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 right answer for the blank is cleavage.
Explanation:
After fertilization the next step is cleavage in which zygote divide mitotically. As the embryo move from oviduct down the fallopian tube it divide into two cell and then four cells. four cell embryo is known as blastomere.
Answer:
Explanation:
An international team of scientists led by the Lomonosov Moscow State University group made a significant step in creating a new type of drug for treatment of autoimmune diseases, such as rheumatoid arthritis and Crohn's diseaase
I pretty sure I answered this and it was B please tell me if I’m wrong
