Step one

• Two gene loci: A & B• What will your first cross be in an experiment

to test for possible meiotic crossing over?

• Hint: what condition do you have to have for crossing to be visible if it happens?

AAbb X aaBB

AaBb

Why is this genotype necessary?What is the next cross, and why?

AaBb X aabb

• Why is this the right cross?• What are the expected outcomes?• In what ratio?

Expected Outcomes

AB Ab aB abab AaBb Aabb aaBb aabb

1/4

250

1/4

250

1/4

250

1/4

250

Observed

AB Ab aB abab AaBb Aabb aaBb aabb

485 20 15 480

What is the null hypothesis here?Do the chi square test on the data.What is the % crossing over?How many map units apart are the loci?

To calculate % crossing over

• % crossing over = (20+15)/(20+15+480+485)• = 35/1000• = 0.035• = 3.5%• = 3.5 map units

Now a triple cross

• Three gene loci: A, B & C• As before, what will your first cross be in an

experiment to test for possible meiotic crossing over?

AABBCC X aabbcc

AaBbCc

Why is this genotype necessary?Draw the chromosomal structure of this result.

What is the next cross, and why?

The Cross

A B C

a b c

a b c

a b cX

The resultsABC 400abc 385Abc 10aBC 18ABc 40abC 35AbC 70aBc 85

Sums

• Non-recombinants = (400 + 385) = 785• AB recombinants = (10+18+70+85) = 183• BC recombinants = (40+35+70+85) = 230• AC recombinants = (10+18+40+35) = 103

Calculations

• % Non-recombinants = 785/1043 = 75.3• AB recombinants = 183/1043 = 17.5• BC recombinants = 230/1043 = 22.1• AC recombinants = 103/1043 = 9.9

B A C

17.5 mu 9.9 mu

22.1 mu

Resulting Map

Assessment• The values yield the correct order of the loci

along the chromosome.• There is a discrepancy between the two values

obtained for the distance between B and C: 22.1 mu and 27.4 mu (17.5+9.1).

• We assume that the greater of the two values is more accurate.– Mapping is more accurate over shorter distances– The values 17.5 and 9.1 will be more accurate

individually than the larger value, 22.1.

Explanation

• The values for long distances along the chromosome are skewed by the frequency of double crossing over.– The greater the distance, the greater the probability

of double crossing over– The standard calculations fail to consider the

contribution of double crossing over– The value 22.1 underestimates the crossing over

frequency between B and C and thus underestimates the BC distance

Logic

• We must correct for the frequency of double crossing over.

• To do so, we must determine which values represent double crossing over .

The Chromosomes

B a C

b A c

B A C

b a c

Before crossing over

After crossing over

The Correction

• Referring to the map, double crossing over occurs between B and C.

• To correct the BC map distance, we must add in the values for the double cross over species– (10+18) = 28– However, because each counted individual in the

population represents a double crossing over event, we must double these values.• 2*(10+18) = 56

Corrected Sums

• Non-recombinants = (400 + 385) = 785• AB recombinants = (10+18+70+85) = 183• BC recombinants = (56+40+35+70+85) = 286• AC recombinants = (10+18+40+35) = 103

Corrected Calculations

• % Non-recombinants = 785/1043 = 75.3• AB recombinants = 183/1043 = 17.5• BC recombinants = 286/1043 = 27.4• AC recombinants = 103/1043 = 9.9

B A C

17.5 mu 9.9 mu

27.4 mu

Corrected Map