MGY377H1 Lecture Notes - Lecture 14: Histidine, Membrane Protein, Repressor

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burgundymule401

1 Feb 2018

School

UTSG

Department

Molecular Genetics and Microbiology

Course

MGY377H1

Professor

William Navarre

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Document Summary

When it binds lactose, it falls off of the dna and rna polymerase then binds, and it creates proteins that can break down lactose allowing the cellll to use lactose as an energy source to produce glucose. If this enzyme is missing in ecoli growing on lactose, the ecoli wont grow. When the repressor loses its lactose, the represses becomes a dna binding protein and binds to the dna. Low ph or lack of oxygen are called 2 component systems because they have 2 components involved. Complex bacteria have lots of 2 component systems. In eukaryotic cells theres many systems that are multiplied and they are 2 component systems. Hk are signaling domains which are anchored in a membrane, they have a sensor kinase domain on the outside and a domain on the inside that gets phosphorylated, and this prohphorylation gets relayed to another protein. The response regulaor gets phosphorylated and it can then bind to cause gene expression.

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In the following points, I have summarized Lac operon in Ecoli. I need to check if this is on the right track? and where does 'cleavage of disaccharides' fit into this picture?

Lactose is a disaccharide and the bacteria need specific genes to manage catabolism

1) The lacl promotor gene synthesis the repressor protein which binds to the operator gene stopping the RNA polymerase from coding the lac operon genes

2)But when lactose is present, it needs the specific genes to catabolise lactose

3) So, When allolactose (a derivative of lactose)is present, it binds to the repressor protein, the repressor releases from the operator gene

4) This allows the RNA polymerase can attach to the DNA control region

5) And drive down the operon for transcription and induction of the enzyme lacZ(beta-galactosidase) lacY and lacA which breaks lactose down into glucose and galactose

6) When the lactose is all gone, the repressor returns to the operator in a feedback loop system

QUESTION 10

If you can drink milk as an adult, it means that you have inherited a mutation in the promoter of your lactase gene (the gene that encodes the enzyme you need to break down lactose). Predict the effect of this mutation:

		The mutation changes the number of domains in the enzyme, which makes it work more efficiently
		The mutation changes the amino acid sequence of the lactase protein
		The mutation increases the number of copies of the lactase gene that will be found in your genome
		The mutation changes whether the lactase sequence is found in an intron or exon
		The mutation affects the expression of the lactase gene

1.2 points

QUESTION 11

A competitive inhibitor is decreasing the activity of an enzyme. Predict the effect of adding more substrate to the reaction.

		The substrate will increase the reaction rate by binding to the allosteric site
		The substrate will increase the reaction rate by competing with the inhibitor for the active site
		The reaction rate will not change unless the inhibitor can be removed
		The enzyme adjusts its shape so that the substrate, but NOT the competitive inhibitor, can bind
		The substrate will bind to the competitive inhibitor and block its ability to bind to the enzyme

1.2 points

QUESTION 12

What determines where in the genome a transcription regulator will bind?

		Transcription regulators bind to the 5' UTR region of a gene
		Regulators bind via complementary base-pairing to certain DNA molecules
		Covalent bonds form between the transcription regulator and the atoms of the DNA backbone
		Every eukaryotic gene has a different transcription regulator that will bind to the 5' end of the gene
		Transcription regulators bind to specific DNA sequences via multiple weak non-covalent interactions

1.2 points

QUESTION 13

What is the basic premise of cell theory?

		DNA -> RNA -> protein
		All cells arise from pre-existing cells
		DNA provides the complete instructions to create a cell
		The identity of a cell is determined through gene expression patterns
		All cells contain the same four basic macromolecules

1.2 points

QUESTION 14

What is the benefit of using BOTH the lac activator and the lac repressor to control gene expression?

		Using both an activator and repressor enables cells to more accurately determine the amount of lactose available in the environment
		Enzymes to digest lactose are only made when energy is low and lactose is available
		The activator can override the inhibition of the lac operon by the repressor
		The repressor can control the enhancer, while the activator can control the promoter
		When neither the lac activator or repressor is present, expression of the lac operon is too high

1.2 points

QUESTION 15

What is the histone code used for?

		Phosphorylation and acetylation of DNA affect its ability to be compacted
		Changes to the sequence of DNA change whether DNA will wrap around histone proteins
		Covalent modifications of histones affect the ability of the transcription initiation complex to form
		Histones provide the codon sequences needed for translation to occur
		The histone code affects which amino acids will get added to a polypeptide

a) You also wish to synthesize a degenerate oligonucleotide primer that could be used eventually to clone DNA that encodes your mutant protein. Design this degenerate oligo, keeping in mind the rules we talked about in class. More information about designing degenerate primers and primers in general are located on the following page.

Write out the sequence of this oligo here

b) If you synthesized the above oligonucleotide primer, how many different oligos would be present in the mixture?

RULES TO FOLLOW for creating primers (Modified from the Thermo Fisher web site):

Designing degenerate primers:

Write out your amino acid sequence, label amino and carboxy termini

For each amino acid, use the genetic code to predict the various nucleic acid codons that can encode this amino acid. For all but Met, you should have redundancy present.

Now you need think about 5â and 3â considerations, and keep in mind that DNA within genes is double-stranded. It is helpful to keep in mind that the 5â end of a gene corresponds to the amino terminus of a protein. Using 5â and 3â labels write out a single strand DNA sequence corresponding to the your amino seq of interest. You may choose to start with only possible codon for each amino acid, or you can do all possibilities using the notation I showed you in class to account for the wobble position of codons.

Now make your DNA double stranded by filling in the opposite DNA strand; label the 5â and 3â ends of this second strand and make sure you have an anti-parallel arrangement of DNA strands.

If you havenât already, now consider the redundancy present on both strands, and make sure you have indicated sequences that account for all possible redundancies.

Realize that when a researched synthesizes a degenerate oligo that the final synthesis contains oligos with every substitution possible. The amount of degeneracy is defined by the number of different primer combinations in the mix. You can determine the degree of degeneracy by multiplying the number of changes present at each position together. If a position has no degeneracy then you multiple by 1 for that position.

Keep in mind that the trade-off between primer specificity and efficiency can be modified by altering the degeneracy of the primer. For example, the more degenerate the primers, the less specific annealing will be; however, decreased degeneracy will allow more potential to identify unknown variants. Anything over 1024 fold degeneracy is at the upper limit of potential usefulness.

Designing Primers:

Good primer design is essential for a successful PCR reaction. There are many factors to take into account when designing the optimal primers for your gene of interest. This information is taken from the Fisher Scientific website and are used when actually designing an experiment. Here are some tips to consider when designing primers.

In general, a length of 18 nucleotides is the minimum necessary for efficient primer binding.

Try to make the melting temperature (Tmm) of the primers between 65Â°C and 75Â°C, and within 5Â°C of each other.

If the Tmm of your primer is very low, try to find a sequence with more GC content, or extend the length of the primer a little.

Aim for the GC content to be between 40 and 60%, with the 3' of a primer ending in C or G to promote binding. This is called a GC clamp.

Typically, 3 to 4 nucleotides are added 5â of the restriction enzyme site in the primer to allow for efficient cutting. Restriction enzymes are ENDO nucleases, so they cut better when their recognition site is not on the end of DNA.

Try to avoid regions of secondary structure, and have a balanced distribution of GC-rich and AT-rich domains.

Avoid intra-primer homology (more than 3 bases that complement within the primer) or inter-primer homology (forward and reverse primers having complementary sequences). These circumstances can lead to self-dimers or primer-dimers instead of annealing to the desired DNA sequences.

MGY377H1 Lecture Notes - Lecture 14: Histidine, Membrane Protein, Repressor

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