Biology 2021 AQA Assessment materials UNOFFICIAL MARKSCHEME

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kate_howard_1
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#1
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#1
Hi! These questions that are released are most likely going to be used on upcoming a level 2021 mini exams however, a mark scheme hasn't been released? Does anyone want to help make an unofficial markscheme?
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svalen123
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Hey is it possible if you could attach the as level markscheme
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ruqayyaha
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AS - BIOLOGICAL MOLECULES

1.1)BDAB
1.2) 18, 36 ,18
1.3) hydrolyse raffinose solution by heating with dilute hydrochloric acid. Neutralise solution by adding sodium hydrogencarbonate and then carry out benedict's test by adding benedict's reagent to the solution and heat in a water bath. Positive reult = orange - red precipitate.

5.1) 1. Cells at the tip of onion root are dividing
2. to form a thin layer so light can pass through and make cells visible.
5.2) 3/32 * 1082 = 101.25 mins
105-101.25/105 * 100 = 3.57%
5.3) cytokinesis
5.4) dont really know

8.1) use biuret test. Add sodium hydroxide into test tube containing sample and add few drops of dilute copper (II) sulfate and mix. should go from blue to purple
8.2) 1. have carbon, oxygen, hydrogen and nitrogen
2. formed through condensation reactions
difference: have different chemical and physical properties
8.3) theyre moving towards negative electrode si both amino acids are positivley charged.

5.3) a competitive inhibitor has a shape similar to the substrate which is complementary to shape of the active site. They bind to the active site so substrate cant bind.

1.1) BAE
1.2) More mitochondria makes more ATP which needed for protein synthesis and trasnport of proteins. Many golgi vesicles needed for vesicular transport of proteins out of cell.

4.2) cell not stained to show nucleas
4.3) S=large vacuole
T= chloroplasts
4.4) transmission electron microscopes can be used

ill do some later, hope this helps
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Bingsuze
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#4
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#4
(Original post by ruqayyaha)
AS - BIOLOGICAL MOLECULES

1.1)BDAB
1.2) 18, 36 ,18
1.3) hydrolyse raffinose solution by heating with dilute hydrochloric acid. Neutralise solution by adding sodium hydrogencarbonate and then carry out benedict's test by adding benedict's reagent to the solution and heat in a water bath. Positive reult = orange - red precipitate.

5.1) 1. Cells at the tip of onion root are dividing
2. to form a thin layer so light can pass through and make cells visible.
5.2) 3/32 * 1082 = 101.25 mins
105-101.25/105 * 100 = 3.57%
5.3) cytokinesis
5.4) dont really know

8.1) use biuret test. Add sodium hydroxide into test tube containing sample and add few drops of dilute copper (II) sulfate and mix. should go from blue to purple
8.2) 1. have carbon, oxygen, hydrogen and nitrogen
2. formed through condensation reactions
difference: have different chemical and physical properties
8.3) theyre moving towards negative electrode si both amino acids are positivley charged.

5.3) a competitive inhibitor has a shape similar to the substrate which is complementary to shape of the active site. They bind to the active site so substrate cant bind.

1.1) BAE
1.2) More mitochondria makes more ATP which needed for protein synthesis and trasnport of proteins. Many golgi vesicles needed for vesicular transport of proteins out of cell.

4.2) cell not stained to show nucleas
4.3) S=large vacuole
T= chloroplasts
4.4) transmission electron microscopes can be used

ill do some later, hope this helps
Can I just say, you’re a life saver omg thank you!
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m4inotn4
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#5
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#5
(Original post by ruqayyaha)
AS - BIOLOGICAL MOLECULES

1.1)BDAB
1.2) 18, 36 ,18
1.3) hydrolyse raffinose solution by heating with dilute hydrochloric acid. Neutralise solution by adding sodium hydrogencarbonate and then carry out benedict's test by adding benedict's reagent to the solution and heat in a water bath. Positive reult = orange - red precipitate.

5.1) 1. Cells at the tip of onion root are dividing
2. to form a thin layer so light can pass through and make cells visible.
5.2) 3/32 * 1082 = 101.25 mins
105-101.25/105 * 100 = 3.57%
5.3) cytokinesis
5.4) dont really know

8.1) use biuret test. Add sodium hydroxide into test tube containing sample and add few drops of dilute copper (II) sulfate and mix. should go from blue to purple
8.2) 1. have carbon, oxygen, hydrogen and nitrogen
2. formed through condensation reactions
difference: have different chemical and physical properties
8.3) theyre moving towards negative electrode si both amino acids are positivley charged.

5.3) a competitive inhibitor has a shape similar to the substrate which is complementary to shape of the active site. They bind to the active site so substrate cant bind.

1.1) BAE
1.2) More mitochondria makes more ATP which needed for protein synthesis and trasnport of proteins. Many golgi vesicles needed for vesicular transport of proteins out of cell.

4.2) cell not stained to show nucleas
4.3) S=large vacuole
T= chloroplasts
4.4) transmission electron microscopes can be used

ill do some later, hope this helps
thank you for this soooo much!! <3
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obj2345
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#6
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#6
(Original post by ruqayyaha)
AS - BIOLOGICAL MOLECULES

1.1)BDAB
1.2) 18, 36 ,18
1.3) hydrolyse raffinose solution by heating with dilute hydrochloric acid. Neutralise solution by adding sodium hydrogencarbonate and then carry out benedict's test by adding benedict's reagent to the solution and heat in a water bath. Positive reult = orange - red precipitate.

5.1) 1. Cells at the tip of onion root are dividing
2. to form a thin layer so light can pass through and make cells visible.
5.2) 3/32 * 1082 = 101.25 mins
105-101.25/105 * 100 = 3.57%
5.3) cytokinesis
5.4) dont really know

8.1) use biuret test. Add sodium hydroxide into test tube containing sample and add few drops of dilute copper (II) sulfate and mix. should go from blue to purple
8.2) 1. have carbon, oxygen, hydrogen and nitrogen
2. formed through condensation reactions
difference: have different chemical and physical properties
8.3) theyre moving towards negative electrode si both amino acids are positivley charged.

5.3) a competitive inhibitor has a shape similar to the substrate which is complementary to shape of the active site. They bind to the active site so substrate cant bind.

1.1) BAE
1.2) More mitochondria makes more ATP which needed for protein synthesis and trasnport of proteins. Many golgi vesicles needed for vesicular transport of proteins out of cell.

4.2) cell not stained to show nucleas
4.3) S=large vacuole
T= chloroplasts
4.4) transmission electron microscopes can be used

ill do some later, hope this helps
please could you do paper 2 xxx
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Mishal__01
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#7
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#7
(Original post by kate_howard_1)
Hi! These questions that are released are most likely going to be used on upcoming a level 2021 mini exams however, a mark scheme hasn't been released? Does anyone want to help make an unofficial markscheme?
I can't see any questions?
Last edited by Mishal__01; 1 year ago
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Maya.kg16
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#8
Report 11 months ago
#8
(Original post by ruqayyaha)
AS - BIOLOGICAL MOLECULES

1.1)BDAB
1.2) 18, 36 ,18
1.3) hydrolyse raffinose solution by heating with dilute hydrochloric acid. Neutralise solution by adding sodium hydrogencarbonate and then carry out benedict's test by adding benedict's reagent to the solution and heat in a water bath. Positive reult = orange - red precipitate.

5.1) 1. Cells at the tip of onion root are dividing
2. to form a thin layer so light can pass through and make cells visible.
5.2) 3/32 * 1082 = 101.25 mins
105-101.25/105 * 100 = 3.57%
5.3) cytokinesis
5.4) dont really know

8.1) use biuret test. Add sodium hydroxide into test tube containing sample and add few drops of dilute copper (II) sulfate and mix. should go from blue to purple
8.2) 1. have carbon, oxygen, hydrogen and nitrogen
2. formed through condensation reactions
difference: have different chemical and physical properties
8.3) theyre moving towards negative electrode si both amino acids are positivley charged.

5.3) a competitive inhibitor has a shape similar to the substrate which is complementary to shape of the active site. They bind to the active site so substrate cant bind.

1.1) BAE
1.2) More mitochondria makes more ATP which needed for protein synthesis and trasnport of proteins. Many golgi vesicles needed for vesicular transport of proteins out of cell.

4.2) cell not stained to show nucleas
4.3) S=large vacuole
T= chloroplasts
4.4) transmission electron microscopes can be used

ill do some later, hope this helps
I believe for question 1.2 it's actually 18,32,16 as 2h20 will have been removed when forming the glycosidic bonds
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Maya.kg16
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#9
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#9
I'm not sure where your answer came from for 5.2, can someone explain? Also some more answers based on what I got:
4.4 a transmission electron microscope allows you to see inside the cell, whereas a scanning electron microscope only allows you to see the surface
4.5 I got 4.71 x 10^5 (that may need checking)
3.1 James Watson and Francis Crick
3.2 DNA Polymerase
3.3 1. Acts as template strand for dna nucleotides to attach to
2. To attach to the template strand to form the second strand of DNA
4.1 both are passive some don’t require ATP
Both involve substance moving down concentration gradient
4.2The concentration of monosaccharides is greater outside the epithelial cells than inside, so the monosaccharides move from the higher concentration (outside) of monosaccharides to a lower concentration (inside), hence the uptake is increasing.
The graph plateaus as the concentration inside compared to outside the epithelial cells has reached equilibrium to there is no net movement of monosaccharides.
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