1. 6- GENE LINKAGE AND GENETIC MAPPING Compiled by Siti Sarah Jumali Level 3 Room 14 Ext 2123

2. LINKAGE AND CROSSING OVER

3. LINKAGE Chromosome is a linkage group Linkage refers to: 1) 2 or more genes may be related on the same chromosome – Physically linked because eukaryotic chromosome contains a single, continuous linear molecule of DNA 2) Genes that are close on the same chromosome tend to be transmitted as a unit – Indicate the linkage has an influence on the pattern

4. Bateson and Punnet discovered 2 traits that did not assort independently They suggested that the transmission of these 2 traits from the parental generation to the F 2 generation was somehow coupled and not easily assorted independently This is due to linkage

6. LINKAGE CONT’D Traits may not assort independently such as in Mendel’s law, where the traits did not segregate showing linkage Therefore it produces bizarre phenotypic ratio than 9:3:3:1 to something like 15:1:1:4 Chi-square (χ 2 ) can be used to distinguish between linkage and independent assortment

7. Trihybrid cross- Lets consider this Gray body, red eyes1159 Yellow body, white eyes1017 Gray body, white eyes 17 Yellow body, red eyes 12 Total2205 Red eyes, long wings770 White eyes, miniature wings716 Red eyes, miniature401 White eyes, long wings318 Total 2205 Nonparental offspring Nonparental Offsprings are called recombinant

8. ? Question Why is the recombinant number so low?

9. Crossing over May produce recombinant phenotypes or known as recombinant or nonparent Because the genetic information recombine during Meiosis 1

10. Chi-square ( χ 2 ) Is used to test goodness of fit between a genetic hypothesis and observed experimental data Must 1 st propose a hypothesis; 2 genes are unlinked, therefore follow Mendel’s law Null hypothesis is said to be null because it is assumed that there will be no difference between experimental data and observed data If χ 2 is low, accept hypothesis, genes assort independently If χ 2 is high, reject hypothesis, genes are linked

11. Chi-square ( χ 2 ) can be used to distinguish between linkage and independent assortment χ 2 =Σ (o-e) 2 / e; – Where o = observed value, e = expected outcome (theory) If the data do not fit, we will reject the idea that the genes assort independently and conclude that the genes are linked

12. Chi-square ( χ 2 ) analysis The larger the chi-square, the smaller the p value P value more than 5% will ensure acception of null hypothesis 5% and below will cause rejection of null hypothesis

13. Experiment: Do the genes for flower color and pollen shape assort independently?

15. Mendel’s law overruled Soon after Mendel's rules were rediscovered, it was found that some loci did not assort independently. The simplest explanation is that the loci lie close to each other on the same chromosome. They are linked on the same chromosome.

16. Thomas Morgan He did experiment on genetic mapping on X chromosome He concluded that genes are located on the same X chromosome, so they are likely to be inherited together Due to crossing over, X chromosome can exchange pieces of chromosomes and create new parental combination of alleles The likelihood of crossing over depends on the distance between 2 genes. I f 2 genes are far apart from each other, crossing over is more likely to occur

17. CROSSOVER VALUE AND GENE MAPPING Determining Map Distance The percent recombination is calculated as before. (# of recombinants ÷ # of offspring) x 100 = % of recombinants 1% recombination = 1 map unit, or 1 centimorgan, in honor of T.H. Morgan, one of the first persons to propose this linkage, and first to win a Nobel prize in genetics. Two phenotypes are in very high frequency have the same phenotypes as the original parents (P 1 ). These are called non-recombinants or parentals. Two phenotypes are in low frequency and combine the phenotypes of the two original parents (P 1 ). These are called recombinants or non-parentals.

18. Three-point cross Determining gene order The pair of phenotypes with the highest frequency is always the non-recombinant group. The pair of phenotypes with the lowest frequency is always the double cross-over group. The probability of a double cross-over is approximately the product of the probability of the single cross- over. compare the wild type class b + pr + c + to the purple double cross-over class b + prc + and we can see that the purple locus does not match indicating that the purple locus is in the middle. Also note that as the parental phenotypes are composed of a gamete from the female that is either b pr c or b + pr + c +,

19. Determining Map Distance The next step is to set-up a table that is titled " number of recombinants between". The percent recombination is calculated as before. # of recombinants 100 x ----------------- = % of recombinants # of offspring thus for the distance from b to pr 887 100 x ----------------- = 5.9% or 5.9 m.u. 15,000 The distance from b to c, the two outside loci, (25.4 m.u.) is the sum of the distance from b to pr (5.9 m.u.) and the distance from pr to c (19.5 m.u.).