D.) Both are circumstantial evidence....
Answer:
B
Explanation:
I learned this 2 years ago
C. membrane bound organelles and no distinct nucleus
Answer:
The foreign gene might be lost
Explanation:
Restriction enzymes have two properties useful in recombinant DNA technology.They cut DNA into fragments of a size suitable for cloning at palindromic sites. Many restriction enzymes make staggered cuts that create single-stranded sticky ends conducive to the formation of recombinant DNA. The foreign might be cleaved and removed from the plasmid. plasmid is an extrachromosomal strand in bacteria.
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
