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AveGali [126]
3 years ago
5

A school aged child is brought to the clinic by her parents. The parents state that the child will be engaged in some activity a

t home and then just stop suddenly for a few seconds and then pick up the activity again as if there had been no break in what she was doing. This child should be assessed for what neurological disorder?
a.) Tonic-clonic seizures
b.) Absence seizures
c.) Myoclonic seizures
d.) Status epilepticus
Biology
2 answers:
TiliK225 [7]3 years ago
8 0

Answer:

Absence seizures

Explanation:

This is a  type of seizure also known as ( Petit Mal seizures).It   is common in children, (between 4-14 years) and  characterised with intent stare at empty space for few seconds(which may occur in 10 -100 times in a day) and  does not caused any significant harm.However,with frequent occurrence it may  affect concentration.It does not exceed  <u>15 seconds</u>

it is caused by repeated electrical signals in the brain due to interference  in the synaptic transmission and also  to alterations in levels of Neurotransmitters .

Staring into empty space,

Fluttering of eyelids

Intermittent stoppage of speech  are some of the symptoms.

Treatment is with medications.,

Ivenika [448]3 years ago
3 0
B.) Absence seizures, or petit-mal seizures, which in English literally means “small bad”, signifying that these seizures are relatively less harmful. Hope this helps!
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In gorillas, the ability to roll the tongue is under the control of 1 gene. The R allele, which confers tongue-rolling ability,
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Answer:

0.4550

Explanation:

In this example, R allele is dominant, so individuals RR and Rr can roll their tongues. If in a population of 1000 gorillas, there are 575 gorillas who can roll their tongues, they will be RR and Rr.

In this case, there is Hardy Weinberg Equilibrium, so the following will be true:

p^{2}+2pq+q^{2}=1

Where:

p^{2}= frequency of RR

2pq= frequency of Rr

q^{2}= frequency of rr

The question is what is the frequency of heterozygotes, or, what is the value of 2pq.

We know that RR+Rr is 575 individuals in a population of 1000, or 0.575.

In other words:

p^{2}+2pq=0.575

So, it is possible to find q^{2}:

q^{2}= 1-2pq-p^{2}  \\q^{2}=1-(0.575)\\q^{2}=0.425

Now, there are two alleles in the population, so the following will be true:

p+q=1

It is possible to find q  (the frequency of allele r) and p (the frequency of allele p):

q=\sqrt{q^{2} }\\ q=\sqrt{0.425} \\q=0.652

Therefore:

p=1-q\\p=1-0.652\\p=0.348

Now, the frequency of heterozygotes or 2pq is:

2pq=2*0.652*0.348\\2pq=0.45

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