Biology 212 General Genetics
Spring 2007
Lecture 10: Linkage Mapping I
Reading: Chap. 4 pp. 122-132
Lecture Outline:
1. Linked genes
2. Two factor linkage maps
3. Genetic maps of chromosomes
Lecture:
1. Linked genes


Genes that are close together on the same chromosome are generally transmitted
together in meiosis
Do not assort independently
First studies of linked genes:
 Laboratory of Thomas Hunt Morgan early 20th century
 Drosophila
Linked genes stay together in meiosis (Fig. 4.1)
For two genes A and B, with 2 alleles that are linked on the same chromosome.
Can have two arrangements of the alleles in the heterozygotes
Alleles will be transmitted together based on their chromosome configuration.
cis configuration (in coupling)
A B
a
trans configuration (in repulsion)
A
b
b
For cis, mostly transmit AB and ab;
Ab and aB are recombinants
a
B
For trans, mostly transmit Ab and aB;
AB and ab are recombinants
Recombination (crossing over) involves exchange of genetic information between
chromosomes.
Fig. 4.4
1
When crossing over occurs, you get:
 two recombinant chromatids
 two non-recombinant (parental) chromatids
2. Two factor linkage maps
a. Importance of recombination frequencies for mapping



genetic distances are defined based on recombination frequencies
% recombination= # recombinant types/all types x 100
1% recombination = 1 map unit = 1 centimorgan (cM)
b. Constructing linkage maps
Drosophila traits (see Fig. 4.5)
White eyes:
w=white eyes
w+=normal red eyes
Diminutive (small) body size:
dm=diminutive
dm+=normal body size
P1 cross
white-eyed females with diminutive males
w dm+/w dm+
w+ dm/Y
F1 progeny
w dm+/w+ dm females and w dm+/Y males
Perform a “testcross”=cross of the progeny with the homozygous recessive type
F1 testcross:
w dm+/w+ dm with w dm/Y males (testcross parent-error in Fig.
4.5)
F2 progeny:
white eyes, normal size
red eyes, diminutive size
red eyes, normal size
white eyes, diminutive size
497
472
19
12
Parental types = 497 + 472 = 969/1000 = 96.9% non-recombinant
Recombinant types = 19 + 12 = 31/1000 = 3.1% recombinant types
Construct a linkage map of the two genes
2
Remember 1% recombination = 1 map unit = 1 cM
Genetic Map of X chromosome:
w
dm
3.1 m.u. (or cM)
3. Genetic maps of chromosomes



For genes far apart on the chromosome, the observed recombination frequency
may provide an underestimate of the map distance.
This is because multiple crossover events can occur that are not all detectable.
Best solution is to determine recombination frequencies over several smaller
intervals and add them together.
Maps of chromosomes are constructed to show the relative positions of genes.
Physical distances on chromosomes can’t be predicted exactly from genetic
distances.
.
3