A philodendron plant and when it is in sun light it produces more bubbles.
O+ 1 in 3 <span>37.4%
</span>O- 1 in 15 <span>6.6%
</span>
O- can receive O-
O+ can receive O+, O-
O can receive<span> O, A, B, AB</span>
The end result is having two identical daughter cells.
Answer:
The enzyme is the pyruvate kinase
Explanation:
The designated enzyme can only grow in pyruvate or in ethanol. They are said to be in anaerobic conditions. If the end product of glycolysis is pyruvate and the enzyme that transforms phosphoenol pyruvate into pyruvate is called pyruvate kinase. In a reaction under anaerobic conditions, the pyruvate is transformed into ethanol. So if the pyruvate kinase enzyme is mutated, pyruvate cannot be transformed into ethanol and Saccharomyces cerevisiae could not survive in that medium.
Answer:
1000 mM
Explanation:
Using V = -60 mV㏒₁₀[K/K'] where V = membrane potential = -60 mV, K = intracellular concentration = unknown and K' = extracellular concentration = 100 mM
So, V = -60 mV㏒₁₀[K/K']
-60 mV = -60 mV㏒₁₀[K/K']
dividing both sides by -60mV, we have
-60 mV/-60 mV = ㏒₁₀[K/K']
㏒₁₀[K/K'] = 1
taking antilogarithm of both sides, we have
[K/K'] = 10¹
multiplying both sides by K', we have
K = 10K'
K = 10 × 100 mM
K = 1000 mM
So, the intracellular Cl- concentration is 1000 mM
