2018 Paper 1 Flashcards

1
Q

Give two pieces of evidence that the cell in figure 1 was undergoing mitosis

A

individual chromosomes are visible because they have condensed
-each chromosome is made up of two chromatids because DNA has replicated

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2
Q

Tick one box that gives the name of the stage in mitosis shown

A

Prophase

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3
Q

Suggest how this procedure moved the chromosomes apart.

A

water moves into the cells by osmosis

-cell gets bigger

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4
Q

Suggest one way the structure of the chromosome could differ along its length to
result in the stain binding more in some areas.

A

difference in base sequences

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5
Q

What is a homologous pair of chromosomes?

A

Two chromosomes that carry the same genes

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6
Q

Give two ways in which the arrangement of prokaryotic DNA is different from the
arrangement of the human DNA in Figure 1.

A

Prokaryotic dna is circular as opposed to linear , not associated with proteins/histones , only one piece of DNA

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7
Q

Describe the method the student would have used to obtain the results in Figure 3. Start after all of the cubes of potato have been cut. Also consider variables he should
have controlled.

A

Method to ensure all cut surfaces of the eight cubes are exposed to the sucrose solution;
Method of controlling temperature; water bath
Method of drying cubes before measuring; filter paper
Measure mass of cubes at stated time intervals;

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8
Q

Is the rate of osmosis per mm2 per minute different between A and B during this time?

A

Yes
A = 7350mm2 = 1.6x10-5 mm-2
B = 14700 mm2 = 1.5x10-5 mm-2

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9
Q

What is meant my the genetic code being degenerate

A

More then one codon codes for single amino acid

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10
Q

change at amino acid 300 had no effect on rate of reaction but change at 279 reduces rate of reaction why?

Suggest reasons for the differences between the effects of these two changes

A

Both negatively charge to positively charged amino acids
Change at amino acid 300 does not change shape of active site
Amino acid 279 may have been involved in a ironic disulfide hydrogen bond and shape of active site changes

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11
Q

Critical value at P=0.05 is 3.841
What does this result suggest about the difference between the observed and expected results and what can the scientist therefore conclude?

A

There is a less than 0.05 probability that the difference occurred by chance
Calculated value > critical value = null hypothesis rejected
Conclude that the proportion of plants that produce 2n gamers does not change from one breeding cycle to the next

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12
Q

Direction selection

A

Scientists selected breeding plants that produce 2n gametes
So there plants passed on alleles
Frequency of alleles for production of 2n gametes increased

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13
Q

Did heat treatment damage phloem

A

Movement in phloem requires ATP
Heat damage in living cells
So transport in phloem throughout only untreated plant
Stops ATP production so transport through phloem old only occur in B

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14
Q

Did heat treatment affect xylem

A

Standard deviations overlap
Not different water contents
Movement in xylem is passive
So unaffected by heat treatment

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15
Q

What can you conclude about the movement of Fe3+ in barley plants

A

Heat treatment has greater affect on young plants than old
Heat treatment damages phloem
Fe3+ moves up leaf
Fe3+ is transported in xylem in older leaf
Younger leaf = some xylem reaches top but mostly transported in phloem in younger leaf

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16
Q

Describe the gross structure of the human gas exchange and how we breath in and out

A

Air into trachea into tracheoles into bronchi bronchioles alveoli.

Breath in = diaphragm contracts. External intercostal muscles contract. Volume increase. Pressure decreases in thoracic. Air moves in.

Breath out = diaphragm relaxes. Internal Intercoatal muscles contract. Volume decrease. Pressure increases in thoracic cavity. Air moves out

17
Q

Compare and contrast triglycerides and phospholipids

A
Both =ester Bonds, glycerol
Fatty acids = saturated or unsaturated 
Insoluble in water 
C, H, O but phospholipids contain P.
T= 3 fatty acids 
P= 2 fatty acids + 1 phosphate 
T= hydrophobic ( non-polar 
P= hydrophobic and hydrophilic regions 
P= phospholipids belayer in H2O
T= dont
18
Q

Describe how lactose is formed and where in the cell it would be attached to a polypeptide to form a glycoprotein

A
Lactose = glucose + galactose 
Condensation reaction 
Glycosidic bond 
Water removed 
Attached in Golgi apparatus
19
Q

Expected
Number of plants that did not produce 2n gametes
Number of plants that did produce 2n gametes

A

Number of plants that did not produce 2n gametes = 52

Number of plants that did produce 2n gametes = 4

20
Q

How many times ROR is greater with enzyme present

A

578/ 3x10-9 = 1.93x10^11

21
Q

ROR of enzyme activity with no lyxose at 2,5 as a % of max rate with lyxose

A

0.44/ 1.4 x100 = 31.4%