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Gene Technology

Genetics as it applies to evolution, molecular biology, and medical aspects.

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Gene Technology

Postby jsmith613 » Mon Apr 09, 2012 12:24 pm

I have attached a question from a past paper that is confusing me.
The answer on the MS is

(white)
blue
lilac
white

I have absolutely no idea why the second and third are correct. I have presumed that with no enzyme there will be no colour change so the petals will remain white (as per the mark scheme)
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Re: Gene Technology

Postby jonmoulton » Tue Apr 10, 2012 3:57 pm

You are correct that if no enzyme is present, the petals will remain white. The interesting part of this question is that there are two enzymes, each catalyzing a step in the biosynthetic pathway for the blue pigment, and that the biosynthetic pathway has a different-colored (lilac) intermediate.

The electrophoresis pattern shown in the top row reveals a single enzyme present and corresponds to white petals observed. If the band was from enzyme 1, the enzyme would have catalyzed the transformation of the uncolored pigment precursor to its lilac form. However, the petals remained white, so we can conclude that the band reveals enzyme 2 and enzyme 1 is missing; if enzyme 1 were present it would have made the lilac pigment from the uncolored precursor but, without enzyme 1 or the lilac pigment present, enzyme 2 can be present in the cell but cannot produce the blue pigment because its substrate (lilac pigment) is not present.

In the second row, both enzymes are present (both bands appeared in the electrophoretic lane). This means enzyme 1 is present and can catalyze the formation of the lilac pigment from its uncolored precursor. Enzyme 2 is also present, so the lilac pigment is transformed into the blue pigment and the petals appear blue.

In the third row, only enzyme 1 is present. It catalyzes the transformation of the uncolored precursor into the lilac pigment. Because enzyme 2 is not present to use up the lilac pigment, that pigment accumulates and gives the petals their lilac color.

In the fourth row, with both enzymes missing from the electrophoretic lane, the uncolored precursor is not changed into colored forms and the petals remain white.

This gives rise to the colors in the answers you reported:
R1 (white)
R2 blue
R3 lilac
R4 white
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