1) DNA from a lac + pur + arg + strain is used to transform a lac - pur - arg -
ID: 61830 • Letter: 1
Question
1) DNA from a lac+pur+arg+ strain is used to transform a lac-pur-arg- bacterial strain. Given the cotransformation data below, determine the gene order.
Exp.1: lac+ pur+, frequency = 0.70
Exp.2: lac+ arg+, frequency = 0.32
Exp.3: pur+ arg+, frequency = 0.11
(I have no clue how to do this so explanation would be appreciated).
gene order answer choices:
a) arg pur lac
b) pur lac arg
c) lac pur arg
d) pur arg lac
2) Along a 15000-bp segment of DNA, approximately how many molecules of H3 protein do you expect to find? Assume there are 150bp between each nucleosome.
answer choices:
a) 1
b) 25
c) 50
d) 100
e) 200
Explanation / Answer
1. (b)
Co-transformation frequency tells us how closely the genes are linked or are far apart. High frequency indicates that genes are closely linked or vice-versa.
If we look closely, lac+ pur+ has highest frequency and pur+ arg+ has lowest frequency.
Thus, we have following gene order: pur+ lac+ arg+
2) (d) 100
The nucleosome core particle consists of approximately 147 base pairs of DNA and a linker
DNA between two nucleosomes. DNA is wrapped in superhelical turns around a histone
octamer consisting of 2 copies each of the core histones H2A, H2B, H3, and H4.
We have a 15000-bp segment of DNA
Linker distance = 150 b.p
DNA wrapped around histone octomer = 147 b.p
So, between 2 nucleosomes we have total DNA = 297 b.p
Dividing total length by total DNA between two nucleosomes we get almost 50 nucleosomes.
We know each nucleosome has two H3 proteins and thus we get total of 100 H3 protein molecules.
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