Answer:
The correct answer is b. C6H12O6 -> 2 C3H4O3+2 H+
Explanation:
Glycolysis occurs in both the condition aerobic and anaerobic so it does not require oxygen. In glycolysis, one glucose molecule is converted into two pyruvate( 2 C3H4O3) and two 2 ATP, 2NADH, and 2 H₂O are produced.
Initially, 2NAD⁺ is produced during glycolysis which is reduced to produce 2NADH and 2 H⁺. Therefore the correct equation is C6H12O6 -> 2 C3H4O3+2 H+.
Then this pyruvate is used in the Kreb cycle which is required for the complete breakdown of glucose into carbon dioxide and water and this process occurs in aerobic conditions. Complete oxidation is important to produce more energy from partially oxidized glucose.
The answer is A, just took the test on edge
Hey! i need point to answer sorryyyy 2
Answer:
i dont know that one im sorry
Explanation:
Answer:
a. True
b. True
c. False
d. True
e. False
f. False
g. True
Explanation:
The homeotic genes refer to evolutionarily conserved genes that modulate the development of different structures in organisms of the same groups (in this case, plants). Moreover, homeobox genes are genes that encode transcription factors involved in the regulation of development in eukaryotic organisms. The knotted1 (<em>kn1</em>) gene is a plant homeobox gene is a member of the <em>kn1</em> homeobox (<em>knox</em>) gene family, which is responsible for maintaining indeterminacy and preventing cellular differentiation. In maize, <em>kn1</em> plays a key role in maintaining the cells of the shoot apical meristem in an undifferentiated state, being mainly expressed in shoot meristems during postembryonic stages of shoot development. It has been observed that maize mutant plants where <em>kn1</em> is ectopically expressed (i.e., in tissues in which this gene is not normally expressed) exhibit proximal-distal patterning defects.
