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
The correct answer should be B. Small regions interspersed with other biomes. This type of a biome is found on all continents and in many countries. It's specific because it's similar to a desert when it comes to yearly precipitation, but it is different insofar that it has a lot of plants and animals that are mostly nocturnal predators.
Answer:
<em>See attached punnet square</em>
Explanation:
Attached is the punnet square that shows the Mendelian assortment of the allele for tongue rolling between a homozygous dominant and heterozygous parents.
Genotype Probability
TT 50%
Tt 50%
Phenotype Probability
Tongue Roller 100%
Answer:
B is the correct answer
Explanation:
All the other answers are opinion based
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