genetics essentials concepts and connections second edition genetics essentials concepts and...

Post on 16-Dec-2015

240 Views

Category:

Documents

3 Downloads

Preview:

Click to see full reader

TRANSCRIPT

• GENETICS ESSENTIALS•Concepts and Connections

• SECOND EDITION

• GENETICS ESSENTIALS•Concepts and Connections

• SECOND EDITION

Benjamin A. Pierce

© 2013 W. H. Freeman and Company

CHAPTER 5Linkage, Recombination, and

Eukaryotic Gene Mapping

Chapter 5 Outline

• 5.1 Linked Genes Do Not Assort Independently, 114

• 5.2 Linked Genes Segregate Together and Crossing Over Produces Recombination Between Them, 116

• 5.3 A Three-Point Testcross Can Be Used To Map Three Linked Genes, 128

• 5.4 Genes Can Be Located with Genomewide Association Studies, 137

The case of male baldness

• X-linked or not?

• Study of linkage of a SNP and the association with male pattern baldness shows that they do not assort independently, they are transmitted together

5.1 Linked Genes Do Not Assort Independently

• Linked genes: genes located close together on the same chromosome-Belong to the same linkage group

Recombination and alleles

Flower color and pollen shape

F1 generation gives purple color and long pollen

• Appears that the purple is dominant to red and long is dominant to short pollen

• What with the F2, it is expected to get 9:3:3:1

• BUT…

There is no 9:3:3:1 ratio; what happened

5.2 Linked Genes Segregate Together and Crossing Over Produces Recombination

Between Them

• Fig. 5.3• Notation for crosses with linkage• Complete linkage leads to nonrecombinant

Gametes and nonrecombinant progeny• Crossing over with linked genes leads to

recombinant gametes and recombinant progeny• Figs. 5.4 & 5.5

Labeling alleles of two genes on a same chromosome

• Cross AA BB x aa bb can be represented byA B a bA B x a b

The F1 will be:A Ba b

Also it can be representedA Ba b

The two test cross cases Linked v.s. not linked

• Linked genes give only nonrecombinant gametes

• Unlinked genes give both nonrecombinant and recombinant

Concept Check 1

For single crossovers, the frequency of recombinant gametes is half the frequency of crossing over because:a. a test cross between a homozygote and heterozygote produces ½ heterozygous and ½ homozygous progeny.b. the frequency of recombination is always 50%.c. each crossover takes place between only two of the four chromatids of a homologous pair.d. crossovers take place in about 50% of meiosis.

What happens in recombination and the gamete formation

5.2 Linked Genes Segregate Together and Crossing Over Produces Recombination

Between Them

• Calculating Recombination Frequency• Recombination frequency = (No. recombinant progeny/Total No. of

progeny) 100%• Fig. 5.6

• Coupling and Repulsion Configuration of Linked Genes• Coupling (cis configuration): wild-type alleles are found on one

chromosome; mutant alleles are found on the other chromosome.

Calculating frequencies

Calculating frequencies

Rec. Frequency = x100%

Rec. Frequency = x100%

8 + 7 55+53+8+7

# of recombinant progenyTotal # of progeny

Rec. Frequency = x100%

15123

Rec. Frequency = 12.2 % OR 0.122

5.2 Linked Genes Segregate Together and Crossing Over Produces Recombination

Between Them

• Coupling and Repulsion Configuration of Linked Genes• Repulsion (trans configuration): wild-type allele and mutant allele

are found on the same chromosome.• Fig. 5.7

• Testing for Independent Assortment • Figs. 5.8 & 5.9

Coupling and repulsionp+ b+p b

p+ bp b+

Coupling and repulsion

• Gene Mapping with Recombination Frequencies:• Genetic maps are determined by recombinant frequency.• Map unit and centiMorgans

• Constructing a Genetic Map with the Use of Two-Point Testcrosses• Fig. 5.10

5.2 Linked Genes Segregate Together and Crossing Over Produces Recombination

Between Them

Recombination frequencies and distances of genes

• Based on recombination frequencies genetic maps are generated

• Actual distances are on physical maps

• Recombination maps are approximations and are measured in map units or cM (centiMorgans)• One map unit equals to 1% recombination frequency

Genetic distances measured with recombination rates are approximately additive: if the distance from gene A to gene B is 5 m.u., the distance from gene B to gene C is 10 m.u., and the distance from gene A to gene C is 15 m.u.

Limitations of gene mapping

• If the genes are far apart and the frequency is 50% we cannot say if the genes are on different chromosomes or just to far on same chromosome

• Two genes that are relatively far will give underestimate of recombination events due to possibility of multiple crosses between the genes.

Example: two point cross to determine map of 4 genes

5.3 A Three-Point Testcross Can Be Used to Map Three Linked Genes

• Constructing a Genetic Map with the Three-Point Testcross• Fig. 5.12• Determining the gene order• Determining the location of crossovers

• Fig. 5.13

Double crossover :three genes

A Three-Point Testcross Can Be Used to Map Three Linked

Genes

• Calculating the recombination frequencies

• Interference and coefficient of coincidence

• Effect of multiple crossovers

5.3 A Three-Point Testcross Can Be Used to Map Three Linked Genes

Calculating the recombination frequencies

• Based on the numbers in the fruit fly testcross for three loci calculate the distances between the loci.• Recombinant progeny with a chromosome that

underwent crossing over between the eye-color locus (st) and the bristle locus (ss) include the single crossovers ( st+ / ss e  and  st / ss+ e+ ) and the two double crossovers ( st+ / ss / e+  and  st / ss+ / e ); Total of 755 progeny; so the recombination frequency between ss and st is:

Calculating the recombination frequencies

• Based on the numbers in the fruit fly testcross for three loci calculate the distances between the loci.• The map distance between the bristle locus (ss) and the

body locus (e) is determined in the same manner. The recombinant progeny that possess a crossover between ss and e are the single crossovers  st+ ss+ / e  and  st ss / e+  and the double crossovers  st+ / ss / e+  and  st / ss+ / e . The recombination frequency is:

Calculating the recombination frequencies

• Based on the numbers in the fruit fly testcross for three loci calculate the distances between the loci.• Finally, calculate the map distance between the outer two

loci, st and e. This map distance can be obtained by summing the map distances between st and ss and between ss and e (14.6 m.u. + 12.2 m.u. = 26.8 m.u.). We can now use the map distances to draw a map of the three genes on the chromosome:

Interference and coefficient of coincidence

• Calculate the proportion of double-recombinant gametes by using the multiplication rule of probability • Applying this rule, the proportion (probability) of gametes

with double crossovers between st and e is equal to the probability of recombination between st and ss multiplied by the probability of recombination between ss and e, or 0.146 × 0.122 = 0.0178.

• Multiplying this probability by the total number of progeny gives us the expected number of double-crossover progeny from the cross: 0.0178 × 755 = 13.4.

• Only 8 double crossovers—considerably fewer than the 13 expected—were observed in the progeny of the cross

Interference and coefficient of coincidence

• Crossovers are frequently not independent events: the occurrence of one crossover tends to inhibit additional crossovers in the same region of the chromosome, and so double crossovers are less frequent than expected.• The degree to which one crossover interferes with

additional crossovers in the same region is termed the interference. To calculate the interference, we first determine the coefficient of coincidence, which is the ratio of observed double crossovers to expected double crossovers:

In our case

• indicates that we are actually observing only 60% of the double crossovers that we expected on the basis of the single-crossover frequencies.

• Calculation of interference• Interference = 1 − coefficient of coincidence

• So the interference for our three-point cross is:• interference = 1 − 0.6 = 0.4

• 40% of recombinations are not observed due to intefrerence

5.4 Genes Can Be Located withGenomewide Association Studies

• Linkage analysis (genetic markers; anonymous markers)• Genome wide association studies• Haplotype• Linkage disequilibrium

top related