Answer:
43 monosaccharides will be joined together to make the complex carbohydrate.
Explanation:
The complex carbohydrates are synthesized by the joining of monosaccharide residues by glycosidic linkages.One glycosidic linkage joins two monosaccharide residues by the elimination of one molecule of water.
In other words it can be said that elimination of one water molecule joins two monosaccharides.By the same way elimination of 42 molecules of water will result in the joining of 43 monosaccaride residues.
Due to the definition of the central dogma, another way of putting it is that the central dogma follows the flow of information from DNA to protein.
Answer:
from glycolysis to electron transport.-glycolytic pathway
NADH and FADH2 are the major electron carrier from glycolysis through the Kreb Cycle to the electron transport chain.
Note-NADH alone transports electron from glycolysis to the Kreb Cycle. while both NADH and FADH2 transport electrons from the kreb'cycle to the electron transport chain.
from citric ac id cycle to the electron transport chain.
as explained above both NADH and FADH2.
The electrons are in the hydrogen atoms, carried by these co-enzymes. When they reached the matrix, the hydrogen atoms are split into protons and electrons(p and e-). it is these electrons that form gradients which are transported as chains in the matrix. The gradients of the electron generated PMF for pumping Hydrogen atoms into the intramembrane of mitochondrial
Note.
Nicotinamide Adenine Di nucleotide Hydrogen(NADH)
Flavin Adenine Dinucleotide Hydrogen(FADH)
Explanation:
Answer:
25%
Explanation:
The possible genotypes and phenotypes are:
- EE or Ee: thick eyebrows
- ee: thin eyebrows
- FF or Ff: free earlobes
- ff: attached earlobes
Tom and Jessica have thick eyebrows and free earlobes. Tom's mother has attached earlobes, so Tom must have inherited an <em>f</em> allele from her, and because he has free earlobes he must be heterozygous<em> Ff. </em>We assume Jessica is homozygous for free earlobes <em>FF</em>.
Tom's mother and Jessica's father have thin eyebrows, so they are both <em>ee</em> and therefore both Tom and Jessica must be heterozygous <em>Ee</em>.
From the above information, Tom has the genotype EeFf and Jessica is EeFF. If you do a Punnett Square of this cross, you'll get that the probability of having a child with free earlobes and thin eyebrows (eeF_) is 25%.
Answer:
what?! my one working breain cell is like" what?!
me no uNdErStanD "
Explanation:
