Question

You are performing a three-point test cross in order to map the order and distance between three genes, designated X, Y, and Z, and you obtain the following results:

Phenotypes      Number Observed

+ Y +                          1

+ + Z                          72

X + Z                           2

X Y +                           70

X + +                          3

+ Y Z                         2

X Y Z                           0

+ + +                          0

Which gene is in the middle?

a) X

b) Y

c) Z

d) There is not enough data to make this determination.

Which is the coefficient of coincidence for this data set?

a) 0

b) 1

c) 8

d) 0.5

e) There is not enough data to make this determination.

Answer 1

1).

Middle gene= Z

2). a). 0

Explanation:

Hint: Always parental combinations are more numbered thant any other combinations. Hence, triple heterozygote would be

XY+/++Z

1). If single cross over occurs between x&y..

Normal = XY/++

After cross over = X+/+Y

X+= 2+3=5

+Y= 1+2=3

Total of this progeny = 8

Total progeny= 150

Recombination frequency = (no. of recombinants / total progeny)100

= (8/150)100

= 5.33%

2).

If single cross over between y&z..

Normal = Y+/+Z

After cross over = YZ/++

YZ = 2

++= 3

Total progeny = 5

RF = (5/150)100

= 3.33%

3).

If single cross over between x&z..

Normal= X+/+Z

After cross over = XZ/++

XZ = 2

++= 1

Total progeny = 3

RF =(3/150)100

= 2%

RF(%)= Distance between genes (m.u.)

X----2--Z---3.33---Y

Expected double cross over frequency = 2% * 3.33% = 0.000666

Observed double cross overs = XYZ &+++

Observed double cross over frequency = 0

Coefficient of coincidence =

Observed double cross over frequency / expected double cross over frequency

= 0/0.000666

=0