AQA A level Biology 2021 assessment materials

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lilyrose270
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#1
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this is publicly available on the AQA website. I wanted to ask whether anyone has done 6.3: a dwarf pink flowered plant was crossed with a heterozygous tall white flowered plant. Complete the genetic diagram to show all the possible genotypes and the ratio of phenotypes expected in the offspring of this cross.
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TSR Jessica
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Sorry you've not had any responses about this. Are you sure you've posted in the right place? Here's a link to our subject forum which should help get you more responses if you post there.
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briteeshbro
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legend.
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lilyrose270
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#4
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what did people get for :
Q6.3 a dwarf pink flowered plant was crossed with a heterozygous tall white flowered plant. complete the genetic diagram to show all the possible genotypes and the ratio of phenotypes expected in the offspring of this cross.
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ruqayyaha
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#5
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(Original post by lilyrose270)
what did people get for :
Q6.3 a dwarf pink flowered plant was crossed with a heterozygous tall white flowered plant. complete the genetic diagram to show all the possible genotypes and the ratio of phenotypes expected in the offspring of this cross.
im stuck on that too
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miakirk02
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(Original post by ruqayyaha)
im stuck on that too
Hi did you find an answer in the end?
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interlanken-fall
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#7
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haveone for ocr? and one for edexcel chem?
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jacob5556
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Anyone manage to answer this question?A scientist produced transgenic zebrafish. She obtained a gene from silverside fish. The gene codes for a growth hormone (GH) She inserted copies of this GH gene into plasmids. She then microinjected these recombinant plasmids into fertilised egg cells of zebrafish. 0.9.1 Describe how enzymes could be used to insert the GH gene into a piasmid. [2 marks] Microinjection of DNA into fertilised egg cells is a frequent method of producing transgenic fish. However, the insertion of the transferred gene into nuclear DNA may be delayed. Consequently, the offspring of transgenic fish may not possess the desired characteristic Suggest and explain how delayed insertion of the GH gene could produce offspring of transgenic fish without the desired characteristic [2 marks] The scientist investigated whether the transferred GH gene increased the growth of transgenic zebrafish. She microinjected 2000 fertilised egg cells with the GH plasmid and left 2000 fertilised egg cells untreated. After 12 months, she determined the mean mass of the transgenic and non-transgenic fish. The results the scientist obtained are shown in Table 3. Table 3 A value of 22 * SD from the mean includes over 95% of the data. Type of zebrafish Mean mass of zebrafish / g (2 SD) Transgenic 1.79 ( 0.37) Non-transgenic 0.68 ( 0.13) 019.3 Using Table 3, what can you conclude about the effectiveness of the GH gene on the growth of zebrafish? [2 marks] 019. Explain how two features of the design of this investigation helped to ensure the validity of any conclusions obtained. Do not include calculating the mean or SD in your answer. [2 marks]
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Sizz
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Has anyone attempted all three sets? I’m gonna start them tomorrow after my chem exam so I don’t mind comparing answers after that
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ruqayyaha
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(Original post by Sizz)
Has anyone attempted all three sets? I’m gonna start them tomorrow after my chem exam so I don’t mind comparing answers after that
this is for the as one

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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Aribob
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#11
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#11
(Original post by ruqayyaha)
this is for the as one

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
Thanks for this! do you think you could help with a- level ones, I need a markscheme for the exchange of substances etc question paper
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Sizz
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#12
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#12
(Original post by lilyrose270)
what did people get for :
Q6.3 a dwarf pink flowered plant was crossed with a heterozygous tall white flowered plant. complete the genetic diagram to show all the possible genotypes and the ratio of phenotypes expected in the offspring of this cross.
hi, hope i’m not too late, just attempted it now.
dwarf pink flowered: ttC^R C^W
tall white flowered: TtC^W C^W
offspring genotypes: TtC^R C^W , TtC^W C^W , ttC^R C^W , ttC^W C^W
phenotypes: tall pink, tall white, dwarf pink, dwarf white
ratio: 4:4:4:4 simplifies to 1:1:1:1
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Sizz
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#13
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(Original post by jacob5556)
Anyone manage to answer this question?A scientist produced transgenic zebrafish. She obtained a gene from silverside fish. The gene codes for a growth hormone (GH) She inserted copies of this GH gene into plasmids. She then microinjected these recombinant plasmids into fertilised egg cells of zebrafish. 0.9.1 Describe how enzymes could be used to insert the GH gene into a piasmid. [2 marks] Microinjection of DNA into fertilised egg cells is a frequent method of producing transgenic fish. However, the insertion of the transferred gene into nuclear DNA may be delayed. Consequently, the offspring of transgenic fish may not possess the desired characteristic Suggest and explain how delayed insertion of the GH gene could produce offspring of transgenic fish without the desired characteristic [2 marks] The scientist investigated whether the transferred GH gene increased the growth of transgenic zebrafish. She microinjected 2000 fertilised egg cells with the GH plasmid and left 2000 fertilised egg cells untreated. After 12 months, she determined the mean mass of the transgenic and non-transgenic fish. The results the scientist obtained are shown in Table 3. Table 3 A value of 22 * SD from the mean includes over 95% of the data. Type of zebrafish Mean mass of zebrafish / g (2 SD) Transgenic 1.79 ( 0.37) Non-transgenic 0.68 ( 0.13) 019.3 Using Table 3, what can you conclude about the effectiveness of the GH gene on the growth of zebrafish? [2 marks] 019. Explain how two features of the design of this investigation helped to ensure the validity of any conclusions obtained. Do not include calculating the mean or SD in your answer. [2 marks]
9.1 restriction enzymes cut plasmid and ligase joins the gene and plasmid

9.2 not sure

9.3 I think we just talk about the standard deviation? It doesn’t overlap so the difference in mass is significant ?

9.4 large sample size therefore representative
and use of a control group for comparisons
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Em174
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#14
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#14
(Original post by Aribob)
Thanks for this! do you think you could help with a- level ones, I need a markscheme for the exchange of substances etc question paper
(Original post by ruqayyaha)
this is for the as one

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
1.2 is 18, 32, 16 because of the condensation reactions so you are losing 2 water molecules and that is 4 Hs and 2 Os
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mhaldermusic
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#15
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?A scientist produced transgenic zebrafish. She obtained a gene from silverside fish. The gene codes for a growth hormone (GH) She inserted copies of this GH gene into plasmids. She then microinjected these recombinant plasmids into fertilised egg cells of zebrafish. 0.9.1 Describe how enzymes could be used to insert the GH gene into a piasmid.
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Wonyo
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#16
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#16
(Original post by mhaldermusic)
?A scientist produced transgenic zebrafish. She obtained a gene from silverside fish. The gene codes for a growth hormone (GH) She inserted copies of this GH gene into plasmids. She then microinjected these recombinant plasmids into fertilised egg cells of zebrafish. 0.9.1 Describe how enzymes could be used to insert the GH gene into a piasmid.
i think you would talk about restriction endonuclaeases cutting up the DNA (gene from the silverside fish) and obtaining the plasmid. and then using DNA ligase to insert this gene into the plasmid
Last edited by Wonyo; 1 year ago
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Raiskhan160
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#17
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(Original post by Sizz)
hi, hope i’m not too late, just attempted it now.
dwarf pink flowered: ttC^R C^W
tall white flowered: TtC^W C^W
offspring genotypes: TtC^R C^W , TtC^W C^W , ttC^R C^W , ttC^W C^W
phenotypes: tall pink, tall white, dwarf pink, dwarf white
ratio: 4:4:4:4 simplifies to 1:1:1:1
Why did you put the arrow between the letters? Should I copy it like that or do it all together, thank you
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mhaldermusic
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#18
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#18
anyone figured out 9.2 yet??? pls help...
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georgia.mb04
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#19
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#19
(Original post by mhaldermusic)
anyone figured out 9.2 yet??? pls help...
Is this something to do with the fact because the restriction endonuclease produces complementary sticky ends for both the gene and plasmid that the DNA Ligase could just join the DMA and plasmid together like they were at the start so the process would have to be repeated until the gene was inserted into the plasmid. I could be completely wrong here but just an idea I had.
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mhaldermusic
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#20
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thanks man. I still unsure about this tbh
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