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WOLFYy WOLFYy
wrote...
Posts: 1
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4 years ago
Part A
Determine the appearance for northern blots of the bacteria listed below. In each case, lane 1 is for mRNA isolated after growth in a glucose-containing (minimal) medium, and lane 2 is for mRNA isolated after growth in a lactose-only medium.
a. lac+ bacteria with the genotype I+P+OCZ+Y+
b. lac- bacteria with the genotype I+P+O+Z−Y+
c. lac- bacteria with the genotype I+P−OCZ+Y+
d. lac+ bacteria with the genotype I−P+OCZ+Y+
e. lac- bacteria with the genotype I+P+OCZ−Y−
f. lac- bacteria with the genotype I+P+O+Z+Y+ and a mutation that prevents CAP-cAMP binding to the CAP site
Drag the appropriate labels to their respective targets. Use black mark for full expression, white mark for low expression, and leave the target empty in case of no expression. Note: Not all targets will be used.
Mine says 2 incorrect. Please give me true answers! Thank you!! Slight Smile
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Answer rejected by topic starter
wrote...
Educator
4 years ago
Hi WOLFYy

The answer is attached below
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wrote...
3 years ago
A. Here, the Oc represents constitutive mutation in operator. Operator is always active and transcription takes place in the presence or in absence of lactose. So you will always get expression of Y. But the expression will not increase in the presence of lactose. Hence we get faint bands in the presence or in absence of lactose.

B. Inducible promoter, Y+ is present and hence a dark band in presence of lactose and no band in absence of lactose.

C. Promoter is mutated so polymerase cannot bind, no expression of genes. No bands I both conditions.

D. I- so constitutive expression, does not respond to lactose presence or absence. Also Oc mutation, so gene expression occurs all the time. light bands in both conditions.

E. No Y+ gene, so no bands in either condition.

F. CAP binding is essential for polymerase binding to promoter. So no expression. No bands in the absence of lactose. But in presence of lactose, faint expression takes place.
wrote...
2 years ago
thank you
wrote...
2 years ago
thank you
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2 years ago
Thank you
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2 years ago
thank you
wrote...
2 years ago
Thank you!
wrote...
2 years ago
Thank you
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2 years ago
Thank you!
wrote...
Educator
A week ago
Try this combination:

A. Light and no band
B. No band and dark
C. No bands for either
D. Light and dark band
E. No bands for either
F. No and light band
Answer verified by a subject expert
lambchop03lambchop03
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Posts: 435
A week ago
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In Part a, the Oc operator prevents repressor binding. Thus, some leaky expression occurs on Minimal media, while full expression occurs when Lactose is added.

In Part b, even though LacZ is mutated, LacY is present and would be transcribed, giving a signal in the presence of Lactose.

In Part c, the Promoter is mutated, preventing any expression.

In Part d, the Oc operator and mutated inducer allow for leaky expression in Minimal Media, while giving full expression in the presence of Lactose.

In Part e, since LacY is absent/mutated, the probe may not bind to the LacY mRNA, giving no expression regardless of the presence/absence of Lactose.

In Part f, a mutation prevents CAP-cAMP binding to the CAP site. Thus in the presence of Lactose, repression will be released but only little expression will occur.
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