The answer would be mitochondria
The primary form of energy that the cell use are come from ATP. The body could produce ATP through different path of metabolism, but the highest amount of ATP produced from electron transport chain. Mitochondria is the main organelle of the cell that produce ATP because it allow the cells to do electron transport chain.
Answer:
0.2404
Explanation:
The genes R/r and E/e are linked and there is 4% recombination between them.
<u>The possible genotypes and phenotypes are:</u>
- RR or Rr: Rh+ blood type
- rr: Rh- blood type
- EE or Ee: elliptocytosis
- ee: normal red blood cells
Tom and Terri each have elliptocytosis (they are E_), and each is Rh+ (they are R_).
Tom's mother has elliptocytosis (E_) and is Rh- (rr), so she has the genotype Er/_r. His father is healthy (ee) and has Rh+ (R_), so he has the genotype eR/e_. Tom must have inherited his E allele from his mother and his R allele from his father, so he has the genotype eR/Er.
Terri's father is Rh+ (R_) and has elliptocytosis (E_), while Terri's mother is Rh- (rr) and is healthy (ee) with the genotype er/er. Terry could only receive the chromosome <em>er </em>from her mother, and because she is heterozygous for both genes the dominant alleles were both received from her father. Terri's genotype is ER/er.
The frequency of recombination is 4%, so 4% of the produced gametes will be recombinant. There are two possible recombinant gametes, so each will appear 2% of the times (a frequency of 0.02).
<u />
<u>Tom will produce the following gametes:</u>
- eR, parental (0.48)
- Er, parental (0.48)
- er, recombinant (0.02)
- ER (recombinant (0.02)
<u>Terri will produce the following gametes:</u>
- ER, parental (0.48)
- er, parental (0.48)
- Er, recombinant (0.02)
- eR, recombinant (0.02)
A child Rh- with elliptocytosis has the genotype rrE_. This can happen from the independent combination of the following gametes from Tom and Terri respectively:
- Er (0.48) × er (0.48) = 0.2304 Er/er
- Er (0.48) × Er (0.02) = 0.0096 Er/Er
- er (0.02) × Er (0.02) = 0.0004 er/Er
And the total probability of having a rrE_ child will be 0.2304 + 0.0096 + 0.0004 = 0.2404
Answer:
Explanation:
Blue-white screening is a method for distinguishing proof of (recombinant bacteria). It depends on the capacity of ( B-galactosidase) to separate lactose. Blue-white tests exploit the molecule called (x-gel)_ which is like lactose in that it is severed by B-galactosidase. When separated, the (5-bromo-4-chloro-indoxyl) (same as past) turns _(_blue). In the event that uncleaved, which implies a non-function B-gal gene, the X-gal remains (white)_. Subsequently, a __(white) bacterial province implies the B-galactosidase gene isn't practical, and in this way there ___lacz__ a recombinant gene embedded into the vector.
