EntranceBasic concepts and laws of chemistry. Basic concepts and laws of chemistry Chemistry and chemical technology
Basic concepts and laws of chemistry.
Molar mass, amount of substance, Avogadro's constant.
Molar volume of gas.
I. Calculation of the mass of a structural unit of a substance.
Determine the number of hydrogen atoms in a 9 g sample of water.
M(H2O) = 18 g\mol
n(H2O) = 9/18= 0.5 mol
From the formula it follows that 1 mole of H2O contains 2 moles of hydrogen atoms, i.e. n(H) = 2×0.5 mol = 1 mol.
Determine the number of hydrogen atoms in a 26.4 g sample of ammonium hydrogen phosphate.
M(NH4)2HPO4 = 132 g\mol
n((NH4)2HPO4) = 26.4/132= 0.2 mol
n(H) =0.2×9= 1.8 mol
Answer: 1.8 mol
The mixture consists of ethanol C2H5OH weighing 46 g and water weighing 72 g. Determine the number of oxygen atoms in the mixture.
n(C2H5OH) = 46/46 = 0.1 mol n(O) =0.1×1= 0.1 mol
n(H2O) =72/18 = 0.4 mol n(O) =0.4×1= 0.4 mol n(O) =0.1+0.4=0.5 mol
The mixture consists of acetic acid CH3COOH weighing 6 g and formic acid HCOOH weighing 9.2 g. Determine the number of oxygen atoms in the mixture.
n(CH3COOH) =6 /60 = 0.1 mol n(O) =0.1×2= 0.2 mol
n(HCOOH) =9.2/46 = 0.2 mol n(O) =0.2×2= 0.4 mol n(O) =0.2+0.4=0.6 mol
Determine the mass of a water sample containing 12.04 1022 water molecules.
n(H2O) = NNA= 12.04∙10226.02∙1023=0.2 mol
m(H2O) = 0.2×18= 3.6 g
Determine the mass of a sample of copper sulfate containing 6.021022 copper atoms.
Answer: 16
Calculate the mass of a sample of ammonium sulfate containing 3.011022 hydrogen atoms.
n(H) =NNA=3.0110226.02∙1023=0.05 mol
there are 8 oxygen atoms in the compound
n(NH4)2SO4 = 0.05 /8 = 0.00625 mol
m = 0.00625 ×132 g\mol = 0.825 g.
Determine the mass of a sample of magnesium hydrogen sulfate if it is known that it contains 3.6121023 oxygen atoms.
n(O) =NNA = 3.612×10236.02×1023=0.6 mol
n Mg(HSO4)2 = 0.6/8= 0.075 mol
m = 0.075×218= 16.35 g
Determine the mass of a solution of sodium sulfate in water containing 30.1 1022 sodium atoms and 6.021024 hydrogen atoms.
n (Na) =NNA =30.1∙10226.02∙1023=0.5 mol
the formula contains 2 sodium atoms, therefore: n(Na2 SO4) = 0.5/2 = 0.25 mol
m substance (Na2 SO4) = 0.25 × 142 g\mol = 35.5 g
n(H) =NNA= 6.02∙10246.02∙1023=10 mol
n(H2O) = 10/2=5 mol
m(H2O) =5×18=90 g
m solution = m substance (Na2 SO4) + m (H2O) = 35.5 + 90 = 125.5 g.
Determine the mass of a solution of ethanol C2H5OH in water containing 12.04 1022 carbon atoms and 24.08 1022 oxygen atoms.
n (C) =NNA= 12.04∙10226.02∙1023=0.2 mol
n (C2H5OH) = 0.2/2 = 0.1 mol
m(C2H5OH) = 0.1 ×46 = 4.6 g
n(O) =NNA =2.405∙10236.02∙1023=0.4 mol
n(H2O) =0.4/1=0.4 mol
m(H2O) =0.4×18= 7.2 g
msolution = m substance + m(H2O) = 4.6 + 7.2 = 11.8 g
Calculate the mass of one sodium atom.
ma= M (Na) / NA = 23/ 6.021023 = 3.8210-23 g.
Find the mass of three calcium atoms.
ma=3 M (Ca) / NA = (3×40) / 6.021023 = 19.910-23 g
Calculate the mass of 7 water molecules.
ma=7 M (H2O) / NA = (7×18) / 6.021023 = 20.910-23 g
Determine the mass of one SO3 molecule
ma= M (SO3) / NA = 80 /6.021023 = 13.3 10-23 g
Calculate the mass (in g) of five formula units of silicon(IV) oxide.
mfe=5×Mr×u=5×60×1.66∙10-27=4.98∙10-22 g.
Calculate the mass:
Sodium atom (g)
ma= M (Na) / NA = 23/ 6.021023 = 3.8210-23 g
ma=ArNa×u=23×1.6610-27 = 3.8210-23 g
Five molecules of white phosphorus P4 (kg)
Answer: 1.02910-24 kg
Ten molecules of rhombic sulfur S8 (g)
Answer: 4.2510-21 g
Three molecules of ozone O3 (in amu)
ma=3×16×3×a.u.m=144a.u.m.
Fullerene molecules C60 (mg)
mm = Mr(C60)×u = 1195.210-27 = 1.19510-24 kg = 1.19510-21 g = 1.195 10-18 mg
Three molecules of caffeine С8Н10O2N4(kg)
Answer: 9.6610-25 kg
Ammonium cation (g)
ma=Mr(NH4+)×u = 18×1.6610-27 = 2.9910-23 g
Anion SO42-(g)
Answer: 1.5910-22 g.
Hemoglobin molecules С2954H4516N780O806S12Fe4 (mg)
Answer: 1.0710-16 mg
Formula unit of sodium chloride (g)
Six formula units of potassium hydroxide (kg)
Three formula units of barium oxide (g)
The mass of a sulfur molecule is equal to the mass of eight oxygen atoms. How many sulfur atoms are in its molecule?
maO=(8×16)×1.6610-27=2.12∙10-26 g
Let us imagine the composition of the molecule as Sx, then
X = mmolecules of atom = 2.12∙10-26Ar∙u = 2.12∙10-2632×1.66∙10-27 = 4 atoms
How many times is the mass of a sucrose molecule C12H22O11 greater than the mass of a molecule of white phosphorus P4 (answer: 2.76 times)
Samples of methane CH4 and oxygen contain the same number of molecules. Find the ratio of the mass of oxygen to the mass of methane (answer: 2)
In the magnesium sample, the number of atoms is three times more number carbon atoms in diamond. Calculate the ratio of the mass of the magnesium sample to the mass of the diamond sample (answer: 6)
II. Calculations using the concept chemical quantity substances.
Calculate the chemical quantity:
Hydrogen in a portion containing 3.01 1024 H2 molecules
n (H2) =NNA=3.01∙10246.02∙1023=5 mol
Nitrogen in a portion of 5.6 dm3 (answer: 0.25 mol)
Sodium sulfate in a sample weighing 14.2 g (answer: 0.1 mol)
The mass of a molecule of a substance is 1.0610-22 g. Calculate molar mass substances.
M= ma×NA=1.06∙10-22×6.02∙1023=64 g\mol
Determine the number of molecules in the indicated portions of substances:
3.25 mol O2 (answer: 1.961024)
11.5 mol H2 (answer: 6.921024)
40 mol NH3(answer: 2.411025)
0.0125 mol H2O (answer: 7.521021)
Calculate the chemical amount in the indicated portions of substances:
3.921023 O2 molecules (answer: 0.651 mol)
14.7 1024 argon atoms (answer: 24.4 mol)
2.451023 formula units Na3PO4 (answer: 0.407 mol)
17.34 1024 H2SO4 molecules (answer: 28.8 mol)
Calculate the chemical quantity for the indicated portions of substances:
5.6 dm3 helium (answer: 0.25 mol)
1.12 dm3SO2 (answer: 0.05 mol)
5 m3 NH3 (answer: 2.23102 mol)
300 cm3 HCl (answer: 1.3410-2 mol)
Determine the volumes of the indicated portions of substances:
3.2 mol H2S (dm3) (answer: 71.7 dm3)
0.05 mol CH4 (cm3) (answer: 1120 cm3)
300 mol O2 (m3) (answer: 6.72 m3)
1.14 mol argon (dm3) (answer: 25.5 dm3)
Calculate the volume (dm3) of a portion of nitrogen containing 1.12 1023 molecules.
n(H2) =NNA= 1.12∙10236.02∙1023=0.18 mol
V (H2) = 0.18×22.4 = 4 dm3
Calculate the volume of a 2 kg portion of oxygen (answer: 1400 dm3)
Calculate the number of molecules in a portion of ammonia with a volume of 3.45 dm3 (answer: 9.271022)
Calculate the mass of a portion of ammonia containing 5.43 1024 molecules. (answer: 153 g).
Besides being memorable, .com domains are unique: This is the one and only .com name of its kind. Other extensions usually just drive traffic to their .com counterparts. To learn more about premium .com domain valuations, watch the video below:
Turbocharge your Web site. Watch our video to learn how.
Improve Your Web Presence
Get noticed online with a great domain name
73% of all domains registered on the Web are .coms. The reason is simple: .com is the where most of Web traffic happens. Owning a premium .com gives you great benefits including better SEO, name recognition, and providing your site with a sense of authority.
Here's What Others Are Saying
Since 2005, we"ve helped thousands of people get the perfect domain name
- Spoke to Ryan over the phone, and he helped me with all my questions! Great customer service and readily available. The process was easy and relatively quick, taking place over the course of only three business days. Would highly recommend for those that are considering buying a domain from hugedomains.com. A third-party gets involved called namebright.com but there is no additional charge and they facilitate the transfer of domains from severs. Cheers - Sammy Lam, 10/14/2019
- Frankly speaking before purchasing domain from HugeDomains.com, i was roo much scared as i read some negative review on some website.. but decided to take a chance. After purchase i was surprised, it was unbelievable my domain name purchase was successful and best part was i transferred it with other registrar and again hugggggggeeee surprise because my domain name get transferred within 30 minutes.. total time taken from purchase to transfer was less than 2 hours.... Too much thanks to HugeDomain.com and NameBright.com - Sandip Rajput, 10/14/2019
- Fast and smooth deal and transfer. Can recommend it! - Tom, 12/10/2019
- More
15. At a given temperature, 2 96 10 3 mol/g of substance is adsorbed from a surfactant solution with a concentration of 0.2 mol/l by some adsorbent. Determine the adsorption capacity of the adsorbent (in mol/g), if the constant a 0 07 mol/l.
[adsorbent capacity 4 00 10 3 mol/g]
16. Using the Freundlich equation, calculate the equilibrium concentration of acetic acid in solution if 1 g of coal
17. Determine the type of adsorption when acetone is dissolved in water, if the concentration of acetone in water is 29 g/l, surface
solution tension - 59 4 |
3 N/m, surface tension |
||||||
water - 73 49 10 |
3 N/m, T |
||||||
[adsorption positive, |
6 mol m2 ] |
||||||
18. With an increase in the concentration of the isobutyric acid solution from 0.125 to 0.250 mol/l, its surface tension decreased from 55.1 to 47.9 mN/m, and that of the isovaleric acid solution from 43.2 to 35.0 mN/m. Compare the adsorption values of substances in a given concentration range at 293 K.
19. Determine whether the amount of surfactant adsorption from aqueous solutions increases or decreases with increasing concentration (T 298 K), if the following experimental data are known:
[adsorption increases from 2 12 10 6 to 9 09 10 6 molm2 with increasing surfactant concentration in solution]
20. How many times will the adsorption of butyric acid from aqueous solution with increasing concentration according to experimental data (T 288 K):
c 102, |
|||||
3 , N/m |
21. Mixed equal volumes of 1% solutions of calcium chloride and sulfuric acid (assuming densities equal to 1 g/ml). Write the formulas of the micelles of the resulting calcium sulfate sol.
[the granule is negatively charged]
22. What volume of a solution of silver nitrate with a concentration of 0.001 mol/l should be added to 10 ml of a solution of sodium chloride with cNaCl 0.002 mol/l to obtain a sol whose granules are positively charged? Write a diagram of the structure of a sol micelle.
[the granule is positively charged, the volume of the AgNO 3 solution must be more than 0.02 l]
23. What minimum volume of ammonium sulfide with a concentration of 0.001 mol/L should be added to 15 ml of a solution of manganese(II) chloride with a concentration of 0.003 mol/L to obtain a sol with negatively charged particles?
24. A granule of Prussian blue Fe 4 Fe CN 6 3 moves to the anode in an electric field. What substance serves as a stabilizer? Write the formula of the micelle.
[stabilizer - K4 Fe CN 6 ]
25. To 100 ml of 0.03% NaCl solution (1 g/ml) added 250 ml of 0.001 Mr AgNO 3 solution. Write the formula for lumicelle sol. Which of the following electrolytes will cause coagulation of this sol with the lowest coagulation threshold: KCl, Ba NO 3 2,
K2 CrO 4, MgSO 4, AlCl3?
[the granule is negatively charged, therefore, the coagulant ions are cations. Lowest coagulation threshold for AlCl3]
26. A silicic acid sol was obtained by reacting solutions of K2 SiO 3 and HCl. Write the formula of a sol micelle and determine which of the electrolytes was in excess if counterions move to the cathode in an electric field?
[excess K2 SiO 3 ]
27. What volume of 0.001 M FeCl 3 solution must be added to 0.03 l of 0.002 M AgNO 3 solution so that the silver chloride sol particles move towards the anode in an electric field? Write the formula of micellesol.
[the volume of FeCl 3 must be more than 0.02 l]
28. The coagulation thresholds of iron(III) hydroxide hydrosol with sodium sulfate and potassium chloride are 0.32 and 20.50 mmol/l, respectively. Determine the sign of the charge of colloidal sol particles. Calculate the coagulating ability of these electrolytes and compare their ratio with the calculated Schulze–Hardy rule.
29. The coagulating ability of electrolytes in relation to a certain sol decreases in the sequence: NH 4 3 PO 4 NH 4 2 SO 4 NH 4 NO3. What is the charge sign of colloidal particles? Give examples of electrolytes whose coagulating ability will be approximately equal to the above.
[sol particles are positively charged, the coagulating ability of ions hardly decreases: PO 3 4 SO 2 4 NO 3 ]
30. The coagulation threshold of the sol with magnesium sulfate is less than that of nitratombarium. How are sol particles charged? What can we say about hecoagulation of the same sol with potassium phosphate?
31. Coagulation of gold sulfide sol with a volume of 1.5 l occurred with the addition of 570 ml of sodium chloride solution with a concentration of 0.2 mol/l. Calculate the coagulation threshold of sodium ion sol.
32. The threshold for coagulation of sol hydroxide and iron phosphate ions is 0.37 mmol/l. What volume of 5% sodium phosphate solution (105 g/ml) is required to coagulate 750 ml of sol?
[volume of Na 3 PO 4 solution is 0.87 ml]
33. Clear coagulation of a 2-liter aluminum hydroxide sol occurred when 10.6 ml of a solution with a K4 Fe CN 6 concentration of 0.01 mol/l was added. Calculate the threshold for coagulation of the sol by hexacyanoferrate ions; write the formula of aluminum hydroxide sol micelles.
34. The threshold for coagulation of gold sulfide sol with calcium ions is 0.69 mmol/l. What volume of solution with a calcium chloride concentration of 0.5 mol/l is required to coagulate 100 ml of sol?
[volume of CaCl2 solution is 0 15 10 3 l]
35. Determine the sign of the charge of colloidal particles of the sol if the following coagulation thresholds (in mmol/l) are obtained during its coagulation with electrolytes: c pc KNO3 300; c pc MgCl2 320; c pc Na3 PO4 0 6.
[positive]
36. Which of the electrolytes Na2 SO4 or MgCl2 will have greater coagulating ability for a silver iodide sol obtained by mixing equal volumes of a solution with a potassium iodide concentration of 0.01 mol/l and a solution with a silver nitrate concentration of 0.015 mol/l?
[coagulating ability of Na2 SO4 is greater]
37. To coagulate a negatively charged sol of 10 ml of AgI, 1.5 ml of a solution with a KNO3 concentration of 1 mol/l or 0.5 ml of a solution with a Ca NO3 2 concentration of 0.1 mol/l or 0.2 ml of a solution with an Al NO3 concentration of 0.01 mol/l is required . Is the rule c p 1 z 6 true?
[the rule is carried out approximately]
38. Write the formula of a micelle of a gold sol stabilized by KAuO2. Which electrolyte - NaCl, BaCl2, FeCl3 - will have a lower coagulation threshold?
[coagulation threshold is lower in case of FeCl3]
39. The density of olive oil at 22 Æ C is 960 kg m3, and the density of water at this temperature is 996 kg m3. Olive oil flows through a viscometer in 21 minutes 15.6 seconds, and the same volume of water
water 22 H2 O 9 58 10 4 N s m2. |
||
[viscosity of olive oil 841 3 |
10 4 N s m2 ] |
|
40. Viscosity of kerosene at 20 |
Æ C is equal to 1 8 |
10 3 Pa s, and vis- |
bone of water under the same conditions - 1 005 10 |
3 Pa s (N s m2). |
|
Determine the density of kerosene if it is known that the flow time of kerosene from the viscometer is 53 s, and the same volume of water is 24 s. The density of water is 998 kg m3.
[kerosene density is 809 4 kg m3]
41. To which electrode will protein particles pI 4 0) move during electrophoresis in an acetate buffer prepared from 100 ml of a solution with a sodium acetate concentration of 0.1 mol/l and 25 ml of a solution with an acetic acid concentration of 0.2 mol/l?
[in this buffer solution the protein is negatively charged and during electrophoresis moves to the anode]
42. The solution contains a mixture of proteins: globulin (pI 7), albumin (pI 4 9) and collagen (pI 4 0). At what pH value can these proteins be separated electrophoretically?
[proteins can be separated electrophoretically at pH 4-9]
43. To which electrode will protein particles move during electrophoresis if its pI is 4 and pH is 5?
[protein particles will move towards the anode]
44. Hemoglobin pI 6 68 was placed in buffer solution with a hydrogen ion concentration of 1 5 10 6 mol/l. Determine the direction of movement of hemoglobin molecules during electrophoresis. It is known that in red blood cells the pH is 7 25. What charge do hemoglobin molecules have at this pH value.
[hemoglobin molecules will move towards the cathode; at pH 7 25 hemoglobin molecules have a negative charge]
45. The osmotic pressure of an aqueous protein solution with a mass concentration of 1 kg m3 at a physiological temperature is 292.7 Pa. Determine the molecular mass of the protein from the average molar mass (the protein molecule is isodiametric).
[relative molecular weight is 87,940]
46. Calculate the average osmotic pressure at 25 centration - 4,176 kg m3;
the molar mass of polystyrene, if Æ C is equal to 120.9 Pa, and the mass con- 1 Pa m6 kg2.
47. Will gelatin (pI 4 7) swell in an acetate buffer with equal contents of components at 0 Æ C? How can you identify the swelling process of gelatin? Explain your answer.
[swelling is minimal near the isoelectric point]
48. When rubber weighing 200 g swelled, 964 ml of chloroform (1 9 g ml) were absorbed. Calculate the degree of swelling of the rubber and the percentage composition of the resulting jelly.
[ 915 8%; rubber 9 84%; chloroform 90 16%]
49. Calculate the average molar mass of the polymer if its characteristic viscosity is 0 126 m3 kg, constant K 5 10 5, 0 67.
50. What mass of polymer must be taken to prepare a solution with a molal concentration equal to 0.0025 mol/kg,
if the mass of the solvent is 1.5 kg? The molar mass of the monomer is 100 g/mol. Degree of polymerization - 100.
51. At what pH value should two enzymes with isoelectric points of 5 and 3 be separated by electrophoresis? How are enzyme particles charged in solutions with pH 4.6?
and 7.9?
[should be separated at pH 4; charges at pH 4 6 " " and " ", at pH 7 9 " " and " "]
52. Determine the molar mass of polymethyl methacrylate using the following data from the viscometric method:
Concentration |
||||||
solution, kg m3 |
||||||
Reduced viscosity |
||||||
IUD solution |
||||||
in benzene |
Constants: K
105 ] |
53. The molar mass of some BMC is 600,000. What is the molar concentration of the solution if the mass concentration of the substance is 6 g/l? What is the osmotic pressure of such a solution at 27Æ C?
54. 1 liter of solution contains 5 g of amylose. The osmotic pressure of such a solution at 27Æ C is equal to 0.15 mm Hg. Art. Calculate the molar mass of amylose.
55. A protein solution with a PrtCl concentration of 0.1 mol/L is placed on one side of the membrane, and a solution with a sodium chloride concentration of 0.2 mol/L is placed on the other side. Calculate the concentration of chloridions on both sides of the membrane when equilibrium is established.
56. To determine the golden number of gelatin, 0.1 ml of a 1% gelatin solution was added to 9.9 ml of red gold sol. Then a series of 10 successive dilutions of the original solution 1 2 was carried out. 1 ml was added to each test tube 10%
sodium chloride solution. In the first five test tubes, no changes were observed, but in test tubes 6–10 the solution acquired a blue tint. What is the golden number?
57. Determine the osmotic pressure at 293 K of an aqueous solution of gelatin having a mass concentration of 2 5 kg m 3 .
The molar mass of gelatin is 104,600, and the coefficient
0 69 Pa m6 |
kg 2. |
58. The molecular weight of polyacrylonitrile at various concentrations in a diphenylformamide solution is 75,000 and 39,100 at 293 K, and the intrinsic viscosity is 0.299 and 0.110, respectively. Determine the coefficients and K in the Mark–Houwink equation.
59. When diagnosing purulent meningitis, the protective number of cerebrospinal fluid proteins is determined. Calculate this number if it is known that to prevent coagulation of 20 ml of AgBr sol under the action of 2 ml of a solution with a mass fraction of NaNO 3 10% required adding to this sol 3 ml of cerebrospinal fluid containing 2 g of proteins in 1 liter.
60. The osmotic pressure of a solution containing 26 g/l of hemoglobin in the iso-eclectic state is equal to the osmotic pressure of a solution containing 0.0117 g/l of sodium chloride. Take the density of solutions equal to 1 g/ml, temperature 25Æ C, the deviation from Van't Hoff's law can be neglected. Calculate the molecular (molar) mass of hemoglobin.
APPLICATION
1. BASIC PHYSICAL CONSTANTS
Avogadro's constant, N A 6 02 1023 mol 1. Universal gas constant, R 8 31 J mol
Boltzmann's constant, k R N A 1 38 10 23 J K. Normal molar volume of a gas, V 0 22 4 l mol. Planck's constant, h 6 63 10 34 J s.
2. FACTORS AND PREFERENCES FOR THE FORMATION OF DECIMAL MULTIPLES AND FRAMEWORKS UNITS AND THEIR DESIGNATIONS
Factor, |
Factor, |
||||||
on which |
on which |
||||||
multiplies |
multiplies |
||||||
main |
main |
||||||
1012 |
|||||||
| meter | |||||||
kilogram |
|||||||
Electric current strength |
|||||||
Thermodynamic temperature |
|||||||
Quantity of substance |
|||||||
Derived quantities |
space and time |
||||||
square meter |
|||||||
cubic meter |
|||||||
Speed |
meter per second |
||||||
Mechanical derivatives |
and thermal quantities |
||||||
Density |
kilogram per |
kg m3 |
|||||
cubic meter |
|||||||
Strength, weight |
|||||||
Pressure |
|||||||
Energy, work, amount of heat, |
|||||||
thermodynamic potential |
|||||||
Entropy |
joule per kelvin |
||||||
Electrical derivatives |
and magnetic quantities |
||||||
The amount of electricity |
|||||||
thermodynamic potential |
|||||||
Electrical voltage |
|||||||
electric potential, |
|||||||
electromotive force |
|||||||
Electrical resistance |
|||||||
Molecular Biology Problems
Problems on the topic “Proteins”
Necessary clarifications:
the average molecular weight of one amino acid residue is taken to be 120
Calculation of molecular weight of proteins:
M min = ----- 100%
B - percentage of component
Task No. 1.
Human blood hemoglobin contains 0.34% iron. Calculate the minimum molecular weight of hemoglobin.
Solution:
M min = 56: 0.34% 100% = 16471
Task No. 2.
Human serum albumin has a molecular weight of 68400. Determine the number of amino acid residues in the molecule of this protein.
Solution:
68400: 120 = 570 (amino acids in an albumin molecule)
Task No. 3.
The protein contains 0.5% glycine. What is the minimum molecular weight of this protein if M glycine = 75.1? How many amino acid residues are there in this protein?
Solution:
M min = 75.1: 0.5% 100% = 15020
15020: 120 = 125 (amino acids in this protein)
Problems on the topic “Nucleic acids”
Necessary clarifications:
the relative molecular mass of one nucleotide is taken to be 345
distance between nucleotides in the chain of a DNA molecule (= length of one nucleotide) - 0.34 nm
Chargaff's rules:
∑(A) = ∑(T)
∑(G) = ∑(C)
∑(A+G) = ∑(T+C)
Task No. 4. On a fragment of one DNA strand, the nucleotides are located in the sequence:
A-A-G-T-C-T-A-C-G-T-A-T.
Determine the percentage of all nucleotides in this gene and its length.
Solution:
we complete the second thread (according to the principle of complementarity)
∑(A +T+C+G)= 24,
Hence: x = 33.4%
∑(G) = 4 = ∑(C)
Hence: x = 16.6%
The DNA molecule is double-stranded, so the length of the gene is equal to the length of one chain:
Task
№5.
Cytidyl nucleotides account for 18% of the DNA molecule. Determine the percentage of other nucleotides in this DNA.
Solution:
C – 18% => G – 18%
The share of A+T is 100% - (18% +18%) = 64%, i.e. 32% each
A and T – 32% each.
Task №6. 880 guanyl acids were found in a DNA molecule
Nucleotides, which make up 22% of total number nucleotides in this DNA.
Determine: a) how many other nucleotides are there in this DNA? b) what is the length of this fragment?
Solution:
1) ∑(G)= ∑(C)= 880 (this is 22%)
The share of other nucleotides accounts for 100% - (22% + 22%) = 56%, i.e. 28% each
To calculate the number of these nucleotides
we make a proportion of 22% - 880
Hence: x = 1120
2) to determine the length of DNA, how many nucleotides are contained in 1 chain:
(880 + 880 + 1120 + 1120) : 2 = 2000
2000 · 0.34 = 680 (nm)
Task
№7.
Given a DNA molecule with relative molecular weight 69,000, of which 8,625 are adenyl nucleotides. Find the number of all nucleotides in this DNA. Determine the length of this fragment.
Solution:
69000: 345 = 200 (nucleotides in DNA)
∑(G+C) = 200 – (25+25)= 150, i.e. there are 75 of them.
2) 200 nucleotides in two chains => in one – 100.
100 · 0.34 = 34 (nm)
Problems on the topic “DNA Code”
Task
№8.
What is heavier: a protein or its gene?
Solution:
Then the mass of this protein is 120x,
The number of nucleotides in the gene encoding this
Protein – 3x
The mass of this gene is 345 3x
120x Answer: a gene is heavier than a protein.
Task №9. The nucleotide sequence at the beginning of the gene that stores information about the protein insulin begins like this:
AAACCCTTGCTTGTAGAC
Write the sequence of amino acids that begins the insulin chain
Solution:
The task is performed using the following table
Genetic code
| First base | Second base | Thirdbase |
|||
| U (A) | C (G) | A (T) | G (C) |
||
| U (A) | Hairdryer Lei | SerSerSer | Shooting Range |
Three | U (A) C (G) A (T) G (C) |
| C (G) | Lei Lei | About About | Gies Gln | Arg Arg | U (A) C (G) A (T) G (C) |
| Ile Meth | Tre Tre | Asn Liz | Ser Arg | U (A) C (G) A (T) G (C) |
|
| G (C) | Shaft Shaft | Ala Ala | Asp Glu | Gli Gli | U (A) C (G) A (T) G (C) |