A physiotherapist will assess the child and will decide whether or not the child will need to have physiotherapy everyday. In some cases, physiotherapy will not be needed everyday but only when the child gets a cold or is experiencing coughing.
Answer:
Small intestine, liver, bile and lipase.
Explanation:
Digestion of fat occurs in the small intestine. Its digestion occurs with the help of bile, that is made in the liver. Bile breaks the fat into small drops that are easier for the lipase enzymes to change it. Lipase enzymes is a type of enzymes that works only on lipids and lipids are broken down into fatty acids and glycerol. These substances are absorbed by our body and used it for producing ATP for the body.
In scientific reasoning, a hypothesis is made before any applicable analysis has been done. A theory, on the opposite hand, is supported by evidence: it is a principle shaped as an effort to clarify things that have already been supported by knowledge<span>.
</span>For example: “It's bright outside.”
Hypothesis: A projected clarification for a development created as a place to begin for additional investigation.
Theory: A well-substantiated rationalization nonheritable through the methodology and repeatedly tested and confirmed through observation and experimentation
The answer is alleles detached from one another during
anaphase of meiosis I, when the homologous pairs of chromosomes separate. During
anaphase I, homologous pairs are drawn apart, and
they go in the direction of the opposites of the cell. Meiosis I finishes
with the manufacture of two haploid daughter cells for the reason that
the homologous pairs of chromosomes have been separated.
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