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beks73 [17]
2 years ago
6

Sort the characteristics as belonging to the first half of the pathway that requires energy or to the second half of the pathway

that produces energy during glycolysis.
Biology
1 answer:
Tpy6a [65]2 years ago
3 0

Answer:

The first half of glycolysis is energy investment phase while second half of glycolysis is energy generation phase.

Explanation:

There are two phases of glycolysis. The first half of glycolysis is energy investment phase while second half of glycolysis is energy generation phase. In the first half of glycolysis, two Adenine Tri Phosphate (ATP) are used or invested for the breakdown of glucose molecule into three carbon atom while in the second half of glycolysis, two NADH and four ATP molecules are formed.

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Sort the statements based on whether they describe DNA replication in eukaryotes or prokaryotes. Replication takes place in the
Ilya [14]
In eukaryotes, <em>replication takes place in the nucleus</em> as prokaryotes do not have a true nucleus and <em>replication takes place in the cytoplasm</em>. The nucleus of the eukaryotes is the location where genetic material (DNA) is found; in prokaryotes, the genetic material is condensed in the cytoplasm called the nucleoid. There are multiple replication forks or <em>multiple origins of replication </em>in eukaryotes in contrast to prokaryotes which only has <em>one origin of replication. </em>Lastly, replication in eukaryotes <em>occurs at multiple points along the chromosome; </em>in contrast with prokaryotes where it <em>occurs at just one point on the chromosome.</em>
8 0
2 years ago
Read 2 more answers
A yellow female Labrador retriever was mated with a brown male. Half of the puppies were brown, and half were yellow. Explain ho
Over [174]

Answer:

The first male was bb Ee, and the second male was bb EE.

Explanation:

In Labradors coat colour is controlled by two genes. Suppose the two genes are B and E. B produces black colour and recessive form bb gives brown colour. Gene E is epistatic over gene B in its recessive form which means that ee will produce yellow colour regardless of the genotype present of B gene.

The first case is possible if the female lab is bbee (yellow) and the male lab is bbEe (brown):

bbee X bbEe

        bE      be

be  bbEe    bbee

So half of the offspring will be brown (bbEe) and half of them will be yellow (bbee)

The second case is possible if the same female bbee mates with a brown male of different genotype which can be bbEE:

bbee X bbEE

         bE

be   bbEe

So all offspring will be brown (bbEe)

Hence, the first male was bbEe and the second male was bbEE.

6 0
2 years ago
Researchers believe that in a fish species, individuals with the recessive genotype aa are predisposed to disease. Homozygous do
Nataly_w [17]

Answer:

The homozygous dominant phenotype is higher than expected, indicating that evolution has occurred.

Explanation:

At the start there are 200 fishes in the pond, 100 of them are AA(50%) and 100 of them are aa(50%). Using the Hardy-Weinberg Equilibrium equation we can say that the gene frequency is

A=0.5

a=0.5

With those frequency, the expected percentage of offspring with dominant genotype will be:

AA= 0.5 * 0.5 = 0.25 = 25%

The number of homozygous dominant found is 35% which is higher than expected  (25%). Higher homozygous dominant frequency than expected means the Hardy-Weinberg Equilibrium is changed. In this case, evolution probably the cause that shifts the gene frequency.

7 0
2 years ago
(c) find the longest interval during which no tree-cutting samples occurred. what is the first year that no tree-cutting samples
MaRussiya [10]

Answer:

Longest interval during which no tree-cutting sample occurred: 9 years (1203 AD - 1212 AD)

First year in that interval: 1203 AD

Explanation:

Lets complete the question first.

Q) A study uses tree rings to determine the year in which a tree was cut. The data is shown in the leaf and stem diagram in the ATTACHMENT.

The three digits number in the left column represents centuries and decades AD. The numbers on the right represents the years.

For example, consider the first input in the diagram

119 | 5   6

which means it represents 2 samples.

1195 AD, 1196 AD.

The longest interval with no tree-cutting samples is found from 1203 AD to 1212 AD.

3 0
2 years ago
One mole of glucose contains 2870 kJ of energy. The hydrolysis of one mole of ATP releases 30.7 kJ of energy. Calculate the perc
Galina-37 [17]

The answer is:

Total used energy is the energy released by the hydrolysis of one mole of ATP and it is 30.7 kJ.  

Total possible energy (used pus released as a heat) is 2870 kJ.

So, we calculate total used/total possible=30.7/2870=0.01069 or 0.017. Expressed in percentage: 1.07% energy is used.


8 0
2 years ago
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