SME
Genetics
Linkage and mapping
Use this for “A test cross gives 418 + 422 parental and 78 + 82 recombinant offspring. How far apart are the genes?”
Diagram
A–B: 16 cM
RF = 16% of the offspring are recombinant: 1% is 1 cM
This is an example. Type your own values to start a new problem.
RF (%)
=
R
N
×
100
d (cM)
=
RF (%)
P
Parental offspring
example
Formulas
N
=
P
+
R
1,000 = 840 + 160
RF
=
100
×
R
N
16
=
100
×
160
1,000
d
=
RF
16 = 16
Assumptions
•
A test cross: every offspring shows which alleles came from the heterozygous parent.
•
The two parental classes are added, and so are the two recombinant ones.
•
Recombinants come from a crossover between the genes; 1% recombinants is 1 cM.
•
Genes past 50 cM assort as if unlinked: RF stops at 50%.
Step-by-step
WE KNOW
P = 840, R = 160
FIND
offspring (N), recombination frequency (RF), map distance (d)
Step 1 · Find offspring (N)
Use N = P + R
Every offspring is parental or recombinant.
N
=
840
+
160
N
=
1,000
Step 2 · Find recombination frequency (RF)
Use RF = 100 × R ÷ N
The share of offspring that are recombinant, as a percent.
RF
=
100
×
160
1,000
RF
=
16,000
1,000
RF = 16%
Step 3 · Find map distance (d)
Use d = RF
One percent recombinants is one centimorgan.
d
=
16
cM
Check
Put the numbers back in. Both sides should match.
1,000 = 840 + 160 ✓
16
=
100
×
160
1,000
✓
16 = 16 ✓
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