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
Answer:
The tryptophan synthase enzyme catalyzes the synthesis of tryptophan. This enzyme is found in all kingdoms except animals.
Tryptophan synthase has a hydrophobic channel that facilitates the passage of an intermediate from one enzyme to the other during the biosynthesis of tryptophan, which makes faster the metabolic process.
The answer is <span>bird droppings that contain seeds from a different location.
</span><span>Gene flow includes a transfer of alleles from one population to another. So imagine a population of plant A on location 1 and another population of plant A on location 2. They cannot interchange their genes. But if birds drop seeds from location 1 and those seeds get to location 2, gene flow may occur.</span>