Your friend is taking a Genetics lab at UT. She needs some help with a lab report so she calls you for some advice—she knows you will be able to help since you are taking Genetics. After analyzing data from a three point cross she has lost her lab manual. The only thing she can find is that the Interference in this cross was 0.68. She remembers a testcross was done with a heterozygous female and that the genes being analyzed were forked bristles, crossveinless wings, and white eyes. She also tells you that there is an equal distance between these genes. What is this distance if she counted 3 double crossover progeny out of 500 total progeny?

The distance between these genes is 0.9375 MU

Explanation:

To calculate the coefficient of coincidence, CC, we must use the next formula:

CC= observed double recombinant frequency/expected double recombinant frequency

Note:  

The coefficient of interference, I, is complementary with CC.

I = 1 - CC

Available data:

•Interference = 0.68

•Equal distance between genes

•3 double crossover progeny

•Total number of individuals in the progeny, N = 500

We know that

CC= observed double recombinant frequency/expected double recombinant frequency

CC = (total number of observed double recombinant individuals/total number of individuals)/( recombination frequency in region I x recombination frequency in region II)

I = 1 - CC

So, all we need to do is to clear each formula with the data given in the problem:

I = 1 - CC

0.68 = 1 - CC

CC = 1 - 0.68  

CC = 0.32

CC = observed double recombinant frequency/expected double recombinant frequency

We already know the value of CC, and we know how to calculate the observed double recombinant frequency. But we need to calculate the expected double recombinant frequency. So,

0.32 = (3/500)/(RF-RI x RF-RII)

0.32 = 0.006 / (RF-RI x RF-RII)

(RF-RI x RF-RII) = 0.006/0.32 = 0.01875

We know that the expected double recombinant frequency is the recombination frequency in region I multiplied by recombination frequency in region II. In this case, it equals 0.01875

We also know that there are equal distances between genes, so now we need to divide 0.01875 by 2

0.01875 /2 = 0.009375  

The genetic distance will result from multiplying that frequency by 100 and expressing it in map units (MU) or centiMorgans (cM). Now we must multiply each recombination frequency by 100 to get the genetic distance

0.009375 x 100 = 0.9375 MU or cM

answer: oof wait

explanation:

the answers are as following:

1. c. double helix, nucleosome, coil, supercoil, chromosome

2. b. g2 phase, preparation for mitosis

3. d. interphase consists of the g1, s, and g2 phases

4. c. metaphase

5. a. chromosomes are duplicated before cell division so that each new daughter cell has a complete set