Answer:
<em>The correct option is B) Missense mutation.</em>
Explanation:
A missense mutation can be described as a type of point mutation in which a mutation in a single nucleotide causes the codon to change and form a different type of amino acid. As in the above scenario, we can see that the code CAU coded for the amino acid, Histidine, before a point mutation occurred. After the point mutation, the codon was changed to UAU instead of CAU. UAU codes for the enzyme, tyrosine, instead of histidine.
Answer: The enzyme is found in different types of cells.
Explanation:
The result that can be concluded is that this enzyme is found in different types of cells. This is the reason that the medicines like aspirin and other non-steroidal anti inflammatory drugs inhibit the prostaglandin pathway and other pathways too.
These medication acts on the enzymes and it will act in the cells where this enzyme will be present. Hence, the enzyme is acting in different types of cells.
It is considered that the child's IQ will be 50% more than parents IQ.
<h3>let's take an example</h3>
If parent's IQ is 100
50% of 100 is 50
Now add these values
100+50=150
So, The child's IQ will be always higher than parents.
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
