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Strike441 [17]
2 years ago
10

Which examples show repair or replacement of damaged cells? 1. A scrape being healed. 2. Scars forming where a cut once belonged

. 3. Skin flaking off your arm in the winter.
Biology
2 answers:
Archy [21]2 years ago
4 0
2 because it shows signs that the cells are getting replace by new cells
galben [10]2 years ago
3 0
2.Scars forming where a cut once belonged.
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A (n) inhibitor has a structure that is so similar to the substrate that it can bond to the enzyme just like the substrate. 2. a
Studentka2010 [4]
The answers are as follows:
1. <span>An inhibitor has a structure that is so similar to the substrate that it can bond to the enzyme just like the substrate: t</span>his is called competitive inhibitor. A competitive inhibitor will compete with the substrate for the active site of the enzyme and bind to the active site, thus incapacitating the substrate from binding to the active site.
2.  An inhibitor binds to a site on the enzyme that is not the active site: this is called non competitive inhibitors. Non competitive inhibitors bind to other site in the enzyme which is not the active site of the enzyme. The binding of the inhibitor changes the conformation of the enzyme as well as the active site, thus making it impossible for the substrate to bind to the enzyme effectively.
3. <span>usually, a(n) inhibitor forms a covalent bond with an amino acid side group within the active site, which prevents the substrate from entering the active site or prevents catalytic activity: this is called irreversible or permanent inhibition. Permanent inhibitors form covalent bonds with the enzyme and prevent substrate from binding to the enzyme.
4. T</span><span>he competitive inhibitor competes with the substrate for the ACTIVE SITE on the enzyme: The active site of an enzyme is the place where the substrate normally bind in order to activate a enzyme. Competitive inhibitors are those inhibitors that compete with the substrate for the active site of the enzyme and prevent the substrate from binding there.
5. W</span><span>hen the noncompetitive inhibitor is bonded to the enzyme, the shape of the ENZYME is distorted. The non competitive inhibitors are those inhibitors that bind to other places in the enzyme instead of the active site. The binding of the non competitive inhibitor usually distort the shape and the conformation of the enzyme thus preventing the substrate from binding to it effectively.
6. E</span><span>nzyme inhibitors disrupt normal interactions between an enzyme and its SUBSTRATE. The principal function of enzyme inhibitor is to prevent the substrate from binding to the appropriate enzyme. This is usually done in the human system in order to regulate the activities of enzymes.</span>
6 0
2 years ago
Which enzymes in the citric acid cycle catalyze oxidative decarboxylation reactions?
dlinn [17]

In the citric acid cycle (also known as Kreb’s Cycle), the enzyme that catalyzes oxidative decarboxylation reactions is α-Ketoglutarate dehydrogenase. The oxidative decarboxylation reaction is the irreversible stage of the citric acid cycle; it generates NADH (equivalent of 2.5 ATP), and regenerates the 4C chain (CoA is excluded).

8 0
2 years ago
Read 2 more answers
What conditions exist for a non-native species that allow them to grow exponentially?
sergejj [24]

Invasive species cause harm to wildlife in many ways. When a new and aggressive species is introduced into an ecosystem, it may not have any natural predators or controls. ... Invasive species can change the food web in an ecosystem by destroying or replacing native food sources.

hope this helps :)

4 0
1 year ago
Assume that there are 8 rats in the cage: 6 trained and 2 untrained. A rat is removed from the cage and it is noted whether or n
mojhsa [17]

Answer:

1. Option A

2. 56

Explanation:

1.

Option A

Option A is not a valid answer for this question as it is given that out of eight rats only two rats are untrained. Thus, an output which shows more that two untrained rats i.s UUTTU is a wrong answer

2.

8C_5 = \frac{8!}{5!(8-5)!} \\8C_5 = \frac{8*7*6*5!}{5! * 3!} \\8C_5 = \frac{8*7*6}{3*2*1} \\8C_5 = 56

Thus , there are 56 outcomes possible for this experiment

6 0
2 years ago
Imagine that you want to breed a species of rose that has a wonderful smell. You have many different types of roses, but none of
Natalija [7]

Answer: if i want to breed a rose and i don't have a certain kind of rose, i have to cross between them until i have a kind of rose that express the phenotype i was looking for. Once i've got it, i'll try to cross it with another rose and generate more species like that to cross with the rose that express the phenotype, this favors to generate a specie that have a pure phenotype.

Explanation: The answer is explain by the Mendel's laws.

Mendel's second law:

if you have two roses that are heterozygous and you cross them, their offspring have a 25% probability of generating a homozygous phenotype for a certain recessive characteristic (such as the rich smell of roses that is not a dominant character), 50% of generate heterozygotes with a dominant phenotype, and 25% generate other homozygotes with a dominant phenotype.

Once the offspring with the recessive homozygous character are obtained, it is possible to start making crosses between those of a recessive nature and there Mendel's first law applies

All the homozygous recessive individuals that are crossed will have a 100% homozygous recessive offspring.

Also, if you cross heterozygotes with another heterozygotes the offspring will be 100% heterozygotes

3 0
2 years ago
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