Marine debris is defined as any man-made solid material that goes into waterways either directly or indirectly. Eighty percent of the marine debris comes from the land-based sources. Examples of marine debris are soda cans, plastic bag, cigarette waste. Anything that ends up in a water body that has a potential in harming the marine life is a marine debris.
Answer:
A, C and D are photosynthetic. They are green stained suggesting the presence of chloroplast and has multiple mitochondria for energy production
Answer:
It decreases the levels of cAMP in the cell, repressing transcription from the lac operon.
Explanation:
When glucose is absent, cAMP serves as coactivator binds to CRP, the catabolite gene activator protein. The CRP-cAMP complex binds to the site near the lac promoter and stimulates the expression of the operon by RNA polymerase many folds.
Catabolite repression refers to inhibition of the synthesis of enzymes of lactose catabolism when glucose is present as an energy source. In the presence of glucose, synthesis of cAMP is inhibited resulting in its lower cellular concentration. The lower cAMP levels do not allow the binding of cAMP and CRP. The result is reduced expressed of lac operon.
Answer:
The correct answers are A. "crenation", B. "hemolysis", C. "hemolysis", D. "crenation" and E. "neither will occur".
Explanation:
0.9% (m/v) NaCl or 5.0% (m/v) glucose are isotonic solutions at which the cells will not suffer any harmful consequence. A solution with a higher concentration than the isotonic conditions would result in the cells crenation, while a solution with a lower concentration would result in the cells hemolysis. Therefore the consequences of putting the red cells to the solutions stated in the question are as following:
A: 3.21% (m/v) NaCl Solution = crenation (higher than 0.9% (m/v) NaCl)
B: 1.65% (m/v) glucose Solution = hemolysis (lower than 5.0% (m/v) glucose)
C: distilled H2O Solution = hemolysis (lower than 0.9% (m/v) NaCl or 5.0% (m/v) glucose)
D: 6.97% (m/v) glucose Solution = crenation (higher than 5.0% (m/v) glucose)
E: 5.0% (m/v) glucose and 0.9%(m/v) NaCl = neither will occur (equal to 5.0% (m/v) glucose and 0.9% (m/v) NaCl)
