LECTURE 6

2.3 Genetic Mapping 2.4 Pedigree

Analysis

OBJECTIVES

At the end of the lesson, students should be able to :• Analyze the example on pedigree

given.• Calculate and map a chromosome’s

genetic loci using recombination data.• Determine the position of genes/loci

along a chromosome based on recombinant frequency data.

• Explain Pedigree analysis.

Genetic Mapping

The proportion of recombinants resulting from a dihybrid test cross is used to calculate the cross over value (COV), a measure of linkage, and if linkage occurs, the distance between genes .

COV is also known as recombination frequency.

Genetic Mapping Formula

COV = total number of recombinant /

total number of offspring x 100

• If the genes are not linked, the expected phenotype ratio of

such a cross is 1:1:1:1 and there is a 50 per cent chance

that alleles on separate chromosomes will be inherited together,

the expected COV is 50 per cent; the lower the value, the closer the

genes.

• Thus the COV can be used to locate the relative positions of genes on chromosomes, a process called chromosome mapping or genetic mapping.

• By convention, one per cent COV is equivalent to one map unit.

• Today the word centimorgan is often used.

Cross over (meiosis)

Homologous chromosome in synapsis

– 4 chromatid (tetrad)

P

Q

p

q

kiasma

r R

Genetic mapping of Drosophila melanogaster genes

Example

P : CCSS x ccss

P : CS/CS x cs/cs

G : CS csF1 : CS/cs x cs/cs

G : CS cs Cs cS cs

x co co

CS/cs cs/cs cS/csCs/cs

480 480 2020

Parent combination

Recombinant

Genotype Individuals

CS/cs 480

cs/cs 480

Cs/cs 20

cS/cs 20

Parent combination

Recombinant (cross over)

Eg : Test cross F1 : CS/cs x cs/cs

Parent combination = (480 + 480)/1000 = 96%

Recombinant = (20 + 20)/1000 = 4% (cov)

If C/c dan S/s genes is far apart, cov > 4%

Jumlah 1000

C S

4 map unit

EXERCISE 1

• P, Q, R dan S genes are linked. • COV for pairs of genes is as follows:

P and Q = 35%P and R = 5%R and Q= 40%Q and S = 10%R and S = 30%

• Map the chromosome.

• Tips: Start with the furthest. (highest cov % )

Solution:R and Q= 40%P and Q = 35%R and S = 30%Q and S = 10%P and R = 5%

R Q

40 map unit

SP5 25 10

Cross over value given for

P and Q = 4%

P and R = 1%

State the cross over value possible for Q and R.

P Q

4

R1 3

Cross over value for Q and R = 3%

P Q41

R

Cross over value for

Q and R = 5%

EXERCISE 2

EXERCISE 3• In tomatoes,

S = smooth s = wrinkledM = red flower m= white flower

• Test cross between smooth, red flower plant with wrinkled, white flower plant resulted in:

Smooth fruit ,red flower = 300Wrinkled fruit , white flower = 300Smooth fruit, white flower = 100Wrinkled fruit, red flower = 100

• Count the map unit for the distance between both S and M genes on its chromosome.

Total progeny = 800

Recombinant = 100 + 100800

= 25%

Map unit for both S and M genes = 25 m.u

S M 25

SM/sm x sm/smSmooth fruit ,red flower = 300 (SM/sm)Wrinkled fruit , white flower = 300 (sm/sm)Smooth fruit, white flower = 100 (Sm/sm)Wrinkled fruit, red flower = 100 (sM/sm)

ConclusionThrough genetic mapping, we can

determine :– Relative distance between two

genes.– Linear sequence of the linked

genes on a chromosome.

Pedigree Analysis

• A pedigree is a family tree that shows genetic interrelationship among parents and offspring across one or more generations.

• Standardized symbols, methods and definitions are used to construct it.

• Eg: Inherited disease in the British Royal family.

I

II

III

1 2

1

1

2 43

Circle = female

Square = male

Roman = generation

Numbers = individuals in generationColored = different phenotype

Eg: “widow’s peak” trait (dominant allele)

W Allele = widow’s peak w allele – no widow’s peak

I

II

III

Color = widow’s peak

Ww ww

ww

WW @ Ww

ww WwWw

Widow’s peak pedigree