Genes on a chromosome

Question 1

Linkage

Answer 1

Coupling of 2 loci leads to a shortage of recombinant classes
Due to physical association of 2 loci on same chromosome

Question 2

Test cross

Answer 2

Cross with a double mutant
Cross two recessive mutants:
P1: purple vestigal x red normal (prpr vgvg x ++ ++)
F1: all red normal (pr+ vg+)

Question 3

Backcross

Answer 3

Cross with a parental type

Question 4

Example of testcross and backcross

Answer 4

Cross two recessive mutants:
P1: purple vestigal x red normal (prpr vgvg x ++ ++)
F1: all red normal (pr+ vg+)
Cross F1 females with purple vestigal males
Expect equal frequency but shortage in recombinant classes

Question 5

Morgan’s Hypothesis

Answer 5

Loci are linked together and are inherited to some extent as a single unit
Deviation from Mendel’s 2nd law due to physical association of mutants

Question 6

Coupling

Answer 6

Two mutants or two wild type in each parental strain

Question 7

Repulsion

Answer 7

Each parent has one mutant and one wild type

Question 8

Example of testcross in repulsion

Answer 8

P1: red vestigal x purple normal (++ vgvg x prpr ++)
F1: all red normal (pr+ vg+)
Shortage in recombinants

Question 9

Mitosis and Meiosis

Answer 9

Homologues pair up + form tetrad in prophase 1
Chiasma
Crossover = recombinants
Sturtevant - Rarer recombinants are closer on the chromosome
Meiosis - reduction division, 1x doubling of DNA, 2x cell division

Question 10

Chiasma

Answer 10

2 homologue DNA break, crossover + exchange genetic info

Question 11

3 point cross

Answer 11

Developed by Alfred Sturtevant
Recombination freq. = no. recombinants/ total
Smaller freq. = fewer chiasmata - closer the loci
Recombination freq. of 1% = 1 map unit

Question 12

Example of 3 point cross

Answer 12

P1: wild type x triple mutant
F1: triple heterozygote
Backcross F1 to triple mutant, if loci aren’t linked expect equal ratio of 8 possible phenotypes (2^3)

Question 13

Drosophila 3 point cross

Answer 13

sc (scute) - loss of thoracic bristles
ec (echinus) - roughened eye
cv (crossveinless wing)
P1: cross wild type and triple mutant
F1: all triple heterozygotes (sc+ ec+ cv+)
Testcross F1 and triple mutant → shortage of 4 recombinants

Question 14

Double crossover

Answer 14

Phenotype with double crossover must be counted twice

Allows order of genes to be established

Question 15

Crossover interference

Answer 15

Interaction between recombinant events, if one chiasma formed has an influence on the chance of another nearby

Question 16

Coefficient of interference

Answer 16

1 - (observed no. recombinants/expected no. recombinants)