Answer:
D-C-B-A
Explanation:
Oldest crust on the bottom
Please help me with this assignment
Answer:
A = parietal labe | B = gyrus of the cerebrum | C = corpus callosum | D = frontal lobe
E = thalamus | F = hypothalamus | G = pituitary gland | H = midbrain
J = pons | K = medulla oblongata | L = cerebellum | M = transverse fissure | N = occipital lobe
Explanation:
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Answer:
22 cells
Explanation:
The concentration of numbers of cells = 6.74 x 10⁶ cells/ml
Dilution stages includes = 1:100 ; 1:100 & 1:3
The consecutive dilution stages can be calculated as:
= (6.74 x 10⁶) × (1/100)(1/100)(1/3)
= 222.42 cells/ml
= 2.22 × 10² cells/ml
So after addition 0.1 ml of the final dilution to a spread plate, the number of CFUs ( Colony forming units) we expect to count will be:
= (2.22 × 10² cells/ml)(0.1 ml)
= 22.2 cells
≅ 22 cells.
Answer:
C. a haploid cell produced by meiosis
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01:29:49
During photosynthesis, the energy from sunlight is used to split water molecules. What happens to the hydrogen ions that are
split from the water molecules?
Answer: hydrogen is incorporated into glucose
Explanation:
6CO2 + 6H2O —> C6H12O6 + 6O2
Answer:
The correct answer is 18 ATP and 12 NADPH and noncyclic photophosphorylation can produce ATP molecule in the absence of cyclic photophosphorylation.
Explanation:
ATP and NADPH molecules are synthesized during light reaction of the photosynthesis which is utilized during the reactions of the dark phases. Dark reaction or Calvin cycle (C3 cycle) is the cyclic pathway of producing glucose triose phosphates (3C) from carbon dioxide and water. This reaction proceeds into 3 phases: carboxylation, reduction and regeneration.
First ATP and NADPH are utilized during the reduction step in the reduction of 3-phosphoglycerate to glucose-3-phosphate by the transfer of phosphate group from 6 ATP to 3-phosphoglycerate and 6 NADPH reduction as it donates an electron. Regeneration step also uses 3 ATP in conversion of G3P to RUBP molecules.
A total of 9 ATP and 6 NADPH are utilized in producing 3C G3P molecule, So, to produce 6C sugar molecule 18 ATP AND 12 NADPH are used.
During chemiosmosis synthesis of ATP, 4 protons produce 1 molecule of ATP which could be able to generate 3 molecules of ATP for each pair of NADPH. so, noncyclic photophosphorylation or Z-scheme will be able to produce ATP in the absence of cyclic photophosphorylation.
Thus, 18 ATP and 12 NADPH and noncyclic photophosphorylation can produce ATP molecule in the absence of cyclic photophosphorylation are the correct answer.
Answer:
Prometaphase (late prophase)
Explanation:
Mitosis is the kind of cell division in which two genetically identical copies of a cell is formed from a cell. Mitosis comprises of stages which include prophase, metaphase, anaphase and telophase. Each phase is characterized by specific occurrences that all work together to achieve the main goal of forming two daughter cells.
According to this question, PROPHASE stage is the stage characterized by the condensation of chromatids into chromosomes, the disappearance of nucleolus and BREAKDOWN of NUCLEAR MEMBRANE.