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
The answer is true. The prosthetic group is present in
several components of the electron transport chain. The complexes one, two and
three can have a presence of prosthetic group in which is being referred to as the
iron sulfur clusters.
The R group should have the ability to form hydrogen bonds.
The R-group take a lot of different chemical forms and can
be acidic, basic, polar or non polar. Amino acids that are either acidic or basic
consequently have either positive or negative charge. These types of R-groups
are stabilized when surrounded by water.
Due to their structure and function they are differ from each other
Answer: The answer is number 2 and number C
Explanation: BECAUSE
