MBIO 1010 Lecture Notes - Lecture 2: Antonie Van Leeuwenhoek, Louis Pasteur, Phylogenetic Tree

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erinchimpanzee657

8 May 2017

School

University of Manitoba

Department

Microbiology

Course

MBIO 1010

Professor

Christopher Rathgeber

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Document Summary

Step 1: dna is collected from a pure culture . Step 2: the ssu rrna gene is amplified using the polymerase chain reaction (pcr) . Pcr a technique used to synthesize many identical copies of a short sequence of dna. Step 4: sequence is aligned with sequences from other organisms . Number of differences is used to calculate evolutionary distance. Phylogenetic tree a graphic representation of the evolutionary distance between organisms. The dna itself is going to be quite large, you want to home into the area you are interested in. How you do is that is a primer who is sequence specific for that rrna area. What happens, is you replicate just that segment. You have all of those copies and put to sequencing techniques so that we can know the nucleotide sequence. Look for the highest degree of similarity to find relatedness. Phylogenetic tree based on 16s or 18s ribosomal dna sequences .

Related Questions

Please answer all

Why does a template-dependent DNA polymerase require a primer to initiate DNA synthesis?

	DNA polymerase require 5' phosphate to add a new nucleotide
	DNA polymerase require 3' hydroxyl group to add a new nucleotide

DNA polymerase require energy by hydrolizing the primers

Which of the following enzymes is template independent?

	Reverse transcriptase
	Klenow fragment

	DNA polymerase
	Terminal deoxynucleotidyl transferase

RNA polymerase

The natural function of a DNA ligase enzyme is:

	Joining together DNA sequences with compatible cohesive ends from two different organisms to yield a recombinant DNA molecule
	Sealing single stranded nicks in double stranded DNA molecules

	Joining together two blunt ended fragments of DNA within a cell
	Joining together two single stranded DNA molecules

The natural function of 3' -> 5' exonuclease activity of a DNA polymerase is to:

	Remove the 5' end of the DNA strand that is being copied
	Remove damaged nucleotide from the template strand

	Remove nucleotides from the end of DNA to generate blunt ends
	Remove incorrect nucleotides from the newly synthesized DNA

Which technique is used to resolve the different sizes of DNA fragments?

	Gene cloning
	PCR

	DNA sequencing
	Gel electrophoresis

Which of the following vectors does not contain bacterial origin of replication (oriC)?

	plasmid
	cosmid

	bacterial artificial chromosome
	lambda vector

Which process of bacteriophage is not used for gene cloning?

	Concatemer formation
	Lytic cycle

	Lysogenic cycle
	viral packaging

What vector would be best suited for creating a contig of bovine (cattle) chromosome 10?

	lambda vector
	Cosmid vector

	P1 vector
	Plasmid

E. coli takes up plasmid DNA by which of the following methods?

	Transduction
	Transformation

	Conjugation
	Transfection

The following times and temperatures are an example of the steps for PCR. You can use the Figure to help you answer the following questions.

Why is the first step is carried out at 94Â°C?

	To degrade the template
	To denature the template

	To activate the polymerase
	To facilitate the primer binding

What happens in the reaction when the temperature shifts to 55Â°C during cycling?

	the DNA polymerase will carry out DNA synthesis by extending the annealed primers
	the DNA polymerase will finish DNA synthesis

	the primers anneal to the single-stranded regions of the DNA
	the primers anneal to the double-stranded regions of the DNA

During cycling, what occurs when the temperature is at 72Â°C?

	the primers anneal to the double-stranded regions of the DNA
	the DNA polymerase will carry out DNA synthesis by extending the annealed primers

	the primers anneal to the single-stranded regions of the DNA
	the DNA polymerase will finish DNA synthesis

You designed a set of primers to amplify 1 kb DNA with polymerase chain reaction. By which cycle of polymerase chain reaction, we would expect to see the first double stranded DNA with expected size?

	The end of cycle 1
	The end of cycle 2

	The end of cycle3
	The end of cycle 4

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.

1. Genomic DNA is digested with a restriction enzyme that has a 4 bp recognition (restriction) site. What would be the expected average size of the fragments that would be produced by digesting DNA of random sequence with this enzyme? A) 8 bp B) 16 bp C) 125 bp D) 256 bp E) 512 bp

2.insertion or removal of one or more nucleotide base pairs in DNA within a gene often results in a ____________ mutation. A) transition B) frameshift C) reversion D) transversion E) suppressor

3. Which ingredient is NOT required in Sanger DNA sequencing? A) Primers for DNA polymerase. B) Reverse transcriptase. C) Dideoxy nucleotides. D) Deoxy nucleotides. E) DNA template.

4. What is the purpose of Taq polymerase in a PCR reaction? a. DNA denaturation b. Primer annealing c. DNA synthesis d. Heating of the reaction e. All of the above are associated with Taq polymerase.

5.You are doing lab work with a new species of beetle. You have isolated lines that breed true for either blue shells and long antenna, or green shells and short antenna. Crossing these lines yields F1 progeny with blue shells and long antenna. Crossing F1 progeny with beetles that have green shells and short antenna yield the following progeny: Blue shell, long antenna 92 Green shell, short antenna 88 Blue shell, short antenna 31 Green shell, long antenna 29 Total 240 Assuming that the genes are linked, what is the map distance between them in cM? A) 75.0 cM B) 25.0 cM C) 12.1 cM D) 12.9 cM E) The genes are actually assorting independently. Genes A and B are linked with 10 map units (cM) between them. A heterozygous individual AB/ab (note that the dominants are in coupling) would be expected to produce gametes in which of the following frequencies?

A) 45% AA; 10% Ab; 45% bb

B) 5% AB; 45% Ab; 45% aB; 5% ab

C) 10% AB; 40% Ab; 40% aB; 10% ab

D) 45% AB; 10% Ab; 10% aB; 45% ab

E) 45% AB; 5% Ab; 5% aB; 45% ab

MBIO 1010 Lecture Notes - Lecture 2: Antonie Van Leeuwenhoek, Louis Pasteur, Phylogenetic Tree

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