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:
Size of the band varied
Explanation:
In the given question, the options are not provided but based on the information provided in the question, it can be predicted.
When the DNA was digested by the HindIII, then the length of the digest came about 1000 bp and the undigested DNA was also found 1000 bp.
When both the bands were analysed by Gel electrophoresis then the bands do not run to the same distance therefore this shows that the size of both the bands was not equal and their size varied.
Thus, the size of the band varied is correct.
Answer:
ICD-10-CM code: Z48.02
Explanation:
Z48. 02 is a billable ICD code for indicating a diagnosis of encounter for sutures removal.
It is an acceptable code for the year 2020 for filing of HIPAA-involved transactions. The ICD-10-CM code Z48.02 also shows conditions or terms such as removal of sutures done.
