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2 years ago

What is chlorine and what is the atomic number

1 answer:

6 0

Chlorine-the chemical element of atomic number 17, a toxic, irritant, pale green gas

atomic number-the number of protons in the nucleus of an atom, which determines the chemical properties of an element and its place in the periodic table.

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Outline the levels of organization starting with the smallest, organisms

k0ka [10]

<span>organelle, cells, tissues, organs, organ systems, organisms, populations, communities, ecosystem, and biosphere. These have been arranged from the smallest to the largest. The biosphere is the largest level of organization while the organelle is the smallest. However, the basic form of life begins with the cell. Many cells together form a tissue and several tissue form an organ.
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3 years ago

How does a blind person know when they are done wiping?

Naily [24]

Answer:

they sniff it probably

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2 years ago

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WARM-UP

Naddik [55]

It’s burning wood ,heating water,Roasting food

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2 years ago

Which activity would help maintain the native biodiversity in a forest ecosystem?

slega [8]

D.

A will harm/kill the native species.

B just create electricity

C destroys the forest ecosystem and the organisms in it

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2 years ago

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Cystic fibrosis is an autosomal recessive disorder. In one population, the frequency of affected individuals (A 2 A 2) is 0.0004

Simora [160]

Hardy Weinberg Equilibrium

Explanation:

Hardy Weinberg Equilibrium formula for allelic frequency is: $p+q=1$

Hardy Weinberg Equilibrium formula for genotypic frequency is: $p^2+q^2+2pq=1$

Here, $p$ =frequency of dominant allele

$q$ =frequency of recessive allele

$p^2$ =frequency of dominant genotype

$q^2$ =frequency of recessive genotype

$2pq$ =frequency of heterozygous genotype

frequency of affected individuals=0.0004(given)

here,affected individuals means the one's having Cystic Fibrosis and since it is recessive disease so we can write frequency of recessive genotype as: $q^2$ =0.0004

If $q^2$ =0.0004 $q$ will be 0.02;frequency of recessive allele will be 0.02

Using the formula $p+q=1$

$p+0.02=1\\p=1-0.02=0.98$

hence,frequency of dominant allele will be 0.98

Frequency of dominant genotype will be $p^2=(0.98)^2=0.9604$
Frequency of heterozygous genotype will be $2pq=2*0.98*0.02=0.0392$

7 0

3 years ago