Question

Which of the following are not mechanisms that regulate transcription?

How/Why is E the answer??

A) The stimulation of a signal transdusction pathway by a hormone that ultimately causes

the activation of very specific transcription factors.

B) The formation of a protein bridge between a silencer and the promoter complex.

C) The formation of a protein bridge between an enhancer and the promoter complex.

D) The availability of transcription factors that bind to the promoter.

E) The addition of the poly-A tail to the 3′ end of the pre-mRNA

You performs a cross between a trihybrid with a tester. The three genes are linked together. The

recombination frequencies between them are 0.1 (10%) and 0.2 (20%). When counting the 1000

offspring, you find 14 double recombinants. What is the interference?

Assume that a cross is made between AaBb and aabb plants and that the offspring occur in the

following numbers: 148 AaBb, 51 Aabb, 49 aaBb, 152 aabb. These results are consistent with

________.

28) _____

A) linkage with 50% crossing over

B) 100% recombination

C) linkage with approximately 33 map units between the two genes

D) linkage with approximately 25 map units between the two genes

E) independent assortment

PLEASE EXPLAIN the WHY??

Answer 1

Answer:

1). E) The addition of the poly-A tail to the 3′ end of the pre-mRNA

Explanation:

Addition of the poly-A tail to eh 3’ end of the pre-mRNA is the post transcriptional modification. It does not occur during transcription.

2). Interference = 0.3

Explanation:

Recombination frequency (%) = Distance between the genes (mu)

a-------10mu------b-----------20mu------c

Expected double crossover frequency = (RF between a & b) * (RF between b & Sc)

= 10% * 20% = 0.02

The observed double crossover frequency = 14 / 1000= 0.014

Coefficient of Coincidence (COC) = Observed double crossover frequency / Expected double crossover frequency

= 0.014 / 0.02

= 0.7

Interference = 1-COC

= 1-0.7 = 0.3

3). D) linkage with approximately 25 map units between the two genes

Explanation:

Hint: Always recombinant progeny are smaller number when the genes are linked.

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

RF = (100/400)100

= 25%

Recombination frequency (%) = Distance between the genes (mu)

So distance between two genes = 25 mu