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Liula [17]
3 years ago
15

Make a claim about how the body's external defenses help to prevent diseases.

Biology
1 answer:
harina [27]3 years ago
6 0

Skin prevents the body from getting infections

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Many mammals control their body temperature by sweating. which property of water is most directly responsible for the ability of
Andreas93 [3]
Sweat evaporates, it is through the process of evaporation that heat is taken away from the body and into the surrounding atmosphere 
3 0
3 years ago
What are the 4 phases of Mitosis?
lozanna [386]

Answer:

During mitosis, when the nucleus divides, the two chromatids that make up each chromosome separate from each other and move to opposite poles of the cell. ... Mitosis actually occurs in four phases. The phases are called prophase, metaphase, anaphase, and telophase.

Explanation:

I hope this helps :))

(This is from google)

7 0
3 years ago
Create a phulogeny using the following taxa: monarh butterfly, rainbow trout, ochre sea star, African elephant, earthworm, nemat
masha68 [24]

Answer:

http://etetoolkit.org/treeview/?treeid=e29a1e3f1e8914f370f5bb2415bbcf32&algid=ce443ed1e53858bf4e11d1e069c7a927

You can view the tree by clicking view tree...

Explanation:

I think that the Ctenophora are the first to diverge from the origin then Calcerea goes for second divergence and it goes straight in the order as provided by your data.

It also shows the organization of evolutionary pathway from simple (Ctenophorea and Calcerea) to complex like mammals (African Elephant)...

Download txt
7 0
2 years ago
Assuming that the average protein in E. coli consists of a chain of 380 amino acids, calculate the maximum number of proteins th
Afina-wow [57]

Answer:

4210.5 proteins

Explanation:

(a)The amount of nucleotide pairs in the DNA molecule calculated using

=molecular weight of DNA/molecular weight of a single pair

molecular weight of DNA

= 3.1 × 10^9g/mol

molecular weight of a single pair

= 0.66 ×10^3g/mol

= (3.1 × 10^9g/mol)/(0.66 × 10^3g/mol)

= 4.7 10^6pairs

Lets, Multiply the number of pairs with the length per pair

Take, length per pair = 0.34 nm/pair

= (4.7 × 10^6pairs)(0.34 nm/pair)

= 1.6 × 10^6nm

= 1.6 mm

= (approx 2mm or 0.002 mm).

Therefore,

the DNA is

(1.6 mm)/(0.002 mm)

= 800

The DNA is 800 times times longer than the cell.

This indivates Tat DNA must be tightly coiled to fit into the cell.

(b) the amount of DNA molecule nucleotide pairs has given

= 4.7 × 10^6 nucleotide pairs,

From solution (a), there must be one-third this number of triplet codons

= (4.7 × 10^6)/3

= 1.6 × 10^6codons

If an individual protein has an average of 380 amino acids, each needs a codon,

the number of proteins that can be coded by E. coli DNA

= (1.6 × 10^6codons) × (1 amino acid/codon)

---------------------------------------------

(380 amino acids/protein)

= 4210.5 proteins

________________________

Here is the complete question

The genetic information contained in DNA consists of a linear sequence of coding units known as codons. Each codon consists of three adjacent DNA nucleotides that corresponto a single amino acid in a protien.

The E.coli DNA molecule contains 4.70 x 10^6 base pairs. Determine the number of codons that can be present.

Assuming that the average protein in E.coli consists of a chain of 400 amino acids, calculate the maximum number of protiens that can be coded by an E.coli DNA molecule

7 0
3 years ago
The first is red cell (erythrocyte)
krok68 [10]

Answer:

leukocyte

Explanation:

leukocyte is the organelle missing in the red blood cell.

3 0
2 years ago
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