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Biology > 2019 Paper 1 > Flashcards

2019 Paper 1 Flashcards

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AP® Biology flashcards Biology 101 flashcards
1
Q

What is Cell wall made from in plants, algae fungi and prokaryotes?

A

Plants & algae = cellulose
Prokaryotes = Murein
Fungi = chitin

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2
Q

A student concluded from Figure 3 that eating an extra 10 g of fibre per day would significantly lower his risk of cardiovascular disease.
Evaluate his conclusion.

A
  1. Negative correlation
  2. Original fibre intake unknown
  3. If current intake is between 5 and 30 gday-1 then decrease in risk
  4. Correlation does not equal causation
  5. Little evidence for higher mass of fibre per day
  6. Large sd at high/ low mass of fibre makes mean less precise
  7. No statistical test
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3
Q

FFQ method advantage and disadvantages

A 
Advantages = over longer period so more representative 
Disadvantages = Relies on long term memory so may not be accurate
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4
Q

How human mass at birth is affected by stabilising selection

A
  1. Most likely to be transferred to special care under 2800g or above 4200g
  2. Extreme mass babies least likely to survive and so less likely to pass on their alleles
  3. Extreme mass at birth decrease in frequency
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5
Q

what Stats test is used to test the following null hypothesis:
‘The presence of KIR2DS1 in the mother’s genome does not affect the frequency of births above 4500 g’

A

Chi squared

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6
Q

The scientists calculated a P value of 0.03 when testing their null hypothesis
What can you conclude?

A

Probability that difference is due to chance is less than 0.05
Reject null hypothesis
Presence of allele des affect the frequency of high birth mass

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7
Q

Describe structure of HIV

A
  1. RNA
  2. reverse transcriptase
  3. Protein capsid
  4. Phospholipid envelope
  5. Attachment proteins
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8
Q

A test sample of 500 mm3 of blood is taken from an HIV controller to determine the viral load.

A

106

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9
Q

Suggest why HIV controllers o not develop symptoms of AIDs

A
  1. All have more T helper cells
  2. Lower viral load to infect helper T cells
  3. Activation of B cells
  4. With B cells more production of plasma cells
  5. More able to destroy other pathogens
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10
Q

Describe and explain the data in table 4

age
percentage of cells undergoing mitosis
percentage of cells undergoing DNA replication

A
  1. Trend of slowing growth from before birth to 21 days

2. Heart growth slowing until fully developed

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11
Q

Describe how BrdU would be incorporated into new DNA during semi-conservative
replication.

A
  1. DNA Helicase
  2. Breaks hydrogen bonds
  3. BrdU complementary to adenine
  4. DNA polymerase joins nucleotides
  5. Phosphodiester bonds form
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12
Q

Cells with BrdU in their DNA are detected using an anti-BrdU antibody with an enzyme attached.
Use your knowledge of the ELISA test to suggest and explain how the scientists identified the cells that have BrdU in their DNA

A
  1. Add antibody anti-BrdU with enzyme attached to cells
  2. Wash cells / DNA to remove excess antibody
  3. Add substrate to cause colour change
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13
Q

Vol of sodium chloride solution (cm3)

Vol of water (cm3)

A

Vol of sodium chloride solution (cm3)= 0.8

Vol of water (cm3) = 19.2

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14
Q

There is a linear relationship between the water potential and the concentration of sodium chloride solution.

A

0.07

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15
Q

One leaf with an original area of 60 cm2 gave a voltage reading of −7 μV
Use Figure 8 (on page 28) and Figure 9 to calculate by how much this leaf increased in area.

A

9 cm2

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16
Q

Why the leaf growth of xerophytic plants would be different from the leaf growth of sunflowers

A

Low growth

Due to smaller number of stomata

17
Q

Use your knowledge of gas exchange in leaves to explain why plants grown in soil with very little water only grow slowly

A

Stomata close

Less CO2 uptake for less photosynthesis

18
Q

Suggest how oxyheamoglobin curves allow the mouse to have a higher metabolic rate than the horse

A

Mouse haemoglobin has a lower affinity for oxygen

More oxygen can be dissociated for respiration

19
Q

Mammals such as a mouse and a horse are able to maintain a constant body temperature.
Use your knowledge of surface area to volume ratio to explain the higher metabolic
rate of a mouse compared to a horse.

A

Mouse = larger SA:V ratio
More heat loss per gram
Faster rate of respiration

20
Q

Gap fill

A 
Allele 
Locus 
Transcribed 
Translated 
Golgi apparatus 
Tertiary

Biology (6 decks)

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