Researchers have identified specific gene variants in the receptors that detect sweetness: TAS1R2 and TAS1R3. There is also high variation in the detection of bitterness. However, the story is more complicated than sweet taste, as we have 25 receptors that detect different bitter molecules
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:
C
Explanation:
From reading what the process is, I think the answer should be C. The definition always says that you get energy from a chemical reaction and the products are CO2 and water. That can only mean that you have a hydrocarbon burning.
Chemically it looks like
C6H12O6 + 6O2 ====> 6CO2 + 6H20
