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babymother [125]
3 years ago
10

Use the Lander and Waterman application of the Poisson distribution to calculate the number of reads that you need to sequence t

o obtain 20X coverage for a genome of size 5 Mb and Illumina reads of 100 bp. How does that compare with the number of reads that you would need with Sanger capillary sequencing (read length 800 bp)
Biology
1 answer:
miskamm [114]3 years ago
3 0

Answer:

120X reads

Explanation:

Data:

The number of the calculated gap of .2% is refered to the genome size in the question. For instance, if we want to sequence 4Mb, the total number of readings will be 800 000. Thus, for a total information, there will be 80 000 000 bp (80 Mbp) which will be 20 times the coverage of the genomes. In fact, there is an insignificant change in the distribution of the genome depending on the size of the reads. Thus, the total information. What is likely to be affected is the general assembly of the genome. A maximum of 25 X on average will yield the about 250 bp. However, for 100 bp, there might be a need of more than 120 X.

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What is the source of energy in a food web?
MatroZZZ [7]

Answer:

A

Explanation:

The sun is what drives most processes, the sun gives energy to the plants which help them grow resulting in creatures consuming the plants.  The sun is the major source so remeber that! :D                

I am sure this is the answer so I hope this helps you!

8 0
3 years ago
Alex recently quit smoking. Now he has trouble paying attention at work and forgets simple things. Which neurotransmitter is mos
joja [24]

Answer:

True

Explanation:

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Thus, true is the correct answer.

8 0
3 years ago
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nydimaria [60]
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5 0
2 years ago
In horses, the gene for black coat color is dominant over the gene for brown coat color. The gene for long manes is dominant ove
Gwar [14]

Answer:

3/16

Explanation:

Given,

Black coat color is dominant over brown coat color.

Long mane is dominant over short mane.

BBLL X bbll =

F1: BbLl = Entire progeny has black color and long mane

BbLl progeny will produce BL, Bl, bL and bl gametes. When they interbreed, the resultant F2 progeny will be in 9:3:3:1 dihybrid ratio. Hence in F2,

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3/16 progeny will have bbLl genotype hence 3/16 progeny will have brown, long mane.

4 0
3 years ago
In a cell with defective chaperones, Question 14 options: proteins would not be able to exit the ribosome. the concentration of
BartSMP [9]

Answer:

the concentration of misfolded proteins would be higher than normal.  

Explanation:

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5 0
3 years ago
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