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Why is the 4.4 kb band barely visible?

Also, what is the mass (ng) of DNA in band X if lane S contains 250 ng of digested lambda phage DNA?

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This question is literally impossible to answer with the information given. What samples are run in each lane? How were they prepared? I assume the left lane is a ladder. What ladder is it? What sample are you adding the HindIII-restricted phage DNA to? –  A. Kennard Nov 11 '13 at 0:25

1 Answer 1

First of all, the 4.4 kb HindIII band is derived from the right end of λ DNA, so during storage it anneals to the 23.1 kb left arm fragment via λ's 12 bp cohesive ends. You can solve this problem by preheating the markers to 65°C for a few minutes then cooling rapidly (i.e. on ice).

In a stained gel the intensity of a band is proportional to the amount of DNA in that band. This is why the heavier bands are more intense. The fragment X in lane 2 is similar in intensity to the 0.5 kb λ fragment so DNA content of the two bands is similar.

The amount of DNA in the 0.5 kb band is 250 ng *(0.5/48.5) where 48.5 is the size of wild-type λ DNA. I'll leave it for you to complete the calculation.

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