Chapter 5: LINKAGE AND GENETIC MAPPING IN EUKARYOTES
5.2 - Genetic mapping in plants and animals
Genetic maps
Purposes
Alfred Sturtevant
Thomas Hunt Morgan’s student
1st genetic map done in Drosophila
Complete problem on the board to determine map distance and interference.
Map distance
Recombination frequency (r. f.) = map distance (m.d.) if <50 cM or mu.
Map distance = (# recombinant offspring / total # of offspring ) X 100
1 mU or cM = 1% r.f.
Interference:
the product rule allows us to predict the likelihood of a double crossover from the individual probabilities of each single
crossover. lower-than-expected value is due to a common genetic phenomenon, termed positive interference
 The first crossover decreases the probability that a second crossover will occur nearby
Determining map distance typically requires that a testcross is performed. Consequently, we expect an equal proportion of offspring
from the testcross!
Calculate map distance
1. Determine gene order
2. Determine recombination frequency for single crossovers
3. Determine recombination frequency for double crossovers
4. Add single and double r.f. X 100 = mU between outside and middle gene.
5. M.D. for outside genes = [single + single + (double X2)] X 100
Calculate Interference
1. Calculate probability of a double crossover using the product rule. Multiply the probability of each single crossover (prob.
of single crossover = r.f. between genes). P (double) = P (single) X P (single)
2. Calculate the expected number of offspring that are produced from a double crossover. P(double crossover) X total = E
3. Calculate the coefficient of coincidence(C). C = observed # of double crossovers/ expected # of double crossovers C=O/E.
4. Calculate interference (I). I= 1=C Interference = % of the expected number of crossovers did not occur
5. If I = 0 then there was no interference and O=E and I=1-1
6. If I=1, then there was complete positive interference
7. If I= negative # then there was negative interference and each or both single crossover enhanced the rate of the other
crossover.
5.3 Genetic mapping in Haploid Eukaryotes
Model organisms such as the sac fungi (ascomycetes) have been particularly useful to geneticists because of their unique style of
sexual reproduction
Exist as haploid or diploid
Easy to manipulate genetically
Short life cycles and cell cycles.
Multicellular or unicellular
Haploid mating produces diploid then sporulation produces haploid offspring.
Gametes in fungi are all viable and symmetric.
We can analyze the offspring from a single meiotic event to determine recombination frequency.
Unordered tetrads or octads
 Ascus provides enough space for the spores to randomly mix together
Ordered tetrads or octads
 Ascus is very tight, thereby preventing spores from randomly moving around
Diploid
animals
and plants
Model
organism(s)
e.g. Drosophila
melanogaster
(fruit flies),
Unordered Saccharomyces
tetrads
cerevisiae
Common
name
Fruit flies;
others
include
tomato, corn
.
Budding
yeast
1. Determine gene order
2. Determine recombination frequency for single
crossovers
3. Determine recombination frequency for double
crossovers
4. Add single and double r.f. X 100 = mU between
outside and middle gene.
5. M.D. for outside genes = [single + single + (double
X2)] X 100
M.D. =
(# of recombinant
offspring/total # of
offspring) X 100
1. Determine the number of parental ditypes (PD), nonparental ditypes (NPD), and tetratypes (TT)
PD = parentals
NPD and TT = non-parentals
M.D. =
[½ TT + 3(NPD)]/
Total ]X 100
2. Use the tetrad analysis tree to determine the map
distance between two genes.
Ordered
Octads
Neurospora
crassa
Mold
Appearance
Map distance
1. Identify offspring that are produced from no crossing over, first
division segregants (FDS). FDS- alleles are separated during the
first division (after meiosis I).
2. The order/arrangement of the spores indicates if recombination
has occurred.
4:4, M1 pattern – FDS
2:4:2 or 2:2:2:2 M2 pattern - SDS
3. Identify offspring that are the result of recombination, seconddivision segregants (SDS). SDS – alleles are separated after the
second division of meiosis (after meiosis II)
Calculate map distance between the gene and its centromere.
M.D. between two
genes.
OXO
Mating
↓
O (Diploid)
Sporulation
↓
O
OO
O
Tetrads
PD, NPD or TT
M.D. = ½
SDS/Total X 100
OOOO0000
M.D. between one
gene and its
centromere.
Ordered octads
A
a
10
20
30
Calculate the map distance between gene A and its centromere.
[½ (SDS)/Total Asci ] X 100 = Map distance
[½ (30)/100] X 100 = 15/100 X 100 = 15 cM or 15 mU
40
Unordered tetrads
 Consider a diploid yeast zygote with the genotype ura+ura-2 arg+arg-3
 ura+ and arg+ = Normal alleles required for uracil and arginine biosynthesis, respectively
 ura-2 and arg-3 = Defective alleles
 Result in strains that require uracil and arginine in their growth medium
100
10
2
Count the number of offspring that represent the PD, TT, and NPD.
Use the tetrad analysis tree to calculate map distance.
Work through problem E26.
E26. His-5 and lys-1 are alleles found in baker’s yeast that require histidine and lysine for growth, respectively. A cross was made between two haploid yeast that are his-5 lys-1
and his+ lys+. From the analysis of 818 individual tetrads, the following numbers of tetrads were obtained:
2 spores: his-5 lys+ + 2 spores: his+ lys-1 = 4
2 spores: his-5 lys-1 + 2 spores: his+ lys+ = 502
1 spore: his-5 lys-1 + 1 spore: his-5 lys+ + 1 spore: his+ lys-1 + 1 spore: his+ lys+ = 312
A. Compute the map distance between these two genes.
Answer The first thing to do is to determine which asci are parental ditypes (PD), nonparental ditypes (NPD), and tetratypes (T). A parental ditype will contain a 2:2 combination
of spores with the same genotypes as the original haploid parents. The combination of the 502 asci are the parental ditypes. The nonparental ditypes are those containing a 2:2
combination of genotypes that are unlike the parentals. The combination of four asci fits this description. Finally, the tetratypes contain a 1:1:1:1 arrangement of genotypes, half of
which have a parental genotype and half of which do not.
Map distance =
312  (6)(4)
818
 20.5 mu
=
T + 6NPD
 0.5  100
Total number of asci