1. [2pt] In E. coli, four Hfr strains donate the following markers in the order
ID: 13892 • Letter: 1
Question
1. [2pt] In E. coli, four Hfr strains donate the following markers
in the order shown with the ?rst marker transferred on the left.
All these Hfr strains are derived from the same F+ strain. What
is the order of these markers on the circular chromosome of the
original F+ strain? Give the order of markers starting with
the ?rst marker transferred by strain 4. Enter the sequence of
markers without spaces or punctuation (e.g. QRSTUVWXYZ).
Strain 1: MZXWC
Strain 2: LANCW
Strain 3: ALBRU
Strain 4: ZMURB
I honestly have been staring at this question for the last three days and honestly have no idea what to make of it...
Explanation / Answer
ZMURBLANCWXZM.
See, it is rather simple. You have to see the four sequences given and match them, remembering that they are present on a circular chromosome(plasmid).
between strain 1 and 2, W and C matches. but they come in reverse order, i.e in strain 1, it is W followed by C and in strain 2, it is C followed by W. So the sequence in strain 2 has to be reversed.
Combining the sequences of strain 1 and 2, we get MZXWCNAL(reverse LANCW---> WCNAL, join the NAL part to sequence of 1.).Let us call this combo.A.
between combo.A and sequence of strain 3, we have AL common. So join the remaining part to the end of combo.A. We get MZXWCNALBRU. Let this be combo.2
between combo.2 and strain 4, we have BRU common, but in reverse. So doing as before, we get the sequence as: MZXWCNALBRUMZ.(combo.3)remember this is on a circle!
The question now gives that the transfer starts from sequence of strain 4, which is ZMURB. So cosidering a circular chromosome, and knowing that the start of the sequence is ZMURB, we can reverse combo.3 and get the FINAL sequence:
ZMURBLANCWXZM.
Hope this helps! :)
P.S.- don't get scared by the long explanation. If u understand it once, it's rather simple! :) :)
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