Tutorial #2
by Ma’ayan Fishelson
Crossing Over
• Sometimes in meiosis, homologous
chromosomes exchange parts in a process
called crossing-over.
• New combinations are obtained, called the
crossover products.
Recombination During Meiosis
Recombinant gametes
Linkage
• 2 genes on separate chromosomes assort
independently at meiosis.
• 2 genes far apart on the same chromosome can
also assort independently at meiosis.
• 2 genes close together on the same chromosome
pair do not assort independently at meiosis.
• A recombination frequency << 50% between 2
genes shows that they are linked.
Two Loci Inheritance
AA
B B
a a
2 b b
1
A a
B b
3
A a
b b
Recombinant
4
5
6
A a
B b
a a
b b
Linkage Maps
•
Let U and V be 2 genes on the same chromosome.
• In every meiosis, chromatids cross over at random along
the chromosome.
• If the chromatids cross over between U & V, then a
recombinant is produced.
The farther apart U & V are  the greater the
chance that a crossing over would occur between
them  the greater the chance of recombination
between them.
Recombination Fraction
• The recombination fraction  between two loci
is the percentage of times a recombination
occurs between the two loci.
•  is a monotone, nonlinear function of the
physical distance separating between the loci
on the chromosome.
(Linkage ) 0    P(Recombinat ion )  0.5 ( No Linkage)
Centimorgan (cM)
• 1 cM (or 1 genetic map unit, m.u.) is
the distance between genes for which
the recombination frequency is 1%.
Interference
• Crossovers in adjacent chromosome regions
are usually not independent. This interaction
is called interference.
• A crossover in one region usually decreases
the probability of a crossover in an adjacent
region.
 observerd # of double recombinan ts 
Interferen ce(I)  1  

expected
#
of
double
recombinan
ts


Building Genetic Maps
• At first: only genes with variant alleles
producing detectably different phenotypes
were used as markers for mapping.
• Problem: the chromosomal intervals between
the genes were too large  the resolution of
the maps wasn’t high enough.
• Solution: use of molecular markers (a site of
heterozygosity for some type of silent DNA
variation not associated with any measurable
phenotypic DNA variation).
Linkage Mapping by
Recombination in Humans.
• Problems:
– It’s impossible to make controlled
crosses in humans.
– Human progenies are rather small.
– The human genome is immense. The
distances between genes are large on
average.
Lod Score for Linkage Testing
by Pedigrees
The results of many identical matings are combined to get
a more reliable estimate of the recombination fraction.
1.
Calculate the probabilities of obtaining a set of results
in a family on the basis of (a) independent assortment
and (b) a specific degree of linkage.
2.
Calculate the Lod score = log(b/a).
A Lod score of 3 is considered convincing
support for a specific recombination fraction.
Question #1
• In rabbits, black (B) is dominant to brown (b), while full
color (F) is dominant to chinchilla (f). The genes
controlling these traits are linked.
• The following cross was made: rabbits heterozygous for
both traits that express black, full color, with rabbits
that are brown, chinchilla. The following results were
obtained:
o 31 brown, chinchilla
o 35 black, full
o 16 brown, full
o 19 black, chinchilla
• Determine the genotype of the heterozygous parents,
and the map distance between the 2 genes.
Question #1-Solution
The information given about the phenotypes of the parents and the dominance
relationships amongst these alleles indicates that this cross was a testcross:
BbFf x bbff
We’re given the numbers for each of the expected phenotypic classes amongst the
offspring:
31 bbff
34 BbFf
16 bbFf
19 Bbff
The latter two classes, which have considerably fewer members, must be the nonparentals. Hence we can calculate the recombination frequency as (16 + 19) /
(31 + 34 + 16 + 19) or 35%. An examination of the genotypes of the parental
classes indicate that the heterozygous parents must have had one chromosome
with the B & F allele on it and the other chromosome with the b & f allele. We
can represent this as BF/bf or BF bf.
Question #2
On chromosome 3 in Drosophila, there are
the following mutations:
• Lyra (Ly) and Stubble (Sb) which are dominant
mutations.
• A recessive mutation with bright red eyes (br).
A female heterozygous for the 3 mutations was
mated to a male homozygous for the bright red
mutation.
Question #2 (cont.)
The following data
was obtained:
Total: 1000
Draw a chromosome
map for the three genes.
Phenotype
Number
Ly Sb br
- - Ly - - Sb br
Ly - br
- Sb Ly Sb - - br
404
422
18
16
75
59
4
2
Question #2 – (solution
highlights)
Phenotype
Number
Ly Sb br
- - Ly - - Sb br
Ly - br
- Sb Ly Sb - - br
404
422
18
16
75
59
4
2
Genes
Ly/Sb
Ly/br
Sb/br
Number of
recombinations
18+16+75+59 =
168
18+16+4+2 =
40
75+59+4+2 =
140
Question #3 - Detecting &
Measuring Linkage in Humans
1
1
2 3
1 2
4
3
5 6
4
7
5
2
8 9 10 11 12 13 14 15 16
Question #3 - (solution
highlights)
Phenotype Diseased Total
OO
1
8
BB
0
0
BO
9
11
AO
0
3