You are caring for a 6-year old child with a swollen, painful deformity to the left forearm. As you communicate with the parents of this child, you should
make sure that they remain aware of what you are doing.
Answer:
Once starch reaches small intestine, it needs to be broken down into smaller units with the help of <u>hydrolysis reaction</u>. Pancreatic amylase catalyzes this reaction which ultimately breaks down glycosidic bonds present in starch molecule to produce simpler bio-molecules like maltose, dextrin etc.
Uses of carbohydrate in body:
(1) It is primary source of energy in the body;
(2) It is required by the body for storage of energy in the form of glycogen;
(3) In plants, it is required for building cellular structures like cellulose.
Explanation:
In order for starch to be digested in the intestine, it needs to be hydrolyzed with the help of pancreatic amylase. The hydrolysis reaction breaks the glycosidic bonds of starch molecule which is a polymer leading to the release of smaller molecules like dextrin, maltose etc. Maltose which is a disaccharide is further broken down into even smaller monomeric units like glucose with the help of enzyme known as maltase.
After conversion into monomeric unit i.e. glucose now it can be absorbed by the blood stream. The blood transports these small glucose molecules to all the cells of the body. After entering the cell, the glucose is used for the production of energy/ATP via cellular respiration.
During cellular respiration, the glucose is first converted into pyruvate via glycolysis. Pyruvate then enters PDH complex wherein it is oxidatively carboxylated into acetyl coenzyme A. The acetyl coenzyme A produced then enters TCA cycle and produce high energy molecules like NADH, FADH₂ and ATP. The NADH and FADH₂ then finally enter electron transport chain (ETC) so as to produce ATPs. These ATPs are then utilized by our body for our day to day activities.
Answer:
Explanation:
using the formula p² + 2pq + 2pr + q²+ 2qr + r² = 1 where p is IA allele is 0.35 and q is IB allele is 0.15.
Thus, the expected frequency of people with type AB blood in this population will be (2pq) = 2 x 0.35 x 0.15 = 0.105
b. The expected frequency of people with type A blood will be p² + 2pr = where r is found using the formular p + q + r = 1 = 1 - (p+q) = 1 - (0.35+0.15) = 0.5
Thus, we have the expected frequency of people with type A blood to be (0.35² + 2x0.35x0.5) = 0.4725.
c. <u> p² + 2pr</u> + <u>2pq </u>+ <u>q² + 2qr</u> +<u> r²</u>
AA AO AB BB BO OO
1850 180 635 2335
frequency of IA- p = (1850 + 180/2) / 5000 = 0.388
frequency of IB- q = (635 + 180/2) / 5000 = 0.145
frequency of i- r = 2335/ 5000 = 0.467.
since r² = 2335, r = √2335 = 48.32
using the formula p + q + r = 1
