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mojhsa [17]
2 years ago
14

What is the role of a receptor in homeostasis?

Biology
1 answer:
Dominik [7]2 years ago
8 0

Explanation:

d. A structure that alerts the brain to make a change

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Which is not a function of the integumentary system?
AleksAgata [21]
The answer is D. the curculitory system transports hormons.

4 0
3 years ago
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
The ultimate source of energy for starting the food chain is: 1glucose 2starch 3ATP 4in cells solar
balu736 [363]
The ultimate souce of energy for the food chain is solar energy from the sun. It allows plants to convert the solar energy to chemical energy for growth. The growing plants in turn allow heterotropic animals to consume them to obtain energy of their own.
7 0
3 years ago
Being interested in native butterflies, you include the native caterpillar host plants of several butterflies in your annual lan
svetoff [14.1K]

Answer:

d. introduce native flowering plants the adult butterflies need for nectar, their main food.

Explanation:

Organisms choose the habitat based on the availability of basic requirements such as food, nutrients, space, etc. in the region. The absence of one or more of these factors makes them choose another habitat. Butterflies feed on nectar made by plants in their flowers. To make the butterflies stay in the landscape, flowering plants adapted to local conditions should be planted. The butterflies would feed on the nectar of these plants and would stay in the landscape.

3 0
3 years ago
Which molecule is active during the last step of DNA replication?
stepan [7]

Correct answer: Option D- DNA ligase

Explanation: In option A, thymine is a nucleotide, so it is present throughout the replication process, wherever it is required. It is added to the newly formed DNA. In option B, Helicase enzyme is active during initiation and elongation stage, as it facilitates the opening of the winded DNA strands. Option C is nucleotidase and it has no role in DNA replication. So, the correct answer is DNA ligase, which is option D.

The okazaki fragments formed during DNA replication are sealed at the end. And in this step, DNA ligase is used. It catalyzes the formation of phosphodiester bond between the nucleotides of okazaki fragments. So it is the last active molecule of the process.

8 0
3 years ago
Read 2 more answers
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