Answer:
Chloroplast
Explanation:
This organelle in cells indicates that an organism can harness energy from the sun and other abiotic factors like carbon dioxide to make their own ‘food’. Chroloplasts have chlorophyl piments that contains photosystems centers that harness energy from the sun for photosynthesis. This light energy from the sun is captured and transferred in chemical bonds of manufactured carbohydrates which are stored in the plants. These plants transfer this energy in an ecosystem when they are consumed by higher organisms in the food chain.
Answer:
option B) The oxygen end of the molecule has a partial negative charge and the hydrogen end has a partial positive charge.
Justification:
The electronegativity of an element accounts for its relative ability to attract electrons.
Being oxygen more electronegative than hdyrogen (the electronegativity of oxygen is 3.44 while the electronegativity of the hydrogen is 2.20), the electron density will be displaced toward the oxygen, letting it with a partial negative charge and the hydrogen with a partial positive charge.
Finally, since the charge is not symmetrical distributed around a center of the molecule, the molecule ends being polar.
Explanation:
Correct Answer: No change in serum glucocorticoid level
Response Feedback:Primary adrenal insufficiency would be indicated when the adrenal cortex fails to produce cortisol when synthetic ACTH is administered. Serum cortisol levelswould increase following ACTH administration if the adrenal cortex is functioning properly. Primary adrenal insufficiency occurs when there is no change in the serum glucocorticoid level.
Answer:
Explanation:
A woman with type A blood (whose father was type O) meaning her genotype is AO mates with
Man that has type O blood (OO genotype)
Both are heterozygous for MN blood group and both also heterozygous for the FUT1 gene controlling the synthesis of the H substance (Hh)- which determines the expression of the A and B antigen.
Cross
A O M N H h
O AO OO M MM MN H HH Hh
O AO OO N MN NN h Hh hh
Type A- 1/2 O-1/2 type M- 1/4 MN-1/2 N- 1/4, type H- 3/4 h-1/4
Type A with M antigen:
1/2*1/4*3/4 = 3/32
Type A with M and N antigens:
1/2*1/2*3/4 = 3/16
Type A with N antigen:
1/2*1/4*3/4 = 3/32
Type O with M antigen:
1/2*1/4*3/4= 3/32
Type O with M and N antigens:
1/2*1/2*3/4 = 3/16
Type O with N antigen:
1/2*1/4*3/4 = 3/32.
The 3/4 value comes from the expression of Hh-3/4 (this determines if the A and B Angie will be expressed).