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Mumz [18]
3 years ago
15

The CRISPR/Cas9 system can cleave genomic DNA at sequences other than the desired target, a phenomenon referred to as off target

cleavage. To avoid off-target cleavag sites targeted for mutation should appear only once in the genome. What is the minimum length of the part of an sgRNA molecule that binds DNA whose base pair sequence would be expected to be unique within the fruit fly genome? (Assume that the size of the fruit fly genome is 1.4 x 108 bp, that the base pair sequence is random, and that GC.) AT- Multiple Choice 14 16 O 234 15 109
Biology
1 answer:
Deffense [45]3 years ago
3 0

Answer:

The minimum length of a sgRNA sequence to avoid off target cleavage by the CRISPR/Cas system in the fly fruit genome is 14 bases

Explanation:

We are trying to use the CRISPR/Cas system to cleavage the genome of the fruit fly (which is 1.4x10^8 bp long). Also we desire the cleavage to be unique. That means we need a target sequence long enough to be able to assume it will only appear once in the genome.

First, we should think that in every position, we can find one out of four different nucleotide (A, C, T, G). So, the probability of getting a sequence of a given length "n" will be (1/4)^n (We are assuming that the probability of finding a nucleotide in the position "i", it's independent of the nucleotide we find in any other position "j").

Also, to know how many times a sequence will appear in a genome (the expected value of occurrence), we must multiply the probability of that sequence to randomly occur by the length of the genome. For our specific example, the number of occurence of a sequence of length "n" is:

nºoccurence=[(1/4)^n]*1.4*10^8

But in this case, what we want is the expected number of times the sequence will appear to be 1, and we want to obtain the length of the target sequence (n).

Given the information above, we know that:

[(1/4)^n]*1.4*10^8 =1

[(1/4)^n]=(1/1.4*10^8)=1.4*10^-8

Then, if we want to calculate n, we can use logarithms and its properties to get:

log[(1/4)^n]=log[1.4*10^-8]

n*log[(1/4)]=log[1.4*10^-8]

n=log[1.4*10^-8]/log[(1/4)] => n=13.29 approximately.

As the sequence needs to have a natural number of elements, <u>we can conclude that using a target sequence of a minimum of 14 bases with the CRISPR/Cas system in the fly fruit genome should be enough to avoid off target cleavage.</u>

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Troyanec [42]

Answer:

C) genetic drift.

Explanation:

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The bottleneck effect occurs when most of the members of a population die because of any natural disaster like flood, fire, etc. This can cause a certain allele to be lost in the remaining population because of losing major populations. So here due to hurricane half of the mammal population gets eliminated which shows it is an example of the genetic drift.

4 0
3 years ago
Suggest a means by which you could separate mrna from the other types of rna in a eukaryotic cell
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A method by which you could separate mRNA from the other types of RNA in a eukaryotic cell is oligo (dT) chromatography.

<h3>What do you mean by mRNA?</h3>

mRNA may be defined as a molecule in cells that holds codes from the DNA in the nucleus to the sites of protein synthesis in the cytoplasm.

The mRNA can be easily isolated from other types of RNA by the most reliable and convincing method of oligo (dT) chromatography which is the magnetic separation method that bounds oligo (dT) molecules on the surface of paramagnetic beads.

Therefore, a method by which you could separate mRNA from the other types of RNA in a eukaryotic cell is oligo (dT) chromatography.

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6 0
2 years ago
A plant has the genotype AAbbCcDDEeff, where each letter represents a different gene. This plant is crossed with another whose g
scoundrel [369]

Answer:

The fraction of the offspring that will show the phenotype produced by the recessive c allele is 1024/4096 = 0.25 A-BbccDdE-Ff = 25%

Explanation:

You can calculate the fraction of the offspring that will show the phenotype produced by the recessive c allele by making the punnet square for each gene and then multipling the phenotypic proportions, like this:

Cross) AAbbCcDDEeff     x      AaBBCcddEEFF

Cross For each gene by separately:

  • Parental )        AA    x    Aa

       Gametes)  A    A         A    a

        F1)             2/4 AA  

                          2/4 Aa

  • Parental )        bb    x    BB

         Gametes)  b    b         B   b

          F1)                   4/4 Bb

  • Parental )        Cc    x    Cc

        Gametes)  C    c         C    c

         F1)             1/4 CC

                           2/4 Cc

                            1/4 cc

  • Parental )        DD    x    dd

       Gametes)  D    D          d    d

        F1)             4/4 Dd

  • Parental )        Ee    x    EE

        Gametes)     E     e      E     E

         F1)             2/4 Ee

                           2/4 EE

  • Parental )        ff    x    FF

         Gametes)     f     f      F     F

          F1)             4/4 Ff

So, fraction of the offspring that will show the phenotype produced by the recessive c allele is:

4/4 A-  x  4/4 Bb   x   1/4 cc   x   4/4 Dd   x  4/4 E-   x   4/4 Ff =  

1024/4096 = 0.25 A-BbccDdE-Ff =25%  

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