Through "cellular respiration" process energy is stored in the form of glucose.
<u>Answer:</u> Option A
<u>Explanation:</u>
The series of metabolic reactions and mechanism take place in organism ranging from microscopic bacteria to large organisms cells in order to transfer biochemical energy from food nutrients (stored in glucose form which is transferred) to adenosine triphosphate (ATP) and then waste product is also released, the whole process is known as "cellular respiration".
The energy required for ATP synthesis extracted from the breakdown of foods and phosphocreatine (creatine phosphate). It is stored inside muscle cells because phosphocreatine is readily available to produce ATP quickly.
Answer:
Deplasmolysis swelling of the cytoplasm of a plasmolyzed cell : reversal of plasmolysis.
Explanation:
One long, single molecule of double stranded, helical, supercoiled DNA. In most bacteria, the two ends of the double-stranded DNA covalently bond together to form both a physical and genetic circle
Really, there are multiple questions all rolled into one.I will try to answer them patiently and systematically.
First summarize data.
A. Neither Olivia nor Marcus have freckles (recessive, ff)
B. Both are heterozygous for the hairline trait (dominant Ww)
C. Neither Marcus nor Olivia can roll their tongues (rr).
D. All four children have dimples (dominant Dx)
E. Both Olivia and Marcus are EE (unattached earlobe trait).
F. Marcus can not detect the bitter taste (pp for PTC gene)Olivia has been found to be able to detect the bitter taste, but she is heterozygous for the trait. (Pp for PTC gene)
A. Freckles, F (Dominant)
"Neither Olivia nor Marcus have freckles" =>
both have genotype ff.
None of the children have freckles (i.e. P(F)=0% for freckles in all children)
B. Widow's Peak, W (dominant)
"Both are heterozygous for the hairline trait"
So both have genotype Ww.
Punnett square
W w
W WW Ww
w Ww ww
Since W is a dominant trait, only ww (25%) will have straight hairline, 75% will inherit the widow's peak.
50% of the children will be homozygous (Ww).
C. Rolling tongues, R (dominant)
"Neither Marcus nor Olivia can roll their tongues"
means that both are homozygous recessive, with genotype rr.As in freckles, all children will have genotype rr, so none of them will roll their tongues.
None will be heterozygous. The whole family's genotype is rr.
D. Dimples, D (dominant)
"D. All four children have dimples"
implies that all children have genotype DD or Dd.
It is likely that at least one parent has genotype DD in order to have 100% of children have DD or Dd.Here are some possibilities
Case 1: DD + DD (both homoozygous dominant)
D D
D DD DD
D DD DD
Phenotype: 100% have dimples
Case 2: DD + Dd (one homoozygous dominant, and other heterozygous)
D d
D DD Dd
D DD Dd
Phenotype: 100% have dimples
Case 3: DD + dd (one homoozygous dominant, and other homozygous recessive)
D D
D DD DD
d Dd Dd
Phenotype: 100% have dimples
Case 4: Dd + Dd (both heterozygous)
D d
D DD Dd
d Dd dd
Phenotype: 75% have dimples, 25 do not.Note: all 4 children could have dimples, with probability 31.6%
Case 5: Dd + dd (Heterozygous + homozygous recessive)
D d
d Dd dd
d Dd dd
Phenotype: 50% have dimples, 50 do not.Note: All four children could have dimples, with probability 6.25%.
Case 6: dd + dd (Both homozygous recessive)
D d
d dd dd
d dd dd
Phenotype: all children have no dimples.
Conclusion:Likely genotypes of parents: DD+DD, DD+Dd, DD+dd
Possible genotypes of parents: Dd+Dd, Dd+dd
Impossible genotype of parents: dd+dd
Therefore we know with certainty that at least one of the parents has dimples.
E. Unattached Earlobe trait, E (dominant)
"Both Olivia and Marcus are EE"
(i.e. unattached earlobe trait).
This means that the whole family will have genotype EE, i.e. all are homozygous dominant, and have unattached earlobes.
F. Bitter taste, P (incomplete dominance)
"Marcus can not detect the bitter taste (pp for PTC gene)
Olivia has been found to be able to detect the bitter taste, but she is heterozygous for the trait. (Pp)"
P p
p Pp pp
p Pp pp
Probability for each single child being able to taste the ptc paper is 1/2.
Probability for all children being able to taste the ptc paper is (1/2)^4=1/16.
If Violet cannot taste the ptc paper, her genotype is pp.
We do not know for sure how many of the children can taste the ptc paper.
The most like situation is only half of them can taste, so do the parents. Therefore, half of the family can taste the ptc paper.
Finally, as to "please answer it correctly", I believe I did. :)
