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:
look below
Explanation:
My species is one of many
In order to solve the problem try it this way
a phylum is a level of classification
taxonomy is the scientific study of naming
All animals have a kingdom
Genus is a taxonomic rank
It is the b. food chain and begins with plant, then animals that eat plants, and finally animals that eat other animals.
Answer: DNA contains the nucleotides guanine (G), cytosine (C), adenine (A), and thymine (T). RNA contains the nucleotides guanine (G), cytosine (C), uracil (U), and thymine (T). The difference between the two is that RNA contains uracil (U) rather than adenine (A).
The base pairs in DNA are adenine with thymine and guanine with cytosine. The base pairs in RNA are uracil with thymine and guanine with cytosine.
