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zalisa [80]
3 years ago
15

Animals get the most of the nitrogen they need

Biology
1 answer:
STALIN [3.7K]3 years ago
8 0
Through the things they eat
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Describe the type of gaseous molecules that are most susceptible to non ideal behavior
const2013 [10]
An ideal gas is a theoretical gas in which inter molecular forces do not exist among the molecules and the molecules do not occupy any space. A gas that does not obey ideal gas law are described as non ideal in behaviour. Gases that are most susceptible to non ideal behaviour are those that have attractive and repulsive forces among their molecules and whose particles have volume. Gases that exhibit non ideal behaviour has compressiblity factor ratio that deviate from 1.
3 0
3 years ago
2. In the citric acid cycle, malate is oxidized to oxaloacetate by the enzyme malate dehydrogenase, which uses NAD+ as an electr
atroni [7]

Answer:

NAD+ act both as coenzyme as well as electron acceptor compound and get reduced to NADH by accepting electron.

Explanation:

NAD+ act as co enzyme of various biological catalyst such as malate dehydrogenase, isocitrate dehydrogenase etc.

             NAD+ can act as co enzyme only in its oxidized form but not in its reduced form called NADH.

         Many reaction needs NAD+ to occur such as conversion of glyceraldehyde 3 phosphate to 1,3 bisphosphoglycerate, malate to oxaloacetate.

              That"s why NAD+/NADH ratio is kept very high because if this ratio bychance get  low then it will hamper the normal redox potential of NAD+/NADH.As a result many biochemical reaction will not take place.

6 0
3 years ago
14). During the early 1700’s, a small group of pacifist Protestants fled Germany to avoid religious persecution. This group, the
abruzzese [7]
A) In the Dunker population, the frequency of IB allele is 0.3 and the frequency of i allele is 0.4. In the general population, the frequency of IB allele is 0.1 and t<span>he frequency of i allele is 0.5.
</span>
If:
I^{A} - <span>the frequency of IA allele
</span>I^{B} - <span>the frequency of IB allele
</span>i - t<span>he frequency of i allele

Then:
</span>I^{A} I^{A} + <span>I^{A} i - the frequency of individuals with A blood type
</span>I^{B} I^{B} + <span>I^{B} i - the frequency of individuals with B blood type
</span>ii <span>- the frequency of individuals with O blood type
</span>
Let's first take a look on the Dunker population:
I^{A} = 0.3
ii=0.16&#10;

<span>Since there is only one possible genotype for O individuals - ii - the frequency of the allele i is square root of the frequency of O individuals:
</span>i= \sqrt{ii}
⇒ i =  \sqrt{0.16}
⇒ i=0.4

Now, we have the frequencies of two alleles (I^{A} and i). To calculate the frequency of I^{B}<span> allele, we will use the formula:
</span>I^{A} + I^{B} + i = 1
⇒ I^{B} = 1- I^{A} - i
⇒ I^{B} = 1-0.3-0.4
⇒ I^{B} = 0.3

Now, in the general population:
I^{A} = 0.4
ii=0.25

<span>Similarly to the work for the Dunker population:
</span>i= \sqrt{ii}
⇒ i = \sqrt{0.25}
⇒ i=0.5

I^{A} + I^{B} + i = 1
⇒ I^{B} = 1- I^{A} - i
⇒ I^{B} = 1-0.4-0.5
<span>⇒ I^{B} = 0.1
</span>


b) A founder effect is a result of geographical separation of a few individuals from the original population. Those founding individuals will form a new population. The Dunker population was not only geographically separated, but also genetically. The group interbreeding was present resulting in increasing those allele frequencies that were the most common in the founding population. In this case, the most individuals from the founding population had B blood type.
6 0
3 years ago
The karyotype of one species of primate has 48 chromosomes. In a particular female, cell division malfunctions, and she produces
Rom4ik [11]

Answer:

E) Either anaphase I or II

Explanation:

Failure of segregation of homologous chromosomes during anaphase I or failure of segregation of sister chromatids during anaphase II leads to the presence of the abnormal number of chromosomes in resultant gametes. In the given example, the egg mother cell with 48 chromosomes (24 pairs) would enter meiosis I but the failure of one pair of homologous chromosomes to segregate from each other followed by normal meiosis II would result in the formation of two gametes with one extra chromosome and two gametes with one less chromosome.

On the other hand, if the nondisjunction occurs at anaphase II of meiosis II, two normal gametes, one gamete with one extra chromosome and one gamete with one less chromosome will be formed. Therefore, nondisjunction at anaphase I or anaphase II would have resulted in the production of eggs with one extra chromosome.

6 0
3 years ago
Economic and social changes during the Gilded Age caused
allsm [11]

Answer:

I think it is D

Explanation:

correct me if I’m wrong

4 0
3 years ago
Read 2 more answers
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