Lecture 12

Question 1

Progeny with the parental type will have

Answer 1

• An intermediate trait score

- High variance

Question 2

Progeny with the recombinant type will be

Answer 2

• Either heigh or low

- Little variance

Question 3

LOD score (Z):

Answer 3

• Logarithm of Odds
• Probability of most likely recombination fraction/probability that they are unlinked
• If numerator <denominator, the log will be negative
• LODs are additive
• Used to answer whether the two genes of interested are linked

Question 4

Mapping markers by recombination:

Answer 4

• Let c be the actual recombination distance between markers
• We estimate c from the observed recombination fraction (r)
• Cest = no. of recombinants/total number

Question 5

Maximum likelihood (ML) approach:

Answer 5

• Write likelihood L (probability of data) as a function of unknown c: L(c)
• CMLE (maximum likelihood estimate) of true c
• L(c) = (1 - c) to the power of 90 x c to the power of 10
  (100 offspring, 90 parentals, 10 recombinants)

Question 6

LOD = 3 means

Answer 6

• Linkage (Ha) is 1000 times more likely than free recombination (Ho) we are confident that we have linkage there.
• This is the standard figure threshold

Question 7

LOD = 0

Answer 7

Ha = Ho (in other words beta base 1 = 0)

Question 8

There are three parameters

Answer 8

• Beta base 1
• Beta base 0
• Variance

Question 9

Beta base 1:

Answer 9

• The mean difference between the two populations

Question 10

Beta base 0:

Answer 10

• The base level trait value

- The trait value that everyone has, eg) height is always 150cm (B0) + x (B1)

Question 11

Trait value:

Answer 11

Y = B1X + B0 + e

Question 12

Single marker analysis:

Answer 12

• T-tests marker by marker

Question 13

Intverval mapping:

Answer 13

• Considers the likelihood of a QTL existing for every point in the linkage map
• Uses LOD analyses

Question 14

Composite Interval mapping:

Answer 14

• Adjusts lokelihood based on the state of QTLs at other parts of the genome

Question 15

Position and effect are somewhat confounded..

Answer 15

• The effect of the causal variant is likely to be underestimated (because recombination between the marker and the causal variant)
• In the causal variant distant with a big effect of closer with a smaller effect?

Question 16

Randomly Amplified Polymorphic DNA (RAPDs)

Answer 16

• Synthesises 10mers, of any sequence, to use as PCR primers
• Low stringency CR with a single primer yields DNA fragments on a gel
• If every individual has it it is no informative
• If only some individuals have it polymorphisms can be identified which provide information about individuals

Question 17

Pros of RAPDS:

Answer 17

• Not resource intensive
• Scattered throughout the genome
• Can use in uncharacterised organisms

Question 18

Cons of RAPDS:

Answer 18

• Low reproducibility
• Bands of different intensity and sharpness
• Very sensitive to slight variations in reagent concentrations
• PCR artefacts
• PCR contamination

Question 19

How do we calculate contribution to variance?

Answer 19

• Compare linkage groups

- Attribute the amount of phenotypic variance each group explains, by comparing LL and HH homozygotes.

Question 20

Monkey flowers, and the YUP gene:

Answer 20

• Controls yellow carotenoid pigments
• Putting lewisii into a cardinalis background and vice versa showed that changing the colour of the flower changes the pollinators (birds or bees)
• Clear evidence of an adaptive trait being mapped and understood.

Question 21

What have we discovered?

Answer 21

• Candidate genes can be identified
• Complementation test to see if natural alleles are complimented by artificial lab alleles
• Positional cloning

Question 22

Positional cloning:

Answer 22

• The closest molecular markers to the trait in the linkage map
• This can be used to identify the region of the physical map

Question 23

Chromosome walking:

Answer 23

1. isolate a series of overlapping clones
2. Determine the order of clones and select clone closest to target
3. Repeat to ‘walk’ along the chromosome bidirectionally

Question 24

QTL mapping to positionally clone, these steps can be replaced by sequencing the genome:

Answer 24

Steps 1 - 6 from last lecture
Step 7: Probe a genomic library with the molecular markers flanking the QTL
Step 8: Chromosomal walk to span the distance between markers
Step 9: Characterise the clones uncovered in the walk
Step 10: Identify and test candidate genes

Question 25

We can test candidate genes by..

5

Answer 25

• Homology with other systems
• Expressed in the correct tissue
• Obvious mutations
• Complementation
• Transgenic test

Question 26

AFLPs (Amplified fragment length polymorphisms):

Answer 26

• Annonymous markers that are robust
• Can still score many polymorphisms
• Good for mapping uncharacterised genomes
• Easily visualised
• Reliale

Question 27

Steps for generating AFLP gels

Answer 27

1. Take genomic DNA
2. Digest with 2 RE
3. Use two dslinkers complementary to the overhanging ends and hook onto the end
4. Perform PCR reaction
5. Determine which bands are polymorphisms - these are informative.