kroki 3 punktowej krzyz

An example of a 3-point cross problem Draw a map of these 3 genes (v, w, and z), showing the distances between all pairs of genes, and then calculate the value of interference. Data: counts of offspring from a test cross vwz 1 v + + 61 + w + 88 v + z 96 + + z 367 +++ 3 + w z 46 v w + 338

Step 1: arrange the data into reciprocal pairs vwz 1 +++ 3 v + + 61 + w z 46 + w + 88 v + z 96 + + z 367 v w + 338 total 1000

Step 2: Determine the parental types (the largest class), and the double crossover types (the smallest class). The other two classes are single crossovers. 2CO

vwz 1

2CO

+++ 3

1CO

v + + 61

1CO

+ w z 46

1CO

+ w + 88

1CO

v + z 96

Parental + + z 367 Parental v w + 338 .

total 1000

Step 3: Determine the order of the genes by comparing the parental types to the double crossovers. The parentals are + + z and v w +. The 2CO's are v w z and + + +. Gene z has switched sides between the parentals and the 2CO's, so z is in the middle. The map order is: v--z--w. Step 4. Determine all 3 map distances. The table below this section classifies each type of offspring as parental (P) or recombinant (R) for each pair of genes. A. v--z distance. Ignore w for now. In the parentals, v and z are on opposite chromosomes; that is, they are in repulsion. So, recombinant offspring, which come from a crossover between v and z, will be v z or + +. There are 1 + 3 + 88 + 96 = 188 offspring that have had a crossover between v and z. The map distance is the number of recombinant offspring divided by the total offspring, times 100. Here: (188/1000) * 100 = 18.8 map units between v and z. B. z--w distance The table below this section classifies each type of offspring as parental (P) or recombinant (R) for each pair of genes. Ignore v for now. In the parentals, z and w are on opposite chromosomes; that is, they are in repulsion. So, recombinant offspring, which come from a crossover between z and w, will be z w or + +. There are 1 + 3 + 61 + 46 = 111 offspring that have had a crossover between z and w. The map distance is the number of recombinant offspring divided by the total offspring, times 100. Here: (111/1000) * 100 = 11.1 map units between z and w. C. v--w distance. Ignore z for now. In the parentals, v and w are on the same chromosome; that is, they are in coupling. So, recombinant offspring, which come from a crossover between v and w, will be v + or + w. There are 61 + 46 + 88 + 96 = 291 offspring that have had a crossover between v and z. The map distance is the number of recombinant offspring divided by the total offspring, times 100. Here: (291/1000) * 100 = 29.1 map units between v and w. Map:

type

phenotype count v--z z--w v--w

2CO

vwz

1

R

R

P

2CO

+++

3

R

R

P

1CO

v++

61

P

R

R

1CO

+wz

46

P

R

R

1CO

+w+

88

R

P

R

1CO

v+z

96

R

P

R

Parental + + z

367

P

P

P

Parental v w +

338

P

P

P

total

1000

Step 5: Determine the interference value. Interference = 1 - (observed 2CO / expected 2CO) The observed double crossovers are found in the data: 3 + 1 = 4. The expected 2CO's are found from the equation: exp 2CO = (interval I distance/100) * (interval 2 distance/100) * (total offspring) Here, the interval I distance is the distance between v and z, 18.8 map units. The interval II distance is the distance between z and w, 11.1 map units. Total offspring counted is 1000. exp 2CO = (18.8/100) * (11.1/100) * 1000 = 20.9. Thus, if there were no interference, you would expect to have seen 20.9 double corssover offspring. Interference = 1 - (obs 2CO / exp 2CO) = 1 - (4/20.9) = 1 - 0.19 = 0.81. Step 6. You're done!