Answer:
New species can appear gradually through small changes in an ancestral species.
Explanation:
The new species that appear are due to hereditary variations that occur in a population. The adaptive variations are said to confer a selective advantage to organisms possessing them. The result of variations is that well adapted individuals are able to survive and reach the reproductive age and pass over their favourable characteristics to their offspring.
Answer:
Null Hypothesis -
The observed frequency is approximately equal to the expected frequency of phenotype.
Explanation:
Pure Breeding Cross - TTww x ttWW
Genotype of offspring in F2 generation - TtWw
Null Hypothesis -
The observed frequency is approximately equal to the expected frequency of phenotype.
The chi square analysis is attached
The degree of freedom for this question is 3
The p value for X^2 estimated through chi square test is 0.5
Hence the null hypothesis is accepted.
Biogenesis, the fir tree began life when fertilization occurred to produce a singlee cell.
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
