Answer:
Explanation:
Normally, under anaerobic condition in yeast, pyruvate produced from glycolysis leads to the production of ethanol as shown below.
pyruvate ⇒ acetaldehyde + NADH ⇒ ethanol + NAD
The pyruvate is converted to acetaldehyde by the enzyme, pyruvate decarboxylase. It should be NOTED that carbon dioxide is released in this step. The acetaldehyde produced in the "first step" is then converted to ethanol by the enzyme alcohol dehydrogenase. It must be noted from the above that the steps are irreversible.
If a mutated strain of yeast is unique because it does not produce alcohol and lactic acid (which is referred to as toxic acid in the question); thus having a high level of pyruvate because of the presence of a novel enzyme. <u>The function of this novel enzyme will most likely be the conversion of acetaldehyde in the presence of carbondioxide back to pyruvate; thus making that step reversible</u>. This could be a possible explanation for the high level of pyruvate present in the yeast.
Answer:
The correct option is : a. diameter
Explanation:
The Kirby–Bauer test or the disk diffusion test, is a method to determine the antibiotic sensitivity of the given bacteria. This test involves the use of antibiotic discs to determine the effect of antibiotics on the bacteria.
In this test, the wafers having antibiotics and the bacteria are placed on the agar plate and incubated. If the antibiotics present stops the growth of the bacteria, there will be an area around wafer with no bacterial growth, such an area is known as the zone of inhibition.
<u>The </u><u>diameter of this zone of inhibition</u><u> is measured to determine the </u><u>antibiotic sensitivity of the given bacteria</u><u>.</u>
Answer:
64
Explanation:
Since the population is in Hardy-Weinberg equilibrium and the equilibrium is defined by the equation
p² + 2pq + q²
where the frequency of the homozygous dominant allele F ( feathers on the leg ) = p²,
F of homozygous ( featherless legs) = q²
F of heterogyous ( carrier of featherless allele) = 2pq
F (q²) = 128/200 = 0.64
q = √0.64 = 0.8
and p + q = 1
p = 1 - 0.8 = 0.2
2pq = 2 × 0.2 × 0.8 = 0.32
number of chickens that are heterozgous for the feathered leg allele = 0.32 × 200 = 64
