Genetic Linkage And Three Point Mapping

Question 1

Meiosis

Answer 1

Meiosis begins with a single mitotic division. Here, the amount of DNA in the parental cell doubles up during interphase. The two daughter cells of the of the division are diploid (2n). These 2 cells divide without doubling their DNA to produce 4 haploid (n) gametes

Question 2

Homologues

Answer 2

2 copies of each chromosome (one paternal and one maternal); homologues pair up and from tetrad in prophase 1

Question 3

Chiasma

Answer 3

2 homologue DNA arms break, crossover and physical exchange of information

Question 4

Crossover

Answer 4

Crossover in between the pr and vg loci generates new combinations of alleles (recombinants)

The rarer the recombinants are, the closer they must be on the chromosome

Question 5

Differences between meiosis and mitosis

Answer 5

Look at booklet an blurt

Question 6

Three point mapping and linkage

Answer 6

Resembles Morgan’s two point cross using drosophila but uses a stock of 3 mutants crossed to the wild type - more efficient

A three point cross (developed by Sturtevant then a student) uses the proportion of recombinants to infer the position of genes on chromosomes

The smaller the frequency the fewer the number of chiasmata, and hence the closer the loci are.

Recombination frequency to 1% is one map unit

Question 7

The three point cross method and analysis

Answer 7

P1 ++ ++ ++ aa bb cc (Triple wild type x triple mutant)

F1 +a +b +c (Triple heterozygotes)

Backcross the F1 flies to the triple mutant aa bb cc stock. If the loci are not linked we expect an equal number of each of the 8 possible phenotypes (two cubed). A 1:1:1:1:1:1:1:1 ratio.
A simplified example from Drosophila. 3 recombinant mutants: sc or scute - loss of thoracic bristles, ec or echinus - roughened eye, cv - crossveinless wing.

P1 scsc ecec cvcv (scute echinus crossveinless) X sc+sc+ ec+ec+ cv+cv+ (wild type)

F1 scsc+ ecec+ cvcv+ (all wild type triple heterozygotes)

Look at numbers in the doc