You are a scientist who is using genomics to currently study a new bacterial species that no one has ever studied before. The following sequence is a piece of DNA within the coding region of a gene that you have recently sequenced.

 

You are using shotgun sequencing to determine the DNA sequence of the genome of this new bacterial species. For one strand of a 30-nucleotide long stretch of DNA, you get the following sequences out of your shotgun sequencing reaction. Assemble the entire 30-nt-long DNA sequence

 

5’-TGGGAGTTCCTCAAACGCGTTGTCACTGAC-3’

You put the DNA sequence that you have assembled into a computer program that tells you that the following piece of DNA, which comes from another bacterium, is a close match to the sequence you have sequenced from your bacterium: 5’-…TGGGCATTTCTCAAGCGGGTTGTAATGGAT…-3’

This 30-nt-long sequence fragment lies in the center of a gene, and that portion of the sequence encodes for this 10-amino acid-long part of a protein:

N-…Trp-Ala-Phe-Leu-Lys-Arg-Val-Val-Met-Asp…-C

You hypothesize that the sequence you have discovered is another bacterial species’ version of the same gene as this previously known gene. To measure how identical the two genes are at the DNA level and/or the two proteins are at the amino acid level, you can calculate a percentage of “identity” for each. This is the percent of nucleotides (for the gene) or the percent of amino acids (for the protein) that are identical between the two sequences.

What is the % identity between the two protein sequences?

#Question id: 9582

#Unit 6. System Physiology – Plant

Electrons ejected from chlorophyll travel through a series of electron carriers organized in the “Z Scheme” shows a current version of the Z scheme, in which all the electron carriers known to function in electron flow from H2O to NADP Presented in the following sequence;

a.) P680 reduce by Yz has received electrons from oxidation of water P680→P680*→pheophytin→ plastoquinones→ cytochrome b6 f complex→ plastocyanin→ in turn reduces (PSI) P700+

b.) P700+→ P700*→ membrane-bound iron–sulfur proteins → A0→ A1 → flavoprotein ferredoxin–NADP reductase (FNR) soluble → ferredoxin(Fd) → NADP+ to NADPH

c.) P700+→ P700*→ A0→ A1→ membrane-bound iron–sulfur proteins→ ferredoxin(Fd)→ soluble flavoprotein ferredoxin–NADP reductase (FNR)→ NADP+ to NADPH

d.) P680 reduce by Yz has received electrons from oxidation of water P680→P680*→ plastoquinones → pheophytin → cytochrome b6 f complex→ plastocyanin→ in turn reduces (PSI) P700+

Which one of the following combinations is correct?

#Question id: 10462

#Unit 6. System Physiology – Plant

During an experiments mutating the conserved cysteine in the LOV1 domain of phot1 does not affect phototropic responsiveness (the seedlings bend toward blue light) whereas the equivalent mutation in LOV2 abolishes the response. What will be the conclusion has been drawn from this experiments?

a)  LOV1 domain is secondarily responsible (primarily responsible is LOV2 domain) for kinase activation and LOV1 is not is coupled with the protein region Jα-helix that’s why it does not affect phototropic responsiveness

b) These studies have demonstrated that LOV2 domain is primarily responsible for kinase activation in response to blue light

c) This is due in part to the position of LOV2 within the phototropin molecule, where it is coupled to a protein region known as the Jα-helix

d) The function of LOV2 domain is thought to play a role in receptor dimerization by accomplishing the kinase activation in response to blue light

Which of the correct conclusion drawn from the above experiment?

#Question id: 31129

#Unit 2. Cellular Organization

#Question id: 12272

#Unit 7. System Physiology – Animal

#Question id: 4349

#Unit 3. Fundamental Processes

RNA Pol IV and Pol V