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MrRa [10]
3 years ago
6

Organelles and other cellular material are held within a cell by which of the following?

Biology
1 answer:
PtichkaEL [24]3 years ago
4 0
The answer is letter A.

<span>Organelles and other cellular material are held within a cell by a cellular membrane. These membranes are what consist of a cell's exterior.

</span>

<span>Thank you for posting your question. I hope you found what you were after. Please feel free to ask me more.</span>

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insens350 [35]

Answer:

by using half to balance the equation

5 0
3 years ago
Which of the following prevent uncontrolled cell growth (cancer)?
Nana76 [90]
Check points throughout the cell cycle that determine when the cell should begin the next phase
5 0
3 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
Which describes the correct paring of DNA?
faust18 [17]

Answer:

what are the options?

may I get brainliest please?

8 0
3 years ago
A performer, seated on a trapeze, is swinging back and forth with a period of 8.90 s. If she stands up, thus raising the center
iren2701 [21]

Answer:

8.800s

Explanation:

When the performer swings, she oscillates in SHM about Lo of the string with time period To = 8.90s.

First, determine the original length Lo, where for a SHM the time period is related to length and the gravitational acceleration by the equation

T = 2π×√(Lo/g)..... (1)

Let's make Lo the subject of the formulae

Lo = gTo^2/4π^2 ..... (2)

Let's put our values into equation (2) to get Lo

Lo = gTo^2/4π^2

= (9.8m/s^2)(8.90s)^2

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

4π^2

= 19.663m

Second instant, when she rise by 44.0cm, so the length Lo will be reduced by 44.0cm and the final length will be

L = Lo - (0.44m)

= 19.663m - 0.44m

= 19.223m

Now let use the value of L into equation (1) to get the period T after raising

T = 2π×√(L/g)

= 2π×√(19.223m/9.8m/s^2)

= 8.800s

7 0
2 years ago
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