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stellarik [79]
3 years ago
13

At the set of gas exchange in the lungs, where does the oxygen move?

Biology
1 answer:
jeyben [28]3 years ago
8 0

Answer:

Gas exchange is the delivery of oxygen from the lungs to the bloodstream, and the elimination of carbon dioxide from the bloodstream to the lungs. It occurs in the lungs between the alveoli and a network of tiny blood vessels called capillaries, which are located in the walls of the alveoli nation:

Explanation:

i hope this helped you!!

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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

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molecular weight of a single pair

= 0.66 ×10^3g/mol

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= 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

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