB
ab
ab
ab
:
aB
1:2:1
3:1
B1
Ab
1
ab
Ab
aB
Ab
ab
aB
ab
1
:
1
Morgan´s laws
1. The genes on chromosomes are in linear
order
2. Number of linkage groups is equal to the
number of pairs of homologous
chromosomes
Strength of gene linkage, p = Morgan‘s number
fr
from B 1 (testcross)
p=
fr + fnr
fr – frequency of recombinant gametes, fnr = frequency of nonrecombinant gametes (phenotypes in B1)
Unit: cM (centimorgan) = % of recombinants =
expression of relative distance of alleles = map unit 1cM =
recombinant frequency 1% (0,01)
0 < p < 0,5
p = 0 complete linkage
p = 0,5 free (independent) assortment
Morgan´s number – from B1
fAb + faB
Phase cis: p =
fAB + fAb + faB + fab
fAB + fab
Phase trans: p =
fAB + fAb + faB + fab
Bateson´s number c: expresses: how more
frequent are inherited non-recombined (parental)
sets of alleles than recombined sets of alleles (from
B1)
Phase cis : c =
fAB + fab
fAb + faB
Phase trans: c =
fAb + faB
fAB + fab
1< c< ∞
for independent assortment c = 1
1. In B1 (testcross) generation of parental cross
AB/AB x ab/ab 902 individuals of phenotype AB, 898 of
ab, 98 of Ab and 102 of aB phenotypes was detected.
Calculate the strength of linkage between genes A
and B (number p).
2. How would you prove that two allelic pairs are linked or
not?
3. Hybrid of parental cross Ab/Ab x aB/aB is crossed
with individual ab/ab in testcross. What is the ratio of
genotypes in progeny comprising 1850 individuals?
The strength of the gene linkage between A and B
genes is 40% of recombinations.
6. Dominant allele D is coding for Rh+ factor, recessive genotype dd
is coding for Rh- phenotype (absence of Rh factor on the surface of
erythrocytes). Elliptic -oval shape of erythrocytes (eliptocytosis) is
coded by the dominant allele E, whereas the homozygously
recessive genotype ee codes for normal round shape of
erythrocytes. Both genes are in genetic distance of 20 cM.
There is a man with the eliptocytosis, whose mother had a
normal shape of erythrocytes and was homozygous for Rh+ and
whose father was Rh- and heterozygous in allelic pair for
eliptocytosis. This man got married to a healthy woman bearing Rhfactor.
a) What is the probability that the first child will be Rh- with
eliptocytosis?
b) What is the probability that the first child will be eliptocytic and
Rh+?
c)If the first child is Rh+, what is the probability that it will suffer from the
eliptocytosis too?
7. Determine the cross ratios in B1 of the hybrid AB/ab,
if c=2 and the number of offsprings in the B1 generation
is 1585.
8. What is the cross ratio of the hybrid Ab/aB in the F2
generation, if c=5?
9.The woman (X DH/X dh) is a carrier (heterozygote) of recessive alleles
for a daltonism (colour-blindness) and a hemophilia. Relative distance
of genes is 10 cM.
a)In which percentual ratio the recombined gametes in the woman
arise? Write their genotypes.
b)What is the probability that woman will have a hemophilic and
daltonic son? What is the probability that she will have a hemophilic but
not daltonic son or a daltonic but not hemophilic son or a healthy son?
You should assume that the father is X DH/Y.
c) The same you should calculate for the case that the mother has got all the
allelic pairs in the trans phase.