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Sphinxa [80]
3 years ago
5

To investigate the possible presence of introns in three newly discovered genes (X, Y, and Z), you perform an experiment in whic

h the restriction enzyme HaeIII is used to cleave either the DNA of each gene or the cDNA made by copying its mRNA with reverse transcriptase. The resulting DNA fragments are separated by gel electrophoresis, and the presence of fragments in the gels is detected by hybridizing to a radioactive DNA probe made by copying the intact gene with DNA polymerase in the presence of radioactive substrates. The following results are obtained:
Source of DNA Number of Fragments After Source Electrophoresis
Gene X DNA 3
cDNA made from mRNA X 2
Gene Y DNA 4
cDNA made from mRNA Y 2
Gene Z DNA 2
cDNA made from mRNA Z 2

Required:
a. What can you conclude about the number of introns present in gene X?
b. What can you conclude about the number of introns present in gene Y?
c. What can you conclude about the number of introns present in gene Z?
Biology
1 answer:
REY [17]3 years ago
6 0

Answer:

a). At least one intron must be present in gene X.

b). At least two intron must be present in gene Y

c). It is impossible to determine whether there are any introns in gene Z.

Explanation:

Introns may be defined as the segments in the RNA molecule or the DNA molecule that does not code for proteins and they interrupts the sequences of each of the molecules.

In the context, we get :

a). In the gene X, at least one of the intron is present. Sometimes more than one introns may also be present. Due to slicing, we get one band less in the cDNA which is made from the mRNA.

b). In dene Y, at least one intron is present. Sometimes more than one introns are present inside the gene Y. In this case we get two bands less in the gene due to slicing.

c). In gene Z, it is not possible to determine the number of introns present in them. In genes X and Y, the Haelll enzyme is present in the intron. Therefore we get less of band after splicing the mRNA to cDNA.

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The CRISPR/Cas9 system can cleave genomic DNA at sequences other than the desired target, a phenomenon referred to as off target
Deffense [45]

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>

3 0
3 years ago
List the 4 abiotic forms of nitrogen cycle and the chemical formula
weeeeeb [17]

Answer:

Four abiotic forms of nitrogen cycle and its chemical formula

Ammonium – NH^{4+}

Ammonia – NH^3

Nitrite – NO^2

Nitrate – NO^3

Explanation:

Ammonium: In the nitrogen cycle ammonium is starting point that is present in soil and are converted to various forms by different process.

Ammonia: In the process of nitrogen fixation, the bacteria having nitrogenase enzymes reacts with the nitrogen as well as hydrogen and produces the ammonia, which is further converted to organic compounds.

Nitrite: The nitrosomonas bacteria present helps in conversion of nitrogen gas into nitrite.

Nitrate: Again the Nitrite is converted into Nitrate by the nitrobacter. During the process of assimilation ammonium and nitrate are absorbed by the plants.

8 0
3 years ago
In order for any animal to grow, repair its body, reproduce and carry out daily activities it must have energy. When an animal d
stealth61 [152]

Answer:

Hungry!

Explanation:

4 0
2 years ago
Does the carbon cycle in this diagram appear to be in balance or out of balance? Use specific evidence from the diagram to suppo
Gennadij [26K]
Yes the diagram is valences
8 0
2 years ago
Research and identify a vestigial structure of an organism of your choosing.
Mariana [72]

Answer:

The one of best example would be appendix in humans.

Explanation:

In the past, the appendix may have helped in the digestion of the plants by producing mucus and prevent any type of infection. Plants with a high content of cellulose are digested with the help of the appendix as reported.

It is not a part of the human digestive system and considers as a vestigial organ. Vestigial organs are the organs that are present in an organism with no particular function and their removal will not lead to any harm. Humans do not need it because our digestive system is more efficient and works better in the first stages of digestion than they used to.

Thus, the correct answer is - appendix in humans.

6 0
2 years ago
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