AS Biology P1 MCQs Preparation Thread

[Soldier313]

Soldier313

^why C and not B?
the compounds in plants would be i think proteins?

umm,,,they'll be converted to ammonium ions first...n then to nitrates...which plants will take,,,

nt sure

didn't understand

 

[N.M]

can u tell me what is respiratory bronchiole...and what is terminal?

see table 11.1 pg 142 biology OCR by mary jones
very helpful table...

 

[Just M]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Biology (9700)/9700_w07_qp_1.pdf

Question 40! Someone please help!! i think it was answered in the thread before but i cant find it!

The ans is C. The ques asked about NET. 4% is not the net. 1.75% is the net productivity. Hence, it's energy passed on divided by the net energy available

 

[Capricedcapri]

Hey do any one have any idea about the n15 nd n14 sumthing related to generations in biology....there Is a question in may June o6 McQ 22 .....cud u help me with that??

 

[Irfan1995]

Aoa!
need help with some more questions plz...
View attachment 12766View attachment 12767View attachment 12768View attachment 12769

19) Lets review the length of spindle fibres during mitosis. During prophase, the two centrioles move to opposite ends of the nucleus and start building up the spindle fibres and it grows until metaphase. During metaphase, they are in their largest because they moved the chromosomes from one end of the cell to the centre. During anaphase, the two centromeres are pulled to opposite ends of the cell, so the spindle fibres get shorter. They keep on getting shorter and shorter until they finally detach from the chromatids, which is at stage D.

30) What you must know is that all four of them are completely true statements. However, only 2 and 4 prove the root pressure model. Both 1 and 3 support the transpiration pull model.

33) I'm pretty new to XPF forums, so I don't know how to draw a table. I'll just try to summarize the features of the lungs here.
Pathway of air: Trachea ---> Bronchus ---> Terminal bronchiole ---> Respiratory bronchiole ---> alveolar duct ---> alveoli
Cartilage: trachea and bronchi only
Goblet cells: trachea and bronchi only
Smooth muscle: trachea, bronchi, and terminal bronchioles
cilia: trachea, bronchi, terminal bronchioles, and respiratory bronchioles

Alveolar duct and alveoli have no features. Both are sites of gas exchange

40) Denitrifying bacteria are obligate anerobes (i.e they CANNOT work in conditions with oxygen). In poorly drained soils, the ground is completely waterlogged, so there is no way for oxygen to enter.

 

[XPFMember]

see table 11.1 pg 142 biology OCR by mary jones
very helpful table...

umm dont have that..

 

[Irfan1995]

Which disease is NOT likely to be passed directly from parent to child?
A. Cholera
B. HIV AIDS
C. malaria
D. Tuberculosis

The answer is malaria...but why cant it be cholera?

Cholera can be passed through food/contaminated water
HIV/AIDS can be passed through breast-feeding
Malaria needs a mosquito to transfer Plasmodium (cannot pass from human to human directly)
Tuberculosis can be passed by sneezing/coughing

 

[Irfan1995]

umm dont have that..

I posted that table in my reply

 

[Irfan1995]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Biology (9700)/9700_w07_qp_1.pdf

Question 40! Someone please help!! i think it was answered in the thread before but i cant find it!

Assume 100 kJ of light enters the system. From the diagram, 1.75 kJ would be the net productivity. Also, the diagram says that 0.75kJ would be incorporated into the primary consumers. So the ratio of the two values would give you the proportion of net productivities. That would be 0.75/1.75 which is C

 

[geek101]

Hey do any one have any idea about the n15 nd n14 sumthing related to generations in biology....there Is a question in may June o6 McQ 22 .....cud u help me with that??

heres the experiment.....the pictures are from the mary jones book.....
so Meselson and his buddy decided to prove the SEMI CONSERVATIVE REPLICATION.
they used E coli (you dont need to know what they grew and what nitrogen source they provided)....just know there are two isotopes of Nitrogen a heavy one, N-15 and a light one N-15.

1) they grew the bacteria in N-15. so the e cloi used the nitrogen - 15 and incorporated it into their DNA. Now the DNA was entirely N-15. They centrifuged it and a band on the lower level of the test tube formed, because N -15 is heavy
(pic 1 on page 70)

2) Then they placed the e cloi in a medium of N -14. because semi conservative replication occurs.....
currently DNA has 2 N-15 strands
when they put it in N-14
The N-15 strand each serves as a template
so a new N-14 strand comes and attaches to it
and 2 DNA molecules are formed each with one N-15 and one N-14 strand.
When this mixture is centrifuged a band is formed in the middle, since theres light and heavy nitrogen.
(pic 2 page 70)

3) Now when the e cloi are left for more generations...in N-14
now....
there are DNA with one N-14 and one N-15 strand.
when this is left for more time in N-14....
each N-15 and N-14 strand serves as a template....
so to the N-15 a N-14 attaches and to the N-14 another N-14 attaches....
this produces a DNA molecule with N-14 and 15 and another molecule with both N-14...
when this is centrifuged There is a band in the middle, and a band of only N14 which is on the top level of the test tube, because its light.
(pic 3 page 70)

4) as the e coli are left for longer in the N14 the band with purely N14 gets wider. because more and more templates of N14 are there to which more N14 strands attaches....
(pic 4 page 70)

the percentage of hybrid DNA becomes half each time....

 

[XPFMember]

19) Lets review the length of spindle fibres during mitosis. During prophase, the two centrioles move to opposite ends of the nucleus and start building up the spindle fibres and it grows until metaphase. During metaphase, they are in their largest because they moved the chromosomes from one end of the cell to the centre. During anaphase, the two centromeres are pulled to opposite ends of the cell, so the spindle fibres get shorter. They keep on getting shorter and shorter until they finally detach from the chromatids, which is at stage D.

30) What you must know is that all four of them are completely true statements. However, only 2 and 4 prove the root pressure model. Both 1 and 3 support the transpiration pull model.

33) I'm pretty new to XPF forums, so I don't know how to draw a table. I'll just try to summarize the features of the lungs here.
Pathway of air: Trachea ---> Bronchus ---> Terminal bronchiole ---> Respiratory bronchiole ---> alveolar duct ---> alveoli
Cartilage: trachea and bronchi only
Goblet cells: trachea and bronchi only
Smooth muscle: trachea, bronchi, and terminal bronchioles
cilia: trachea, bronchi, terminal bronchioles, and respiratory bronchioles

Alveolar duct and alveoli have no features. Both are sites of gas exchange

40) Denitrifying bacteria are obligate anerobes (i.e they CANNOT work in conditions with oxygen). In poorly drained soils, the ground is completely waterlogged, so there is no way for oxygen to enter.

JazakAllahu Khairen!!!!!!!!!!!!!!!!!!!

that was really helpful....

 

[Capricedcapri]

heres the experiment.....the pictures are from the mary jones book.....
so Meselson and his buddy decided to prove the SEMI CONSERVATIVE REPLICATION.
they used E coli (you dont need to know what they grew and what nitrogen source they provided)....just know there are two isotopes of Nitrogen a heavy one, N-15 and a light one N-15.

1) they grew the bacteria in N-15. so the e cloi used the nitrogen - 15 and incorporated it into their DNA. Now the DNA was entirely N-15. They centrifuged it and a band on the lower level of the test tube formed, because N -15 is heavy
(pic 1 on page 70)

2) Then they placed the e cloi in a medium of N -14. because semi conservative replication occurs.....
currently DNA has 2 N-15 strands
when they put it in N-14
The N-15 strand each serves as a template
so a new N-14 strand comes and attaches to it
and 2 DNA molecules are formed each with one N-15 and one N-14 strand.
When this mixture is centrifuged a band is formed in the middle, since theres light and heavy nitrogen.
(pic 2 page 70)

3) Now when the e cloi are left for more generations...in N-14
now....
there are DNA with one N-14 and one N-15 strand.
when this is left for more time in N-14....
each N-15 and N-14 strand serves as a template....
so to the N-15 a N-14 attaches and to the N-14 another N-14 attaches....
this produces a DNA molecule with N-14 and 15 and another molecule with both N-14...
when this is centrifuged There is a band in the middle, and a band of only N14 which is on the top level of the test tube, because its light.
(pic 3 page 70)

4) as the e coli are left for longer in the N14 the band with purely N14 gets wider. because more and more templates of N14 are there to which more N14 strands attaches....
(pic 4 page 70)

the percentage of hybrid DNA becomes half each time....

Thank u soooo much!!

 

[confused123]

i thi

are u sure solute potential and water potential are same???????solute potential mean concn of solute bt water potential means potential of water nt solute.isn't it???

i think more the negative solute potential in number the more solute content it has.

 

[confused123]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Biology (9700)/9700_w06_qp_1.pdf
MCQ 15 PLS
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Biology (9700)/9700_s06_qp_1.pdf
MCQ 6,20,29 ,30

q.30 anyone?

 

[XPFMember]

q.30 anyone?

Aoa wr wb!
well what happens, between 1 and 2 , as well as between 3 and 4...both the valves are close...

at 1 ....the av valves close...at 2 semilunar valves open...between that both are closed...
at 3 the semilunar valves close....but it's only at 4 when the av vlves will open...hence between 3 and 4 too, both valves are closed...

so calculate the total time..

P.S. Due to difficulty reading the graph...this question was removed...according to the er

 

[Casablanca]

heres the experiment.....the pictures are from the mary jones book.....
so Meselson and his buddy decided to prove the SEMI CONSERVATIVE REPLICATION.
they used E coli (you dont need to know what they grew and what nitrogen source they provided)....just know there are two isotopes of Nitrogen a heavy one, N-15 and a light one N-15.

1) they grew the bacteria in N-15. so the e cloi used the nitrogen - 15 and incorporated it into their DNA. Now the DNA was entirely N-15. They centrifuged it and a band on the lower level of the test tube formed, because N -15 is heavy
(pic 1 on page 70)

2) Then they placed the e cloi in a medium of N -14. because semi conservative replication occurs.....
currently DNA has 2 N-15 strands
when they put it in N-14
The N-15 strand each serves as a template
so a new N-14 strand comes and attaches to it
and 2 DNA molecules are formed each with one N-15 and one N-14 strand.
When this mixture is centrifuged a band is formed in the middle, since theres light and heavy nitrogen.
(pic 2 page 70)

3) Now when the e cloi are left for more generations...in N-14
now....
there are DNA with one N-14 and one N-15 strand.
when this is left for more time in N-14....
each N-15 and N-14 strand serves as a template....
so to the N-15 a N-14 attaches and to the N-14 another N-14 attaches....
this produces a DNA molecule with N-14 and 15 and another molecule with both N-14...
when this is centrifuged There is a band in the middle, and a band of only N14 which is on the top level of the test tube, because its light.
(pic 3 page 70)

4) as the e coli are left for longer in the N14 the band with purely N14 gets wider. because more and more templates of N14 are there to which more N14 strands attaches....
(pic 4 page 70)

the percentage of hybrid DNA becomes half each time....

So let's say they left it for another generation, would we get two bands that are equally as thick? Btw, your explanation helped me a lot. I never understood this before. Thanks.

 

[geek101]

So let's say they left it for another generation, would we get two bands that are equally as thick? Btw, your explanation helped me a lot. I never understood this before. Thanks.

step 3 will have two bands with equal widths...

 

[XPFMember]

aoa ..
need help with this

 

[Soldier313]

Can someone please help with my qns
http://www.xtremepapers.com/communi...-preparation-thread.17120/page-38#post-336173

Thanx a lot!

 

[Babri]

aoa ..
need help with this

View attachment 12819

When p53 is present it binds with damaged DNA so it doesn't replicate & no problems arise. However when p53 is absent damaged DNA will no longer be inhibited from replicating so this would cause problems aka cancer. (damaged DNA replication could lead to cancer as it could have mutated to multiply uncontrollably)