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3 years ago

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There are 81 amino acid residues in the precursor molecule of bovine insulin. What is the least number of bases that must be in

the DNA chain that codes for this molecule?

2 answers:

Usimov [2.4K]3 years ago

8 0

If you count the methionine expressed for the start codon as part of the precursor protein, then (81+1)*3. (The start codon is expressed, but the stop codon isn't.)

Andreyy893 years ago

5 0

Answer:

246 amino bases

Explanation:

An amino acid is coded by three bases. However, in a DNA molecule, there is coding that takes place. In DNA replication, the start and stop codons are not expressed during the formation of the double helix. Hence, an extra amino acid will be needed. In fact, this can be expressed mathematically like this:

$Number of bases = (n+1) * 3$

where n is the number of amino acid residues. Therefore, taking 81 amino acid residues yields:

N = $(81+ 1) * 3\\246$

Therefore, the least number of bases needed will be 246

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Determine Z and V for steam at 250°C and 1800 kPa by the following: (a) The truncated virial equation [Eq. (3.38)] with the foll

makvit [3.9K]

Answer:

Explanation:

Given that:

the temperature $T_1$ = 250 °C= ( 250+ 273.15 ) K = 523.15 K

Pressure = 1800 kPa

a)

The truncated viral equation is expressed as:

$\frac{PV}{RT} = 1 + \frac{B}{V} + \frac{C}{V^2}$

where; B = - $152.5 \ cm^3 /mol$ C = -5800 $cm^6/mol^2$

R = 8.314 × 10³ cm³ kPa. K⁻¹.mol⁻¹

Plugging all our values; we have

$\frac{1800*V}{8.314*10^3*523.15} = 1+ \frac{-152.5}{V} + \frac{-5800}{V^2}$

$4.138*10^{-4} \ V= 1+ \frac{-152.5}{V} + \frac{-5800}{V^2}$

Multiplying through with V² ; we have

$4.138*10^4 \ V ^3 = V^2 - 152.5 V - 5800 = 0$

$4.138*10^4 \ V ^3 - V^2 + 152.5 V + 5800 = 0$

V = 2250.06 cm³ mol⁻¹

Z = $\frac{PV}{RT}$

Z = $\frac{1800*2250.06}{8.314*10^3*523.15}$

Z = 0.931

b) The truncated virial equation [Eq. (3.36)], with a value of B from the generalized Pitzer correlation [Eqs. (3.58)–(3.62)].

The generalized Pitzer correlation is :

$T_c = 647.1 \ K \\ \\ P_c = 22055 \ kPa \\ \\ \omega = 0.345$

$T__{\gamma}} = \frac{T}{T_c}$

$T__{\gamma}} = \frac{523.15}{647.1}$

$T__{\gamma}} = 0.808$

$P__{\gamma}} = \frac{P}{P_c}$

$P__{\gamma}} = \frac{1800}{22055}$

$P__{\gamma}} = 0.0816$

$B_o = 0.083 - \frac{0.422}{T__{\gamma}}^{1.6}}$

$B_o = 0.083 - \frac{0.422}{0.808^{1.6}}$

$B_o = 0.51$

$B_1 = 0.139 - \frac{0.172}{T__{\gamma}}^{ \ 4.2}}$

$B_1 = -0.282$

The compressibility is calculated as:

$Z = 1+ (B_o + \omega B_1 ) \frac{P__{\gamma}}{T__{\gamma}}$

$Z = 1+ (-0.51 +(0.345* - 0.282) ) \frac{0.0816}{0.808}$

Z = 0.9386

$V= \frac{ZRT}{P}$

$V= \frac{0.9386*8.314*10^3*523.15}{1800}$

V = 2268.01 cm³ mol⁻¹

c) From the steam tables (App. E).

At $T_1 = 523.15 \ K \ and \ P = 1800 \ k Pa$

V = 0.1249 m³/ kg

M (molecular weight) = 18.015 gm/mol

V = 0.1249 × 10³ × 18.015

V = 2250.07 cm³/mol⁻¹

R = 729.77 J/kg.K

Z = $\frac{PV}{RT}$

Z = $\frac{1800*10^3 *0.1249}{729.77*523.15}$

Z = 0.588

3 0

3 years ago

What is the difference between 2 ml of salt and 2.0 ml of salt?

faltersainse [42]

Answer:

The 2 ml and 2.0 ml is the same thing.

8 0

3 years ago

In terms of bonds, what would the molecule C₆H₁₂ be classified as?

Colt1911 [192]

Answer:

Alkene

Explanation:

5 0

3 years ago

Which of the following is true?

Tomtit [17]

Answer:

a. A reaction in which the entropy of the system increases can be spontaneous only if it is endothermic.

Explanation:

The change in free energy (ΔG) that is, the <u>energy available to do work</u>, of a system for a constant-temperature process is:

$ΔG = ΔH - TΔS$

When ΔG < 0 the reaction is spontaneous in the forward direction.

When ΔG > 0 the reaction is nonspontaneous. The reaction is

spontaneous in the opposite direction.

When ΔG = 0 the system is at equilibrium.

If <u>both ΔH and ΔS are positive</u>, then ΔG will be negative only when the TΔS term is greater in magnitude than ΔH. This condition is met when T is large.

3 0

3 years ago

Be sure to answer all parts.

sergey [27]

The moles of O2 are needed to react completely with 1.00 mol of C₂H₂ is 5 mol , 0.23 moles of C₂H₂ are needed to form 0.46 mol of CO₂ .

<h3>What is a Balanced Reaction ?</h3>

A balanced reaction is a reaction in which the number of atoms present in the reactants is equal to the number of atoms in the products.

It is given in the question that

the combustion of Acetylene is given by

2 H―C≡C―H + 5 O₂ → 4 CO₂ + 2 H₂O

Given moles of Acetylene = 1 mol

Moles of oxygen needed = 5 mol

as the mole fraction of Acetylene to Oxygen is 1 :5

The mole of Acetylene needed to form 0.46 mol of CO₂ is ?

mole fraction of CO₂ to Acetylene is 4:2

Therefore 0.23 moles of CO₂ is required.

To know more about Balanced Reaction

brainly.com/question/14280002

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5 0

2 years ago