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Biology P1 > Unit 1 =) > Flashcards

Unit 1 =) Flashcards

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AP® Biology flashcards Biology 101 flashcards
1
Q

Similarities between the structures of starch and cellulose

A
  1. Polymers
  2. Contain glucose (C, H, O)
  3. Glycosidic bonds
  4. Have 1-4 links
  5. Hydrogen bonds (within structure)
  6. Joined by condensation
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2
Q

Differences between starch and cellulose

A

Starch:
1. α glucose
2. branched (not straight)
3. 1,6 bonds
4. glucoses same way up
5. no H-bonds between molecules
6. no fibres/ fibrils

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3
Q

Explain how cellulose molecules are adapted for their function in plant cells. (3)

A
  1. Long, straight chains
  2. Linked tgt by many hydrogen bonds -> fibrils
  3. Provide strength (cell wall)
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4
Q

Explain how a non-competitive inhibitor would decrease the rate of reaction catalysed by an enzyme. (3)

A
  • binds to sites other than active site
  • causes change in shape of active site
  • substrate no longer able to bind to a.s.
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5
Q

E. coli has no cholesterol in its cell-surface membrane. Despite this, the cell
maintains a constant shape. Explain why. (2)

A
  1. Cell unable to change shape;
  2. (Because) cell has a cell wall;
  3. (Wall is) rigid / made of peptidoglycan / murein.
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6
Q

Hydrogen bonds are important in cellulose molecules. Explain why. (2)

A
    • holds chains tgt
    • forms crosslinks betw chains
    • forms microfibrils
  2. Hydrogen bonds strong in large numbers
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7
Q

What is meant by the tertiary structure of a protein? (1)

A

The way the whole protein is folded.

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8
Q

Explain how heating an enzyme results in it being denatured. (2)

A

↑ ke
Bonds holding tertiary structure break

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9
Q

Describe the role of enzymes of the digestive system in the complete breakdown of starch. (5)

A
  1. Amylase
  2. Starch –> Maltose
  3. Maltase
  4. Maltose –> Glucose
  5. Hydrolysis
  6. Breaking of glycosidic bond
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10
Q

Explain why each new DNA strand is synthesised in opposite directions. (4)

A

1.DNA has antiparallel strands
2. Shape of the nucleotides is different
3. Enzymes have active sites w/ specific shapes
4. Only substrates w/ complementary shapes (3’ end) can bind to active site of DNA polymerase

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11
Q

Cholesterol increases the stability of plasma membranes, by making them less flexible.

Suggest one advantage of red blood cells containing more cholesterol than cells lining the ileum. (1)

A
  • rbc free in blood (not supported by other cells)
  • cholesterol helps maintain shape
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12
Q

The higher the altitude, the lower the mean temperature.

Explain how the lower temperature at high altitude reduces growth of plants. (4)

A
  • ⬇️ enzyme activity
  • ⬇️ rate of photosynthesis so⬇️ carbohydrates formed
  • ⬇️ respiration
  • ⬇️ rate of nutrient uptake
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13
Q

Give differences between DNA and RNA.

A
  1. double stranded vs single stranded
  2. thymine vs uracil
  3. longer vs shorter
  4. deoxyribose vs ribose
  5. stays in nucleus vs leaves
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14
Q

A polypeptide had 51 amino acids in its primary structure.
The actual number of bases in the gene for this polypeptide is more than the minimum number of DNA bases required to code for the amino acids. Explain why. (1)

A

Some regions of the gene are non-coding (introns).

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15
Q

A student carried out an investigation into the mass of product formed in an enzyme-controlled reaction at three different temperatures. Only the temperature was different for each experiment.
Explain why the curves for 27 °C and 37 °C level out at the same value. (2)

A
  • All substrate changed into product
  • Same amount of product formed as same initial conc. of substrate
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16
Q

What is an enzyme? (2)

A
  • Protein
  • Catalyst
  • for a specific substrate
  • Lowers activation energy
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17
Q

ATP is an energy source used in many cell processes.
Give two ways in which ATP is a suitable energy source for cells to use. (2)

A
  1. Immediate energy source
  2. Releases relatively small/ manageable amounts of energy
  3. Phosphorylates other compounds, making them more reactive
  4. Rapidly resynthesised
  5. x leave cells (x lost)
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18
Q

A precipitate is produced in a positive result for reducing sugar in a
Benedict’s test.
A precipitate is solid matter suspended in solution.
A student carried out the Benedict’s test. Suggest a method, other than
using a colorimeter, that this student could use to measure the quantity of
reducing sugar in a solution. (2)

A
  • filter and dry
  • find mass
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19
Q

Use of a colorimeter in this investigation would improve the repeatability of
the student’s results. (instead of determining which solution is darker in colour in a Benedict’s test)
Give one reason why. (1)

A
  • quantitative
  • standardises method
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20
Q

When bread becomes stale, the structure of some of the starch is changed. This changed starch is called retrograded starch.
Scientists have suggested retrograded starch is a competitive inhibitor of amylase in the small intestine.

Assuming the scientists are correct, suggest how eating stale bread could help to reduce weight gain. (3)

A
  • ↓ HYDROLYSIS of starch
  • to maltose
  • ↓ absorption of glucose
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21
Q

The sequence of bases on one strand of DNA is important for protein synthesis.

What is its role? (1)

A
  • determines amino acids
  • specific protein produced
  • mRNA formation
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22
Q

Give one advantage of DNA molecules having two strands. (1)

A
  • stability
  • protects bases
  • replication
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23
Q

Describe how a saturated fatty acid differs in molecular structure from an unsaturated fatty acid. (2)

A
  • absence of double bond
  • in hydrocarbon chain
  • saturated w/ H
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24
Q

Where is amylase produced?

A

Pancreas –> maltose

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25
Q

Where is maltase produced?

A

EPITHELIUM of small intestine –> glucose

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26
Q

What is a polymer?

A

Molecule made up of many identical monomers.

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27
Q

Explain why a buffer solution is added to an enzyme-substrate mixture. (2)

A
  • constant pH
  • change in pH will slow rate of reaction/ denature enzyme
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28
Q

Explain how sulphur-containing amino acids help to give keratin (protein) molecules their characteristic strength. (2)

A
  1. disulphide bridges (chemical bonds) formed between sulphur-containing groups
  2. bind chains to each other
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29
Q

Explain why differences in primary structure result in keratins with different properties. (2)

A
  • diff. sequences of a.a.
  • bonds in diff. places so give diff. shapes
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30
Q

Humans and grasshoppers have very similar percentages of each base in their DNA but they are very different organisms.

Use your knowledge of DNA structure and function to explain how this is possible. (2)

A
  • diff. genes
  • bases in diff. seq
  • diff. seq of a.a. –> diff. proteins
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31
Q

HSV infects nerve cells in the face (line 1).
Explain why it infects only nerve cells. (3)

A
  • antigens outside of virus
  • shape complementary to receptors in membrane of cells
  • found only on membrane of nerve cells
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32
Q

Explain what it means when a virus is described as inactive inside the body. (2)

A
  • no more cells infected
  • ∵ virus X replicating
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33
Q

This microRNA binds to some of the nerve cell’s own mRNA molecules. These mRNA molecules are involved in programmed cell death of nerve cells.

The scientists concluded that production of this microRNA allows HSV to remain in the body for years.

Explain how this microRNA allows HSV to remain in the body for years. (4)

A

microRNA binds to mRNA
- by specific base pairing
- prevents mRNA being read by ribosomes
- prevents translation/ production of proteins
- that cause cell death

34
Q

Suggest one advantage of programmed cell death (in the context of viral infection). (1)

A

Prevents replication of virus

35
Q

Use your knowledge of enzyme action and DNA replication to explain why new nucleotides can only be added in a 5’ to 3’ direction. (4)

A
  • active site of DNA polymerase
  • specific
  • only complementary to 3’ end
  • 3’ end & 5’ end have diff. shapes
36
Q

Describe how amino acids join to form a polypeptide so there is always
NH2 at one end and COOH at the other end. (2)

A
  • one amine group joins to carboxyl group –> peptide bond
  • each a.a. is orientated in the same direction in the chain (–> a free NH2 group at one end, a free COOH group at the other)
37
Q

Explain how the active site of an enzyme causes a high rate of reaction. (3)

A
  • lowers activation energy
  • induced fit causes active site to change shape
  • E-S complex causes bonds to break/ form
38
Q

The secondary structure of a polypeptide is produced by bonds between amino acids.
Describe how. (2)

A
  • H bonds
  • between NH & C=O groups –> β pleated sheets / α helix
39
Q

Two proteins have the same number and type of amino acids but different tertiary structures.
Explain why. (2)

A
  • diff. seq. of a.a.
  • ionic/ H/ disulphide bonds in diff. positions
40
Q

Formation of an enzyme-substrate complex increases the rate of reaction.
Explain how. (2)

A
  • reduces activation energy
  • ∵ bending bonds
41
Q

Explain how proteins are suited to their roles as receptor molecules. (3)

A
  • many diff. types of proteins
  • diff. seq of a.a.
  • diff. tertiary structure
  • shape of a.s.- where substrate fits
42
Q

Explain why glucose and maltose both taste sweet but starch does not. (1)

A

glucose & maltose- soluble vs starch- insoluble

43
Q

Saccharin, cyclamates and sucrose are chemically different but they all taste sweet.

Suggest why. (2)

A
  • similar molecular shape
  • ∴ bind to receptors
44
Q

In terms of your knowledge of the way in which enzymes work, explain why it is necessary to pass the milk through the reactor several times to reduce the amount of lactose sufficiently. (3)

A
  • enzyme conc= limiting
  • substrate in excess
  • saturation of a.s.
  • some substrate goes thru unchanged
    //
  • low temp
  • ↓ke
  • ↓ collisions w/ enzyme
  • ↓ rate of reaction
45
Q

Decreasing Substrate vs time reaction:

Explain the change in rate of reaction with time. (2)

A
  • ↓ substrate
  • ↓ chance of collision w/ enzyme –> E-S complexes
46
Q

investigating effect of acid on digestion of starch:

Starch could have been used instead of amylase in converting starch into glucose.

Explain why it is not used. (1)

A
  • x specific
  • forms by-products

(acid hydrolyses starch)

47
Q

Why would the rate of reaction change if the temperature fell by 10oC? (2)

A
  • molecules have ↓ ke
  • move slower
    ↓ collisions
48
Q

Why would the rate of reaction change if pH changed substantially? (3)

A
  • changes shape
  • changes a.s. (tertiary)
  • substrate x longer fit
49
Q

Suggest two reasons why a test for glucose in urine based on glucose oxidase and peroxidase might be preferred to one using Benedict’s reagent. (2)

A
  • quantitative (conc)
  • SPECIFIC to glucose (benedict’s x)
  • ↑ sensitive
50
Q

Suggest how you could use Benedict’s test to compare the amount of reducing sugar in
two solutions. (2)

A
  • standardise specific feature
    – eg. amounts used; time heated; temp
  • compare colour/ amount of ppt./ time taken to get colour
51
Q

Humans synthesise more than their body mass of ATP each day. Explain why it is necessary for them to synthesise such a large amount of ATP. (2)

A
  • ATP X be stored
  • only releases a small amount of energy at a time
52
Q

State and explain the property of water that can help to buffer changes in temperature. (2)

A
  • high specific heat capacity
  • needs ↑ energy change temp
  • √ gain/ lose a lot of heat w/o changing temp
53
Q

Give two properties of water that are important in the cytoplasm of cells.
For each property of water, explain its importance in the cytoplasm. (4)

A
  1. polar
    – solvent
    ∴ metabolic reactions faster in solution
  2. reactive
    – for hydrolysis
54
Q

Describe how ATP is resynthesised in cells. (2)

A
  • ATP synthase
  • from ADP + Pi
  • in respiration/ photosynthesis
55
Q

Each person in the control group was given 50 g of lactose containing the same amount of radioactive carbon. All the products of lactose digestion were absorbed into their blood. The concentration of glucose was measured in mg per 100 cm3 of blood.

Explain why the variation in the results may be due to differences in body mass. (2)

A
  • ↑ mass ↑ blood ∴ ↓ blood glucose conc.
  • ∵ same amount of glucose absorbed
56
Q

High GI foods raises blood glucose concentration after 0.5 hours of consumption.

Explain how a sports drink could provide an energy boost when running. (3)

A
  • sports drinks contain carbs–> high GI–> raises blood glucose conc rapidly
  • contains salt–> glucose absorbed ↑ rapidly
  • ↑ glucose for resp. in muscles
  • faster resp. –> faster energy release
57
Q

Chitin is a nitrogen-containing polysaccharide.
Name one chemical element present in a phospholipid which would not be present in chitin. (1)

A

Phosphorus

58
Q

Suggest why glucagon is able to bind to liver cells but not to cells in other parts of the body. (1)

A
  • the receptors are only found on liver cells
  • x other body cells
59
Q

Heating may affect the tertiary structure of a protein.
Explain how. (1)

A

(- too much ke)
- breaks bonds in tertiary structure
- shape changed–> denatured

60
Q

Breakdown of chitin leads to “death by osmosis” of fungi attacking the grain.
Explain how. (2)

A
  • water enters fungus by osmosis–> ↑ Pa inside
  • cell wall x present to withstand it
61
Q

Describe how a phosphodiester bond is formed between two nucleotides within a DNA molecule. (2)

A
  • condensation
  • between phosphate & deoxyribose
  • by DNA polymerase
62
Q

Describe a biochemical test that could be performed on a sample of food to determine whether it contained triglycerides. (2)

A
  • dissolve in ethanol–> mix w/ water
  • if trig. present–> white emulsion
63
Q

Contrast the structures of ATP and a nucleotide found in DNA to give two differences. (2)

A

ATP vs nucleotide:
1. 3 phosphate groups vs 1
2. ribose vs deoxyribose
3. only adenine vs diff. bases

64
Q

NMO is a disease that leads to damage to nerve cells in the spinal cord. A person with NMO produces anti-AQP4 antibody that attacks only these nerve cells.
Explain why the anti-AQP4 antibody onyl damages these cells. (4)

A
  • antibody has a specific tertiary structure
  • binding site only complementary to one antigen
  • complementary antigen is only found on these nerve cells
  • bind to form an antigen-antibody complex
65
Q

Describe how the structure of glycogen is related to its function. (4)

A
  1. helical- compact
  2. branched- ↑ ends for faster hydrolysis
  3. polymer of glucose- easily hydrolysed
  4. releases glucose- provides respiratory substrate for energy release
  5. insoluble- x affect Ψ
66
Q

Fatty acids are used to:
1. make plasma membranes
2. respired for energy
3. converted to other fatty acids

Explain 2 ways in which fatty acids are important in the formation of new cells. (4)

A
  • f.a. used to make phospholipids
  • pplipids in mem
  • ↑ pplipids ↑ mem made

//
- f.a. respired to release energy
- ↑ triglycerides ↑ energy released
- energy used for cell prod.

67
Q

Describe how a triglyceride molecule is formed. (3)

A
  • 1 glycerol + 3 fatty acids
  • condensation + removal of 3 molecules of water
  • ester bond formed
68
Q

Describe the induced-fit model of enzyme action and how an enzyme acts as a catalyst. (3)

A
  • substrate binds to a.s. –> E-S complex
  • a.s. changes shape –> complementary to s
  • ↓ a.e.
69
Q

Describe the structure of DNA. (5)

A
  • polymer of nucleotides
  • each nucleotide- deoxyribose, nitrogenous base, phosphate group
  • phosphodiester bonds between nucleotides
  • double helix held by H bonds
  • H bonds between A+T, C+G
70
Q

Describe how an enzyme can be phosphorylated. (2)

A
  • attachment of Pi
  • released from hydrolysis of ATP
71
Q

State 3 differences between DNA in the nucleus of a plant cell and DNA in a prokaryotic cell. (3)

A
  1. ass. w/ histones vs x
  2. linear vs circular
  3. x vs plasmids
  4. introns vs x
  5. longer vs shorter
72
Q

Describe how amino acids join to form a polypeptide so there is always NH2 at one end and COOH at the other end. (2)

A
  • 1 amine group joins to a carboxyl group –> peptide bond
  • in chain, each a.a. is orientated in same direction (–> free NH2 at 1 end + free COOH at the other)
73
Q

A biochemical test for reducing sugar produces a negative result with raffinose solution.
Describe a biochemical test to show that raffinose solution contains a non-reducing sugar. (3)

A
  1. heat w/ acid + neutralise
  2. heat w/ Benedict’s
  3. red ppt
74
Q

Describe how an ester bond is formed in a phospholipid molecule. (2)

A
  • condensation
  • between glycerol + f.a.
75
Q

Describe how the results from a colorimeter can identify the fruit juice containing the higher sugar content. (1)

A

↑ absorbance ↑ sugar

!!

76
Q

The student controlled variables in the test using Benedict’s solution.
Give 2 variables the student controlled. (2)

A
  1. temp of water bath
  2. vol of Benedict’s
  3. conc. of Benedict’s
  4. duration of heating
77
Q

Describe how monomers join to form the primary structure of a protein. (3)

A
  • condensation betw. a.a.
  • peptide bonds formed
  • creates specific seq. of a.a.
78
Q

Explain how theorganic bases help to stabilise the structure of DNA. (2)

A
  • H bonds between base pairs hold 2 strands tgt
  • many H bonds- provide strength
79
Q

Give an equation of the reaction catalysed by ATP synthase. (1)

A

ADP + Pi –> ATP + H2O

80
Q

A student investigated the effect of different sugars on the rate of respiration in yeast. She first used glucose solution.
The student repeated the experiment using yeast in maltose solution. She found the rate of carbon dioxide production was slower than with yeast in glucose solution.
Suggest why. (2)

A
  • maltose = disaccharide
  • needs to be hydrolysed

x broken down!

Biology P1 (34 decks)

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