Week 2

Question 1

Information

Answer 1

the letters used to write the sequences

Genomes in a double or single stranded states

Question 2

What is information used for?

Answer 2

we use information in DNA seqeunce to align sequences

Question 3

Contig

Answer 3

Summary sequence of overlapping DNA fragments

Question 4

Why can we align DNA sequences

Answer 4

How long does a sequence have to be for us to expect it to be unique?

Information theory and Genetics

Question 5

The restriction enzyme HaeIII cuts the sequence GGCC. What is the probability that any random four base pair sequence is a HaeIII site?

Answer 5

1/4 chance of getting either base

1/4 x 1/4 x 1/4 x 1/4

(1/4)^n

n=length of the genome

Question 6

How many distinct permutations in sequence can a stretch of n bases long have?

Answer 6

How many sequences (permutations) are possible for a giving length of a DNA strand (n-bases long) if the bases occur at an equal frequency in the genome
, how many permutations?

how many peices of distinct information can a stretch of n bases have (potential)

4^n

Question 7

We can expect a sequence to be unique in a genome when the # of permutations is greater than the # of sequences of n bases in the genome

Answer 7

Bacterial genome = 4x10^6 bases

Human genome = 3x10^9 base (single stranded)

What is the # of sequences of n bases in the genome?

Question 8

What is the number of seqeunces of n bases in the genome?

Answer 8

because genomes are so long the number of sites = the length of the genome if the genome is single stranded

the number of sites in double stranded DNA = 2x the length of the genome

Palindromic sequences are an exception.

Question 9

We can expect a sequence to be unique in a genome when the # of permutations is greater than the number of sequences of n bases in the genome.

Answer 9

when is 4^n > 8x10^6 (bacteria)

n>=12 in bacteria is unique

when is 4^n > 6x10^9 (humans)

n>=17 in humans is unique

Question 10

Say I have 9 genomes made of random sequence of 3 X 109 bases. And I search each of these with ATAGACATAGGATACAT. How many would you expect to have the sequence?

Answer 10

6 X 10^9 / 17 X10^9 = 1/3
Therefore, 1/3 X 9 = 3.
I expect about 3 of these random genomes to have this sequence

Question 11

How much information is there in DNA?

Answer 11

A,T,G,C each letter can arbitrarily represented as:

A=00
G=01
C=10
T=11

the information content at any position in a DNA sequence is 2 bits

smallest unit of information is a bit

The total bits of a sequence n is the addition of the bits at each position.

Question 12

Information theory

Answer 12

trying to pick out an identified object

we have eight objects each of the objects are assigned a binary number, each object is identified by a distinct binary number

when do i have sufficient information to identify the object

if you only have one bit of information when you search through the object you are going to identify half

Question 13

Minimum amount of information

Answer 13

Imin = log2 N

I>= Imin to be sufficient information

N is the number of objects/length of the genome

12 base sequence = 24 bits > log2 8 X 10^6 = 22.9

17 base sequence = 34 bits > log2 6 X 10^9 = 32.5

Question 14

BLAST Basic local alignment search tool

Answer 14

Genbank has about 10^13 bases of sequence in the database. Therefore, Imin= 44 and a sequence of 23 bases or more could be expected to be unique, more often than not, in the database.

Blast takes your query and breaks it into short words (5 sequence words), they take the database and break into words as well (5 base word) these words are indexed with position information.

If so the blast program creates an aligment to align the query to the sequences in the database

Question 15

Two statistics of a BLAST

Answer 15

Score: higher the score the better.

e value: lower the e value the better.

e value is telling you what the random chance of your alignment occurring when searching the database. Low e value low chance.

Question 16

Cost of DNA sequencing

Answer 16

Has fallen over the past 20years

large change occurs after massive parallel sequencing is introduced

Question 17

Sanger

Answer 17

reads per/run: 1

Length of read: 1,000 bases

basic method: termination

expensive/accurate

Question 18

Illumina

Answer 18

length of read: 100 bases
# of reads/per run: 10^9-10^10
basic method: real time incorporation
cheap/accurate

Question 19

Pacific bioscience

Answer 19

length of read: 15,000 bases

reads/run: 10^6

real time incorporation

cheap/low accuracy

Question 20

nanopore

Answer 20

read length: 100,000
#reads per run: 10^6
current changes
cheap/low accuracy

Question 21

Real time incorporation: Illumina

Answer 21

A series of rounds of synthesis, the nucleotides are modified to have a fluorescent group and a blocked 3’OH

after cycle one the 3’OH and the fluorescent group is removed

the cycle is repeated, continue on and on and take successive pictures to determine the sequence of DNA

Question 22

Real time incorporation: Pacbio

Answer 22

the cell is set up so you can see the DNA polymerase, and DNA polymerase incorporated labeled dNTPs, the label is at the very end of the phosphate, what you observe is the read output, labeled dNTPs are incorporated in the chain, this fluorescence is detected, and diffuses away shortly after incorporation

Question 23

Nanopore

Answer 23

the sequence that is in the pore alters the amount of current that can flow through the nanopore in a specific manner depending on the sequence.

as each base goes through we see a current change

a set of bases is read and is being read, the current we’re looking at is the G coming out naf the T going in

Question 24

Illumina basic steps

Answer 24

DNA molecule has adaptors added.

Sticks to oligos on a glass slide.

Amplified to make a microdot on the slide.

First sequence read.

Reorientation of the DNA molecule.

Second paired sequence read.

Bridge amplification to close fragments

Question 25

Pacific Biosciences sequencing video

Answer 25

DNA molecule has adaptors added. DNA isolated Label adpators to DNA creating a circular template, primer and polymerase is added to the library DNA and DNA polymerase in a SMRT cell. Smart cell contains Zero-Mode Waveguides (ZMWs) A single molecule of DNA is immobilized at the bottom of the ZMW (smallest detection unit) As polymerase incorporates nucleotides, light is emitted (after which the fluorescent tag is cleaved) Nucleotide light emission is used to measure incroporation As each ZMW is illuminated from below the wavelength of the light is too large to allow it o pass through the waveguide Attenuated light form the excitation beam penetrates the lower 20-30 nm of each ZMW DNA sequence read. Circular concensus sequencing mode to produce highly accurate long reads (Hifi) Use continuous long read sequencing mode to generate long as possible reads

Question 26

Nanopore sequencing video

Answer 26

DNA molecule has adaptors added. Motor molecule binds Adaptors are important for the binding of a motor molecule, the DNA is released over membraens tha thave a single pore over them. The DNA molecules guided to the nanopore with the help of tethers that will grab onto the motor and help position it over nanopore. Sequencing directly enables the detection of base modifications and methylation Motor protein recognizes adaptor sequence bound to DNA and guides DNA to the nanopore Tether guides DNA through the nanopore DNA passes through causing a disruption of ionic current measured in signal trace. Intact DNA sequences are sequenced in real time by the nanopore, regardless of their length (no runtime, fragments) DNA enzyme complexes approach the nanopore and the single stranded DNA is pulled through aperture of the nanopore, the enzyme ratchets the the DNA through the nanopore one nucleotide at a time. The enzyme bands to a single stranded leader at the end of a single strand, unzipping the double stranded DNA. Speed of the enzyme can be controlled. Hair pin structure ensures both strands of DNA can be read in one read. A nanopore

Question 27

Contig

Answer 27

Assembling a summary sequence from fragments of DNA

Question 28

Genome sequecing starts with

Answer 28

All genome sequencing starts with random fragmentation. In the tube you have a random set of fragments from different genomes of different sizes, starting at different positions. If you pick enough DNA to sequence such that you would have sequenced six random sequences of 11 bases in length, every nucleotide once. Picking them randomly is not a complete genome sequence

Question 29

Coverage

Answer 29

how many times has a base been sequenced = bases sequenced/length of genome.

Question 30

Why do we have gaps?

Answer 30

This is called the zero class the proportion that will be zero = e^-m where m is the expected mean = # of times you seqeunced the genome e^-m X total length of the genome (ss) Repetitive sequence also creates gaps in assembly. High information content is necessary to create a sequence

Question 31

Scaffold

Answer 31

A scaffold is an ordered array of contigs with gaps remaining.

Question 32

How do we generate scaffolds?

Answer 32

Paired end reads are used to create a scaffold. Sanger and Illumina DNA fragment of some length, what we can determine a read of DNA sequence at one and and a read at the other end. The reverse complement. Distance between them. These two sequences are linked by a defined distance

Question 33

Is the human genome completely sequenced?

Answer 33

8% is missing sequences around the centromere sequences of tandem rDNA (rRNA genes) repeats Sequences of areas with repeats Other genomes may remain incompletely sequenced due to the lack of interest/funds.

Question 34

Reference genomes

Answer 34

Drosophila: genome determined from a specific fly stock. Human: Was supposed to be determined from a number of individuals but ended up with most from a single white male. References genome allows rapid assembly of genome sequence for a single individual. So its easy once you have a reference genome just to align your sequence reads with the reference genome NO. Comparison with other Human genomes a large amount of sequence is missing. A more sophisticated Human reference needs to be developed.