Answer:
The Laura tried to grow micro-organisms in culture broth medium in Petri-plates but the microbial growth was not observed even on the fourth day of culture which could be due to:
1. Laura could have not provided the correct broth medium maintaining the with pH and salt concentrations.
2. Microorganisms did not get optimal growth conditions.
3. The bacterial source could have died after the collection which could have lysed after their death and thus no longer visible.
Answer:
yes it uses energy to reproduce inside a host cell
Explanation:
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.
The answer is B. (an increase in inbreeding) hope this helped !!
Answer:
See the answer below
Explanation:
Let the disorder be represented by the allele a.
Since the disease is an autosomal recessive one, affected individuals will have the genotype aa and normal individuals will have the genotype Aa or AA.
Since the four adults are carriers, their genotypes would be Aa.
Aa x Aa
Progeny: AA 2Aa aa
Probability of being affected = 1/4
Probability of being a carrier = 1/2
Probability of not being affected = 3/4
(a) The chance that the child second child of Mary and Frank will have alkaptonuria = 1/2
(b) The chance that the third child of Sara and James will be free of the condition = 3/4
(c)
(d) If someone has no family history of the disorder, their genotype would be AA.
AA x aa
4 Aa
<em>The chance that a child with alkaptonuria will have an offspring with alkaptonuria if the child's mate has no family history </em>= 0
(e)
(f) <em>The chance that a child with alkaptonuria will have an offspring with alkaptonuria if the child's mate has no family history</em> = 0
