Lecture 9: Linkage Mapping

Question 1

crossover frequency is a function of the

Answer 1

distance between two loci

Question 2

how can we mathematically prove if a test cross result is due to linkage or not?

Answer 2

perform a chi-square test, which assesses the likelihood that a deviation from expectation is due to chance

Question 3

why is the chi-square test necessary to prove linkage?

Answer 3

• deviations from 1:1:1:1 ratios can represent chance events OR linkage
• the chi-square test accounts for sample size

Question 4

chi-square test: framing a hypothesis

Answer 4

null hypothesis - observed values are not different from the expected values. for linkage studies, this is no linkage where you expect a 1:1:1:1 ratio of gametes

alternative hypothesis - observed values are different from expected values. for linkage studies, this is that the genes are linked and you expect a significant deviation from 1:1:1:1 ratio

Question 5

4 steps to conducting a chi-square test

Answer 5

1. formulate null hypothesis: genes are not linked (1:1:1:1 ratio of phenotypes
2. compute: x^2: Σ (O-E)^2/E
3. determine degrees of freedom (number of independent measurements)
4. consult x^2 chart of critical values

Question 6

how do we know whether to accept or reject the null hypothesis?

Answer 6

if p<0.05, then we reject the null hypothesis and accept the hypothesis that genes are linked to

Question 7

what does p<0.05 mean?

Answer 7

less than 5% of the time, we are incorrectly rejecting the null

Question 8

what did Alfred Sturtevant do?

Answer 8

he postulated that the frequency of crossovers between two genes is a function of their distance apart on the chromosome; he also created the first genetic linkage map

Question 9

recombination frequency (RF) =

Answer 9

number of recombinants/total number of progeny

Question 10

one map unit (m.u.)=

Answer 10

one centiMorgan (cM) = 1% recombination
- 1 product of meiosis out of 100 is recombinant

Question 11

in any given cross, the recombination frequency (RF) cannot be greater than

Answer 11

50% (=unlinked)

Question 12

steps in analysing a three-point test cross

Answer 12

1. anticipate and identify 8 kinds of meiotic products
2. identify pairs of reciprocal products
3. identify parental types as the most frequent pair of products
4. identify double crossover products as least frequent pair of products
5. compare parental and double crossover to deduce locus order
6. compute map distances (distances between loci) by breaking down the results for each interval - RF = SXO+DXO/total

Question 13

define interference

Answer 13

crossover in one region interferes with simultaneous crossing over in adjacent regions

Question 14

how can we calculate interference?

Answer 14

1. expected frequency of DCO = product of frequency crossovers in two regions
2. coefficient of coincidence = observed DCO/expected DCO
3. interference = 1- coefficient of coincidence

Question 15

genetic vs physical distances on a chromosome

Answer 15

• recombination and physical maps are co-linear (syntenic) but are frequently misaligned as recombination distances (cM) are frequently not the same as physical distances (bp of DNA)
• that is, one cannot say that X bp of DNA sequence always equals Y cM on a linkage map
• this is because recombination varies across the length of a chromosome

Question 16

do traits need to be phenotypic for linkage mapping?

Answer 16

no - we can also use molecular markers - any polymorphism can be mapped

Question 17

why can only female drosophila be used in linkage experiments?

Answer 17

In male Drosophila melanogaster, meiosis does not involve recombination, a phenomenon known as achiasmy