AQA qualification support AS/A-level Biology: Preparing to Teach Resources A

BOOKLET 2

Published date: Spring 2015 version1

Permission to reproduce all copyright materials have been applied for. In some cases, efforts to contact copyright holders have been unsuccessful and AQA will be happy to rectify any omissions of acknowledgements in future documents if required.

Contents Page

Specification at a glance and assessment objectives 5

Question papers and mark schemes 10

Extended response (structured free response)

A-level Paper 1, question 11 10 AS Paper 2, question 10 14

Comprehension questions A-level Paper 2, question 10 17 AS Paper 1, question 9 21

Assessment of practical skills at A-level A-level Paper 3, question 5 24

New topics A-level Paper 2, question 3 29 A-level Paper 2, question 7 32

Essay questions A-level Paper 3, question 6 35

Command words 41

Useful links 45

4

Specifications at a glance

5

6

Assessment objectives

7

8

9

Messenger RNA (mRNA) is used during translation to form polypeptides. Describe how mRNA is produced in the nucleus of a cell.

[6 marks]

1 1 . 1

[Extra space]

10

Describe the structure of proteins. [5 marks]

Question 11 continues on the next page

1 1 . 2

[Extra space]

11

Describe how proteins are digested in the human gut. [4 marks]

1 1 . 3

[Extra space]

12

Question Marking Guidance Mark Comments

11.1 1. Helicase;

2. Breaks hydrogen bonds;

3. Only one DNA strand acts as template;

4. RNA nucleotides attracted to exposed bases;

5. (Attraction) according to base pairing rule;

6. RNA polymerase joins (RNA) nucleotides together;

7. Pre-mRNA spliced to remove introns;

6 max

11.2 1. Polymer of amino acids;

2. Joined by peptide bonds;

3. Formed by condensation;

4. Primary structure is order of amino acids;

5. Secondary structure is folding of polypeptide chain due to hydrogen bonding;

6. Tertiary structure is 3-D folding due to hydrogen bonding and ionic/disulfide bonds;

7. Quaternary structure is two or more polypeptide chains;

5 max

5. Accept alpha

helix/pleated sheet

11.3 1. Hydrolysis of peptide bonds;

2. Endopeptidases break polypeptides into smaller peptide chains;

3. Exopeptidases remove terminal amino acids;

4. Dipeptidases hydrolyse/break down dipeptides into amino acids;

4

13

When a vaccine is given to a person, it leads to the production of antibodies against a disease-causing organism. Describe how.

[5 marks]

1 0 . 1

[Extra space]

14

Describe the difference between active and passive immunity.

[5 marks]

END OF QUESTIONS

1 0 . 2

[Extra space]

15

Question Marking Guidance Mark Comments

10.1 1. Vaccine contains antigen from pathogen;

2. Macrophage presents antigen on its surface;

3. T cell with complementary receptor protein binds to antigen;

4. T cell stimulates B cell;

5. (With) complementary antibody on its surface;

6. B cell secretes large amounts of antibody;

7. B cell divides to form clone all secreting/producing same antibody;

5 max

10.2 1. Active involves memory cells, passive does not;

2. Active involves production of antibody by plasma cells/memory cells;

3. Passive involves antibody introduced into body from outside/named source;

4. Active long term, because antibody produced in response to antigen;

5. Passive short term, because antibody (given) is broken down;

6. Active (can) take time to develop/work, passive fast acting;

5 max

16

Read the following passage carefully.

A large and growing number of disorders are now known to be due to types of mitochondrial disease (MD). MD often affects skeletal muscles, causing muscle weakness. We get our mitochondria from our mothers, via the fertilised egg cell. Fathers do not pass on mitochondria via their sperm. Some mitochondrial diseases 5 are caused by mutations of mitochondrial genes inside the mitochondria. Most mitochondrial diseases are caused by mutations of genes in the cell nucleus that are involved in the functioning of mitochondria. These mutations of nuclear DNA produce recessive alleles. One form of mitochondrial disease is caused by a mutation of a mitochondrial 10 gene that codes for a tRNA. The mutation involves substitution of guanine for adenine in the DNA base sequence. This changes the anticodon on the tRNA. This results in the formation of a non-functional protein in the mitochondrion. There are a number of ways to try to diagnose whether someone has a mitochondrial disease. One test involves measuring the concentration of 15 lactate in a person’s blood after exercise. In someone with MD, the concentration is usually much higher than normal. If the lactate test suggests MD, a small amount of DNA can be extracted from mitochondria and DNA sequencing used to try to find a mutation.

Use information in the passage and your own knowledge to answer the following questions.

Mitochondrial disease (MD) often causes muscle weakness (lines 1–3). Use your knowledge of respiration and muscle contraction to suggest explanations for this effect of MD. [3 marks]

1 0

1 0 . 1

[Extra space]

17

Two couples, couple A and couple B, had one or more children affected by a mitochondrial disease. The type of mitochondrial disease was different for each couple. None of the parents showed signs or symptoms of MD.

• Couple A had four children who were all affected by an MD. • Couple B had four children and only one was affected by an MD.

Use the information in lines 5–9 and your knowledge of inheritance to suggest why: • all of couple A’s children had an MD • only one of couple B’s children had an MD. [4 marks]

Question 10 continues on the next page

1 0 . 2

[Extra space]

Couple A

Couple B

18

Suggest how the change in the anticodon of a tRNA leads to MD (lines 10–13).

[3 marks]

If someone has MD, the concentration of lactate in their blood after exercise is usually much higher than normal (lines 15–17). Suggest why.

[3 marks]

A small amount of DNA can be extracted from mitochondria and DNA sequencing used to try to find a mutation (lines 18–19).

From this sample: • how would enough DNA be obtained for sequencing? • how would sequencing allow the identification of a mutation?

[2 marks]

1 0 . 3

[Extra space]

[Extra space]

1 0 . 4

1 0 . 5

19

Question Marking Guidance Mark Comments

10.1 1. Reduction in ATP production by aerobic respiration;

2. Less force generated because fewer actin and myosin interactions in muscle;

3. Fatigue caused by lactate from anaerobic respiration;

3

10.2 Couple A, 1. Mutation in mitochondrial DNA/DNA of

mitochondrion affected;

2. All children got affected mitochondria from mother;

3. (Probably mutation) during formation of mother’s ovary/eggs;

Couple B, 4. Mutation in nuclear gene/DNA in nucleus affected;

5. Parents heterozygous;

6. Expect 1 in 4 homozygous affected;

4 max

10.3 1. Change to tRNA leads to wrong amino acid being incorporated into protein;

2. Tertiary structure (of protein) changed;

3. Protein required for oxidative phosphorylation/the Krebs cycle, so less/no ATP made;

3

10.4 1. Mitochondria/aerobic respiration not producing much/any ATP;

2. (With MD) increased use of ATP supplied by increase in anaerobic respiration;

3. More lactate produced and leaves muscle by (facilitated) diffusion;

3

10.5 1. Enough DNA using PCR;

2. Compare DNA sequence with ‘normal’ DNA;

2

20

Read the following passage.

Herpes simplex virus (HSV) infects nerve cells in the face, including some near the lips. Like many other viruses, HSV can remain inactive inside the body for years. When HSV becomes active, it causes cold sores around the mouth.

Human cells infected with a virus may undergo programmed cell death. While HSV is inactive inside the body, only one of its genes is transcribed. This gene is the latency-associated transcript (LAT) gene that prevents programmed cell death of an infected nerve cell.

Scientists have found that transcription of the LAT gene produces a microRNA. 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.

Use information from the passage and your own knowledge to answer the following questions.

5

10

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

9

0 9 . 1

[Extra space]

21

Turn over

HSV can remain inactive inside the body for years (lines 2–3). Explain why this virus can be described as inactive. [2 marks]

Suggest one advantage of programmed cell death (line 4). [1 mark]

The scientists concluded that production of this microRNA allows HSV to remain in the body for years (lines 10–12). Explain how this microRNA allows HSV to remain in the body for years. [4 marks]

END OF QUESTIONS

0 9 . 2

0 9 . 3

0 9 . 4

[Extra space]

22

Question Marking Guidance Mark Comments

09.1 1. Outside of virus has antigens/proteins;

2. With complementary shape to receptor/protein in membrane of cells;

3. (Receptor/protein) found only on membrane of nerve cells;

3 Accept converse argument

09.2 1. No more (nerve) cells infected/no more cold sores form;

2. (Because) virus is not replicating;

2

09.3 Prevents replication of virus; 1

09.4 MicroRNA binds to cell’s mRNA (no mark)

1. (Binds) by specific base pairing;

2. (So) prevents mRNA being read by ribosomes;

3. (So) prevents translation/production of proteins;

4. (Proteins) that cause cell death;

4

23

Scientists investigated the effect of a mycorrhizal fungus on the growth of pea plants with a nitrate fertiliser or an ammonium fertiliser. The fertilisers were identical, except for nitrate or ammonium. The scientists took pea seeds and sterilised their surfaces. They planted the seeds in soil that had been heated to 85 oC for 2 days before use. The soil was sand that contained no mineral ions useful to the plants.

Explain why the scientists sterilised the surfaces of the seeds and grew them in soil that had been heated to 85 oC for 2 days.

[2 marks]

Explain why it was important that the soil contained no mineral ions useful to the plants.

[1 mark]

The pea plants were divided into four groups, A, B, C and D. • Group A – heat-treated mycorrhizal fungus added, nitrate fertiliser • Group B – mycorrhizal fungus added, nitrate fertiliser • Group C – heat-treated mycorrhizal fungus added, ammonium fertiliser • Group D – mycorrhizal fungus added, ammonium fertiliser The heat-treated fungus had been heated to 120 oC for 1 hour.

Explain how groups A and C act as controls. [2 marks]

5

0 5 . 1

0 5 . 2

0 5 . 3

24

Question 5 continues on the next page

After 6 weeks, the scientists removed the plants from the soil and cut the roots from the shoots. They dried the plant material in an oven at 90 oC for 3 days. They then determined the mean dry masses of the roots and shoots of each group of pea plants.

Suggest what the scientists should have done during the drying process to be sure that all of the water had been removed from the plant samples.

[2 marks]

The scientists’ results are shown in Table 3.

Table 3

Treatment

Mean dry mass / g per plant (+ standard deviation)

Root Shoot A – heat-treated fungus and nitrate fertiliser

0.40 (±0.05)

1.01 (±0.12)

B – fungus and nitrate fertiliser

1.61 (±0.28)

9.81 (±0.33)

C – heat-treated fungus and ammonium fertiliser

0.34 (±0.03)

0.96 (±0.26)

D – fungus and ammonium fertiliser

0.96 (±0.18)

4.01 (±0.47)

0 5 . 4

25

What conclusions can be drawn from the data in Table 3 about the following? [4 marks] The effects of the fungus on growth of the pea plants.

The effects of nitrate fertiliser and ammonium fertiliser on growth of the pea plants.

Question 5 continues on the next page

0 5 . 5

26

The scientists determined the dry mass of the roots and shoots separately. The reason for this was they were interested in the ratio of shoot to root growth of pea plants. It is the shoot of the pea plant that is harvested for commercial purposes. Explain why determination of dry mass was an appropriate method to use in this investigation.

[2 marks]

Which treatment gave the best result in commercial terms? Justify your answer. [2 marks]

0 5 . 6

0 5 . 7

27

Question Marking Guidance Mark Comments

05.1 1. To kill any fungus/bacteria on surface of seeds or in soil;

2. So only the added fungus has any effect;

2

05.2 So that only nitrate or ammonia/type of fertiliser affects growth;

1

05.3 1. So that effects of nitrate or ammonium alone could be seen;

2. So that effects of fungus can be seen;

2

05.4 1. Weigh samples at intervals during drying;

2. To see if weighings became constant (by 3 days);

2

05.5 With live fungus – showing effects of the fungus: 1. Fungus increases growth of roots and shoots in

both;

2. Produces greater growth with nitrate;

With heat-treated fungus – showing effects of fertiliser: 3. Similar dry masses for roots and shoots;

4. (Probably) no significant difference because SDs overlap;

4

05.6 1. Dry mass measures/determines increase in biological/organic material;

2. Water content varies;

2

05.7 1. Fungus with nitrate-containing fertiliser gave largest shoot: root ratio;

2. And largest dry mass of shoot;

3. 6.09:1 compared with ammonium-containing fertiliser 4.18:1;

2 max

28

In fruit flies, the genes for body colour and wing length are linked. Explain what this means. [1 mark]

A scientist investigated linkage between the genes for body colour and wing length. He carried out crosses between fruit flies with grey bodies and long wings and fruit flies with black bodies and short wings. Figure 2 shows his crosses and the results. • G represents the dominant allele for grey body and g represents the

recessive allele for black body. • N represents the dominant allele for long wings and n represents the

recessive allele for short wings.

Figure 2 Phenotype of parents grey body, × black body,

long wings short wings Genotype of parents GGNN ggnn Genotype of offspring GgNn

Phenotype of offspring all grey body, long wings These offspring were crossed with flies homozygous for black body and short wings. The scientist’s results are shown in Figure 3.

Figure 3

GgNn crossed with ggnn

Grey body, long wings

Black body, short wings

Grey body, short wings

Black body, long wings

Number of offspring 975 963 186 194

0 3 . 1

29

Use your knowledge of gene linkage to explain these results. [4 marks]

If these genes were not linked, what ratio of phenotypes would the scientist have expected to obtain in the offspring?

[1 mark]

Which statistical test could the scientist use to determine whether his observed results were significantly different from the expected results? Give the reason for your choice of statistical test.

[2 marks]

0 3 . 2

0 3 . 4

0 3 . 3

[Extra space]

30

Question Marking Guidance Mark Comments

03.1 (Genes/loci) on same chromosome; 1

03.2 1. GN and gn linked;

2. GgNn individual produces mainly GN and gn gametes;

3. Crossing over produces some/few Gn and gN gametes;

4. So few(er) Ggnn and ggNn individuals;

4

03.3 (Grey long:grey short:black long:black short)

=1:1:1:1 1

03.4 1. Chi squared test;

2. Categorical data;

2

31

Osmoreceptors are specialised cells that respond to changes in the water potential of the blood. Give the location of osmoreceptors in the body of a mammal. [1 mark]

When a person is dehydrated, the cell volume of an osmoreceptor decreases. Explain why. [2 marks]

Stimulation of osmoreceptors can lead to secretion of the hormone ADH. Describe and explain how the secretion of ADH affects urine produced by the kidneys. [4 marks]

7

0 7 . 1

0 7 . 2

0 7 . 3

[Extra space]

32

The efficiency with which the kidneys filter the blood can be measured by the rate at which they remove a substance called creatinine from the blood. The rate at which they filter the blood is called the glomerular filtration rate (GFR). In 24 hours, a person excreted 1660 mg of creatinine in his urine. The concentration of creatinine in the blood entering his kidneys was constant at 0.01 mg cm–3.

Calculate the GFR in cm3 minute–1. [1 mark]

Creatinine is a breakdown product of creatine found in muscle tissues. Apart from age and gender, give two factors that could affect the concentration of creatinine in the blood. [1 mark]

0 7 . 4

0 7 . 5

Answer =

1

2

33

Question Marking Guidance Mark Comments

07.1 Hypothalamus; 1

07.2 1. Water potential of blood will decrease;

2. Water moves from osmoreceptor into blood by osmosis;

2

07.3 1. Permeability of membrane/cells (to water) is increased;

2. More water absorbed from/leaves distal tubule/collecting duct;

3. Smaller volume of urine;

4. Urine becomes more concentrated;

4

07.4 115.2/115.3 (cm3 minute–1); 1

07.5 Any two of the following for 1 mark;

Muscle/body mass

Ethnicity

Exercise

Kidney disease – do not accept ‘health’

1

34

Section B

Answer one question.

Write an essay on one of the topics below.

EITHER

The importance of movement in living organisms.

[25 marks] OR

The importance of receptors in living organisms.

[25 marks]

6

0 6 . 1

0 6 . 2

35

Question 6 Level of response marking guidance Level of response marking instructions Level of response mark schemes are broken down into five levels, each of which has a descriptor. The descriptor for the level shows the average performance for the level. There are five marks in each level. Thus the descriptor for the level represents the mid mark in that level. Before you apply the mark scheme to a student’s answer, read through the answer and annotate it (as instructed) to show the qualities that are being looked for. You can then apply the mark scheme. Step 1 Determine a level Start at the lowest level of the mark scheme and use it as a ladder to see whether the answer meets the descriptor for that level. The descriptor for the level indicates the different qualities that might be seen in the student’s answer for that level. If it meets the lowest level then go to the next one and decide if it meets this level, and so on, until you have a match between the level descriptor and the answer. With practice and familiarity you will find that for better answers you will be able to quickly skip through the lower levels of the mark scheme. When assigning a level you should look at the overall quality of the answer and not look to pick holes in small and specific parts of the answer where the student has not performed quite as well as the rest. If the answer covers different aspects of different levels of the mark scheme you should use a best fit approach for defining the level and then use the variability of the response to help decide the mark within the level. i.e. if the response is predominantly level 3 with a small amount of level 4 material it would be placed in level 3 but be awarded a mark near the top of the level because of the level 4 content. Step 2 Determine a mark Once you have assigned a level you need to decide on the mark. The descriptors on how to allocate marks can help with this. The exemplar materials used during standardisation will help. There will be an answer in the standardising materials which will correspond with each level of the mark scheme. This answer will have been awarded a mark by the Lead Examiner. You can compare the student’s answer with the example to determine if it is the same standard, better or worse than the example. You can then use this to allocate a mark for the answer based on the Lead Examiner’s mark on the example. You may well need to read back through the answer as you apply the mark scheme to clarify points and assure yourself that the level and the mark are appropriate. Indicative content in the mark scheme is provided as a guide for examiners. It is not intended to be exhaustive and you must credit other valid points. Students do not have to cover all of the points mentioned in the indicative content to reach the highest level of the mark scheme. An answer which contains nothing of relevance to the question must be awarded no marks.

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Question 6 Level of response marking guidance

21–25

Extended abstract

Generalised beyond specific context

Response shows holistic approach to the question with a fully integrated answer which makes clear links between several different topics and the theme of the question.

Biology is detailed and comprehensive A-level content, uses appropriate terminology, and is very well written and always clearly explained.

No significant errors or irrelevant material.

For top marks in the band, the answer shows evidence of reading beyond specification requirements.

16–20

Relational

Integrated into a whole

Response links several topics to the main theme of the question, to form a series of interrelated points which are clearly explained.

Biology is fundamentally correct A-level content and contains some points which are detailed, though there may be some which are less well developed, with appropriate use of terminology.

Perhaps one significant error and, or, one irrelevant topic which detracts from the overall quality of the answer.

11–15

Multistructural

Several aspects covered but they are unrelated

Response mostly deals with suitable topics but they are not interrelated and links are not made to the theme of the question.

Biology is usually correct A-level content, though it lacks detail. It is usually clearly explained and generally uses appropriate terminology.

Some significant errors and, or, more than one irrelevant topic.

6–10

Unistructural

Only one or few aspects covered

Response predominantly deals with only one or two topics that relate to the question.

Biology presented shows some superficial A-level content that may be poorly explained, lacking in detail, or show limited use of appropriate terminology.

37

May contain a number of significant errors and, or, irrelevant topics.

1–5 Unfocused

Response only indirectly addresses the theme of the question and merely presents a series of biological facts which are usually descriptive in nature or poorly explained and at times may be factually incorrect.

Content and terminology is generally below A-level.

May contain a large number of errors and, or, irrelevant topics.

0 Nothing of relevance or no response.

38

Question 06.1: The importance of movement in living organisms The following are suitable topic areas from the specification that could obviously be used to demonstrate the importance of movement. Movement can include any movement from the molecular level to whole organisms. Please note that to obtain full credit, students must use information to show the importance of movement, not just write about topics that include movement. In order to fully address the question and reach the highest mark bands students must also include at least five topics in their answer, to demonstrate a synoptic approach to the essay.

Specification Reference Topic Area

3.1.4.2 Enzyme-catalysed reactions 3.1.5.2 DNA replication 3.1.6 ATP 3.2.2 Cell division 3.2.3 Transport across membranes 3.2.4 Immune response 3.2.2 Gas exchange 3.3.3 Digestion and absorption 3.3.4.1, 4.2 Mass transport 3.4.2 DNA and protein synthesis 3.4.3 Meiosis 3.5.1 Photosynthesis 3.5.2 Respiration 3.6.1 Survival and response 3.6.1.2 Receptors 3.6.1.3 Control of heart rate 3.6.2.1 Nerve impulses 3.6.2.2 Synapses 3.6.2.2 Synaptic transmission 3.6.3 Skeletal muscle 3.6.4.2 Control of blood glucose concentration 3.6.4.3 Control of blood water potential 3.7.3 Evolution (population isolation and movement between) 3.8.2.2 Regulation of transcription and translation 3.8.2.3 Gene expression and cancer

Students may be able to show the relevance of other topics from the specification. Note, other topics from beyond the specification can be used, providing they relate to the title and contain factually correct material of at least an A-level standard. Credit should not be given for topics beyond the specification which are below A-level standard.

39

Question 06.2: The importance of receptors in living organisms The following are suitable topic areas from the specification that could obviously be used to demonstrate the importance of receptors. Receptors can include any receptors from the molecular level to whole organs. Please note that to obtain full credit, students must use information to show the importance of receptors, not just write about topics that include receptors. In order to fully address the question and reach the highest mark bands students must also include at least five topics in their answer, to demonstrate a synoptic approach to the essay.

Specification Reference Topic Area

3.1.4.2 Enzymes 3.2.1.2 Structure of prokaryotic cells and of viruses 3.2.3 Transport across cell membranes 3.2.4 Cell recognition and the immune system 3.3.4.1 Mass transport in animals 3.4.2 DNA and protein synthesis 3.5.1 Photosynthesis 3.5.2 Respiration 3.6.1.1 Survival and response 3.6.1.2 Receptors 3.6.1.3 Control of heart rate 3.6.2.1 Nerve impulses 3.6.2.2 Synaptic transmission 3.6.3 Skeletal muscles 3.6.4.1 Principles of homeostasis 3.6.4.2 Control of blood glucose concentration 3.6.4.3 Control of blood water potential 3.8.2.2 Regulation of transcription and translation 3.8.2.3 Gene expression and cancer

Students may be able to show the relevance of other topics from the specification. Note, other topics from beyond the specification can be used, providing they relate to the title and contain factually correct material of at least an A-level standard. Credit should not be given for topics beyond the specification which are below A-level standard.

40

Command words

Command words are the words and phrases used in exams and other assessment tasks that tell students

how they should answer the question.

The following command words are taken from Ofqual's official list of command words and their meanings

that are relevant to this subject. In addition, where necessary, we have included our own command words

and their meanings to complement Ofqual's list.

Analyse

Separate information into components and identify their characteristics

Annotate

Add notation or labelling to a graph, diagram or other drawing

Apply

Put into effect in a recognised way

Argue

Present a reasoned case

Assess

Make an informed judgement

Calculate

Work out the value of something

Comment

Present an informed opinion

Compare

Identify similarities and/ or differences

Complete

Finish a task by adding to given information

41

Consider

Review and respond to given information

Construct

NA

Contrast

Identify differences

Criticise

Access worth against explicit expectations

Debate

Present different perspectives on an issue

Deduce

Draw conclusions from information provided

Define

Specify meaning

Describe

Give an account of

Design

Set out how something will be done

Determine

Use given data or information to obtain an answer

Develop

Take forward or build upon given information

42

Discuss

Present key points

Distinguish

List the differences between different items

Draw

Produce a diagram

Estimate

Assign an approximate value

Evaluate

Judge from available evidence

Explain

Give reasons

Explore

Investigate without preconceptions about the outcome

Give

Produce an answer from recall or from given information

Identify

Name or otherwise characterise

Justify

Support a case with evidence

Label

Provide appropriate names on a diagram

43

List

List a number of features or points without further elaboration

Name

Identify using a recognised technical term

Outline

Set out main characteristics

Predict

Give a plausible outcome

Relate

Give a technical term or its equivalent

Show

Provide structured evidence to reach a conclusion

Sketch

Draw approximately

State

Express in clear terms

Suggest

Present a possible case

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Useful links Visit our website to find the latest resources and information on A-level planning, teaching and assessment resources which includes the Specifications and Practical handbook: http://www.aqa.org.uk/subjects/science/as-and-a-level/biology-7401-7402 Specifications launch webcast: Part 1 - Science main changes and rules http://aqa.adobeconnect.com/p3w2qz5jdya/ Specifications launch webcast: Part 2 - Biology content http://aqa.adobeconnect.com/p84ur3oncxd/

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