Answer:
2 ATP
Explanation:
2 ATP is required to get the glycolysis started. The glycolysis process is divided into two parts, the first part is called the preparatory phase in which energy is required and the second part is called the payoff phase in which energy is produced.
During the first phase in the first step during the conversion of glucose to glucose six phosphate one ATP is consumed and in the third step during the conversion of fructose 6 phosphate to fructose 1,6 biphosphate one more ATP is consumed and in the second phase, 4 ATP is produced.
Therefore the net gain is 2 ATP because 2 ATP are added to get glycolysis started.
The nonselective, passive process performed by the glomerulus that forms blood plasma without blood proteins is called filtration.
<span>Glomerular filtration is the first step in making urine which includes the<span> separation of the liquid part of the blood (plasma) from the blood cells</span>. Glomerulus of the kidneys (a tuft of blood capillaries)<span> filters excess fluid and waste products out of the blood into the urine collecting tubules of the kidney and thus eliminates it out of the body.</span></span>
Answer:
Ventilation and mixing outdoor air with indoor air
Explanation:
The recessive phenotype is used first to determine the q squared value because it is only expressed when h0m0zygous.
<h3>What is a recessive phenotype?</h3>
A recessive phenotype is a phenotype which expresses the recessive trait of a gene.
The recessive phenotype is always h0m0zygous for that triat to be expressed.
In the Hardy-Weinberg equilibrium equation, the allele and phenotype frequencies are given as follows:
where;
- p^2 is the h0m0zygous dominant phenotype frequency
- 2pq is the heterozygous phenotype frequency
- q^2 is the h0m0zygous recessive phenotype frequency
Therefore, the recessive phenotype is used first to determine the q squared value because it is only expressed when h0m0zygous.
Learn more about recessive phenotype at: brainly.com/question/22117
Answer:
DDWw x Ddww
DdWw x DDww
DDWw x DDww
Explanation:
D for tallness
d for dwarfness
W for violet
w for white
For Parental Plants that has
tall, violet x tall, white
The representation will be
DDWW × DDww (We the apply the mendelian crossing)
This will yield offsprings with the following trait
DDDD × DwDw
DDDD × DwDw
DWDW x WwWw
DdWw x ddww
Ddww x ddWw
DdWw x DdWw
DDWw x ddww
From the above Offsprings
1/2 tall, white and 1/2 tall, violets offspring are
DDWw x Ddww
DdWw x DDww
DDWw x DDww
