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Tasya [4]
3 years ago
6

Please help me! asap im trying to figure this out

Biology
2 answers:
gayaneshka [121]3 years ago
6 0

Answer:

you have it right I think

irina [24]3 years ago
4 0

Answer:

\boxed {\tt 2nd \ Answer}

Explanation:

An enzyme is a protein that speeds up a chemical reaction. Enzymes are <u>shape specific</u>. The active site is a specific shape, which means it can only break down specific substrates.

Certain conditions can <u>denature</u> enzymes. For example, extreme temperatures or pH will cause the enzyme to deform and change shape, which means it cannot perform its job.

Therefore, the best choice is the second one: <em>enzymes work best in specific conditions and only break down specific substrates.</em>

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9. Which of the following is considered a Local Histone Mobilization (circle all the correct ones and explain each) a. Ejection
kati45 [8]

Answer:

Phosphorylation

Methylation

Explanation:

Histone mobilization deals with the covalent bonding of several functional groups to the free nitrogens in the R-groups of lysine in the N-terminal tail.

DNA transcription can be controlled or altered via 'methylation' on the histones. Methylation is the mode of transferring methyl group to amino group of histone proteins that form the nucleosomes, which result to chromosomes formation through wrapping of DNA double helix around it. Histones methylation can either increase or decrease genes transcription. Chromatin modification is known to stimulate neural pathways that are essential for long memories and learning.

Histones phosphorylation is associated transcriptional activation. It deals with the transfer of phosphate group to amino acids of the proteins. Phosphate group known to be negatively charged, when transferred to amino acids of the histone proteins, it creates a repulsive force with the negatively charged phosphates of the DNA backbone. It can occur in serine, tyrosine and threonine. The enzymes kinases activate phosphorylation, while protein phosphatases remove phosphate groups.

6 0
3 years ago
burning fossil fuels like gasloine relases harmful gases into the air. why don’t the cars in the world solar challenge release t
Alex787 [66]

Answer: the cars run on pure solar electricity and are driven by motors so there is no use or out put of gasoline of any kind

Explanation:

8 0
3 years ago
What is the heroic part of the Philosophical Breakfast Club’s story? What is the flip side of this story? Explain the details of
pav-90 [236]

Answer:

Before the advent of modern biology, study of life and organism was in the realm of natural philosophy and was never in the club of classical sciences like physics, chemistry and engineering. Today biology is a science that studies life and living organisms.

Explanation:

A biologist can be categorized as a scientist because all the processes that supports scientific investigations are carried out in the study of biology- data collections, critical thinking, observation and collection of result for the purpose of public consumption.

On the other hand forensics is the mainly the aplication of science-mostly biology- in resolving criminal activities. It involves the collection of samples, anaylsing and presentaion of result for decison making.

6 0
4 years ago
A giraffe's neck is an example of ______.
weeeeeb [17]
An example of an adaptation 
6 0
4 years ago
In a study of the Hopi, a Native American tribe of central Arizona, Woolf and Dukepoo (1959) found 26 albino individuals in a to
Keith_Richards [23]

Answer:

0.9342

Explanation:

The Hardy-Weinberg equation states that p² + 2pq + q² = 1,

where p is the frequency of the dominant 'normal' (n) allele and q is the frequency of the recessive 'albino' (a) allele in the population, while q² represents the frequency of the homo-zygous albino genotype (aa), p² represents the frequency of the homo-zygous normal genotype (nn) and 2pq represents the frequency of the heterozygous genotype (na).

In this case, the frequency of individuals in the population that have the genotype aa (q²) is equal to 26/6000 = 0.004333. In consequence, q is equal to √ 0.004333 = 0.0658.

Moreover, the allele frequency of the normal (n) allele p is equal to 1 - q = 1 - 0.0658 = 0.9342, so p² (nn) = (0.9342)² = 0.8727.  

Finally, the frequency of the heterozygous genotype (na) is 2pq = 2 x 0.9342 x 0.0658 = 0.123.

3 0
3 years ago
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