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:
D.-
Explanation:
A control group is needed to carry out the experiment, that control group is the one that is not injected with anything. Ethanol and pesticide need to be injected into separated groups to document whether it is one or the other that is harmful
Lasting genetic similarities make the two bears biological sister species, more closely related to one another than to any others. ... But Ursus maritimus has also evolved to match its polar environment, growing a longer snout than brown bears' and larger, paddle-like paws, both of which aid polar bears in hunting seals.
The new organism should be placed with the bird group based upon the homology obtained by the genomic data. This would imply a shared common ancestor. The trait shared with bats but not birds could be an analogous trait.
Survival would be correct
