Tutorial #2by 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

Recombinant

2 1A AB B

a ab b

A aB b 3 4

a ab b

A ab b 5 6

A aB 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

Linkage) No(5.0)ionRecombinat(0)Linkage( P

• 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.

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.

tsrecombinan double of # expected

tsrecombinan double of # observerd1ce(I)Interferen

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.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:

o31 brown, chinchillao35 black, fullo16 brown, fullo19 black, chinchilla

• Determine the genotype of the heterozygous parents, and the map distance between the 2 genes.

Question #1-SolutionThe 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 bbff34 BbFf16 bbFf19 Bbff

The latter two classes, which have considerably fewer members, must be the non-parentals. 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 redmutation.

Question #2 (cont.)

The following data was obtained:

Total: 1000

Draw a chromosomemap for the three genes.

PhenotypeNumber

Ly Sb br404- - -422Ly - -18 - Sb br16Ly - br75

-Sb- 59Ly Sb- 4

- -br2

Question #2 – (solution highlights)

PhenotypeNumber

Ly Sb br404- - -422Ly - -18 - Sb br16Ly - br75

-Sb- 59Ly Sb- 4

- -br2

GenesNumber of recombinations

Ly/Sb18+16+75+59 = 168

Ly/br18+16+4+2 = 40

Sb/br75+59+4+2 = 140

Question #3 - Detecting & Measuring Linkage in

Humans

1 2

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

1 2 3 4 5

Question #3 - (solution highlights)

PhenotypeDiseasedTotal

OO18

BB00

BO911

AO03