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TRANSCRIPT
7 7' (.t,,' :::.. T .._, --THE KENYA NATIONAL EXAMINATIONS COUNCIL Kenya Certificate of Secondary Education a 231/2 BIOLOGY Paper 2
Nov. 2017 - 2 hours
Name . ·-............................................... . Index N umber . .......................... .......................... ..
Candidate's Signature .... ........ ... .......... Date ........................... .......................................... ..
Instructions to candidates
(a) Write your name and index number in the spaces provided above. {b) Sign and write the date of examination in the spaces provided above. (c) This paper consists of two sections; A and B. (d) Answer all the questions in section A in the spaces provided. (e) In section B answer question 6 (compulsory) and either question 7 or 8 in the spaces prov,ded a~er
question 8. {f) This paper consists of 12 printed pages.
(!]) Candidates should check the question paper to ascertain that all the pages are printed as indicat.d and that no questions are missing.
(h) Candidates should answer all the questions in English.
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For Examiner's Use Only
Section Question Maximum Candidate's Score Score
1 8
2 8
A 3 8
4 8
5 8
6 20
B 7 20
8 20
Total Score
0 2017 The Kenya National Examinations Council 231/ 2
Tum over
2
SECTION A (40 maria)
Al'ISl#l'llll tJw fllOlloNI In this s«llon In the 1poca provided.
1. Th- cftS au lnlOM ,epa n j?'II • nucleus.
E F
(a) Name the structun:s labelled E and F. (2 matks)
Ci> E ... N..~!.~.~;J .. ,2.L.~. &F../1 .. 1 .• 2._§ .... 4/ ... b..1f:J1J··f ·~-r1~.-:j. ............. . F ... N..f!.f-J#..Y.. .. /r.!:/b./ .JJ.~/ l!.r..~; ......................... · ····· ······ · ·· · ......... . State the function off. ( l mark)
J..(!, __ l:!,/4dµ ... ./141!J/.l.~ .. 4 ... /'d,d.r~ ..... l.~ ... ~ .. ~!... o.f: ..... ~ ....... n..~ .;. ... j:, .... f!f..!.l ... fo.. ...... tl.~~ .. !. ...... .. ..
············································--··················································· .............................. . (iii") W'db reference to the nucleus, llalc oae difference between ;m animal and a
bacterial cell ( 1 mark)
Af p,t,lJ.~ ..... f!d.P.:~ ... l..~ .... ~ ...... l:?..~ ...... Cl.-fd ...... ~ .... flA:.
~ .rl..P..!.:Y/. ..... .w.ii.L ..... ,4 ... ~ .b;-!f.l.l!...l/!w..f..f!-[l!?.f1.i ...... ~~ . .,u__
I ~-··~~.J'.All. .... .it. .... .1a ... ~~.iLI/./R.':f.!fd!:fJ'--t..f ... ;. ......... .
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(hi N11111t lltt plnul,, II ttt1H111t lk
Ill 111111 •tu11 , 1 hl11111pliyll ( I mark)
I ( ('II I j, 1,., I • f/, 1,1,,.,.,1(.,,; '· AJ' .. j~!!.~." !.!..~~!'..~ ................. . , ( ., 1 a1111u, ~ , . , ,'/.:,.... , ,-1. ,u•••••••··r ;.J
I HUii I 1111 U•i II I •• , •• u, II lfltlttt•••••f• II• o l• • •• • •• • ••• u••••• 1 ••• ••••• •
0 0 •• ••• •••••• ••'
0 0 ••••• •• ••• ••••••
( I mark)
.. 1,,.lfJ.1 ti. r.!.1,1(, ... ~ .... ,1: .. ;:.l.1:.r.:.{.) ................................ _ ................................. a... Slnh• hn1 11111111 lt1111·l 11111~ ol lite vuctmlc III lhe umoeba. (2 marks)
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t , ,, ••• ,...,.1. •••• r., .~, .•• r, .... , .•• ...... , .. ,................................................ .--,
2. 'I h<1 h1h11, Ix-low "howH vn11ntio11M in U,c form carbon (IV) oxide is transported in the blood at
1t11 •ml d11ri11K l)hy11ic11I e1tcrci11C.
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C'11rbo11 (IV) oxide transport in blood plasma at rest and during exercise
1-'u1 w uf U'UJ.l.5J)J.lll l>iltlll1lvcd curbon (IV) oxide B1c11rb111111lc 11111 ( 'aroon (IV) oxide bound to protein
Total corbon (IV) oxide in pla.,mn
pll ofhlood
Rest (Mol/Q 0.52
12.34 0.26
13.12
7.42
Exercise (Mol/\) 0.97
13.68 0.16
14.81
7.09
(a) Explain why more carbon (JV) oxide is transported in the fonn of bicarbonate ion. (2 marks)
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.. ~y ... Jl. ......... !ui.111.r:.1.i.r.., . .<J •.•• (16 ····(9::r.£m.l1.Jt). .. ~d.e. ..... -t.i ..... L.A/.~ ........ .
(l. •. ~ ,.c.. .... ~~/; ..... i~l.w.~ ..... d.iJI.c!~.i: .... _. .. ~k. .... iy-t..~#. {:pcylc,t1n,...& fer,/ Ht 0
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(b) Account for the high to({1J pla: 1111 cootcnt of carbon (JV) Oltidc dunng cxcrci~I!-,.
/I t 'J I ·J .,&J!., lilN! < ... { :7";1;<3marlcs) cr~ ....... <'LP..~.I. ... ~ ... ;. !.':.<.1:.f.f,{, ..... ,1. d.i.t ... rf ) ... ~ ... n.t.r. ... . ·-!if .kt ....... :· ....... 2 ·.!/ ,x d, 11 ..•..• .• ?. .«.I I. (..M . .t ..... ~J ,.,, •... .1.,,,.UkflJ.,., ..... "I¼. .. : t ':/··
&.!!Y.~ ..... C'. ..... /.. ... ~./..tf)./. ... ~.t..\.d.!!..(.1..,,,, ... tk ... l.,1r.(!(.I. ..... .J.,j. l.rY.L.t1i .; .. ... :.l. (c) State how one's involvement rn lhc exercises affects blood pH. (2 marks)
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Y \-f'- (d) Name the protein re~ponsible for u(e transport of carbon (IV) oxide in t6e blood. I
3.
11n.s.s
( I mark) .:i..
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TI1e diagram below illustrates the appearance of a plant cell after it had been put ma certain solullon.
(a) Explain the appearance of the cell at the end of the trea~ent (3 marks)
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(b) Explain the results obtained if a red blood cell is subjected to the same treatment. (3 marks)
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t<. .W..~ .... ll1.. ..... 4 .t-a.!::/r..!M.!.1.4.t.~ ..... t'.(tt1f:.t .1.k1.v.,/kJ .. Ol1!:1.rJ.~ . .l.f .... l.1!:f§. .... J.!!.. 1 ,,,. I I ( ,~ ~ . ·'.) · 1 J::2r.nfa1r1 r.C. •.. M.1.4. d.i.. f1.f!1 ·; 1 .t'f .t !Y: !:'1. ..(.i. J./ ... li.:>dl:!\, l.. TC .. .t .. 'k.,/,,?. 1. I. J.. ~1 .... (/.~
(c) Exp~ y~~~f~~~istilled water is not recommended for a dehydrated patient. (2 marks)
1l. ..... h .!{ .. t.t~1iof ..... ~ l.(I. ....... lc!.o.t.~l .... 4'!:.r.:t.t/;J:Y.-:.; .... r.t.M. ... i. .... i'~~ :::J qf ....... P,-1 M.<mc. ~ .... jhr.t. f St,(H..lfm ... IJ. f · .. :f{..e. ..... /,. f .t: 11 ,:{ . ...... , b,;. .t. .. b! ..... ll .t.: Y.: r:1# $4-' ;
................................................................................................................................... ~·~ .. :,.,~~ "- ~ ~ .... ,v~,i 'F ..
4. (a) Explain how the sex of a male child is determined in human beings.~ (2 marks) Y
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.-,...,v) (i) Define the term dip loidy. .J ( I mark)
~ ~f~'\'.J,. S.r.~ .. P.f. .... k~~ .... -:k..v. ....... AY. .... /J ..... ttii,~~:s.1 .. ~J ...... t:::· ... 9 ~~ ~~ )J J~ ~f.l./c .. MtJ.;.(r!.f ... ~ ... l!ff :tM.1.1.':1) ... ,t.r.~ ... u .~ .Cr:tl.4!.~ ... ~~,..,,
-;;-, frr .... uA.J ........................................................................................................... .
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' (ii) Name the type of ce!J division that gives rise to diploid cells. (1 mark)
Mlfe.1it.;.. . . . . .. .. . . .... .. . .. :1 :\ Name the type of cells in which the process named in (b) (ii) above occurs .
.. /3..a~ ... Wllj.SR..MD.J,..c. ... c.at; ......... ~~.7 ... h.:":J/ !i~:;'._ w; if..,
Kenya~ of Secondary Education, 2017 231/2
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(c)
(iv) SIDie the significance of d1plo1dy. (2 marks)
l .t/UiA.\.'.°P .. L ... l~l ...... JL~ ..... lh.&.l! .. ?:1~S.tM- .. r./.:p Y.!!%.'. ....... (~:~L.~11
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Nffl{t't:1~~it':t!li,,~;f ;~;tJtrh~:t!,.~,,. io human males. (1 mark)
Answer
6.
r ' ..... k. ! .. t .... t , .. ~ .c t.'I. ~. ; .... ........ .i?J~ .. ~ .')!. ~ · .>.f.~ ;\( ! .~ ....................... ....................... ... . .
~ S. In beans, the gene for purple colour is dominant over the gene for wh.ite colour. A pure breeding ~ ~
bean plant with purple colour was crossed with a heterozygous bean planL ·~ ..J~\ r; • (
(a) Using the letter P to represent the gene for purple colour, work out the genotypic ratio oft ~ ~ tbeoffspring. ('17/ (Smades) --: ,tll--
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/ rr . o ............ f. .. \ ......................... P.P. Pb P o o.P • , ~.:~ \1~_ ~ ... ." I 1p; Ct~~ CC- ·c .Q. p ... ;·~ ··· .. ·······i··:···j··; ··· ...... l...1.. ...... 1.1 ..... .,.. ................ : ... J~i:- Y;"'.,. "-"" •.
1: ; (b) State two ad~ tage'Tcftusmg gfnetichl~f modified Jarieties in bean fanning. (2 mark:s)l"'>
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.f .1ik.~.1. .... l(..f.11.fa.~,.~ ... ,h ... & l~-Y..~~.1;!.J.f.,< .. / ............................................. .
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Y.) ........... •••••• •••••• / ····· ................... .
·······•··············•·········•··········· ····•··•············ ··················. ····· .......... ······· .................................... .
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SECTION B (40 marks)
Answer question 6 (co,npulsory) and either question 7 or 8 in the spaces provided after question 8
6. In an investigation, two potted plants G and H belonging to the same species were exposed to increasing light intensities at different temperatures. 30 °C and 20°c respectively. The rate of photosynthesis was measured for each plant and results recorded as shown in the table below·
Light intensity (in arbitrary units) I 2 3 4 5 6 7 8
Rate of photosynthesis for plant G at 30°C 0 84 148 196 232 260 284 296
Rate of photosynthesis for plant H at 20 •c 0 72 ll 5 148 170 186 204 216
(a) On the same axis, plot graphs of rate of photosynthesis against light intensity for plants G and H. (8 marks}
I
917755 K-.,a ~ o/S«»ndary Education, 2017
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Kenya Certificate of Secondary Education, 1017 231/2
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{h} Stare 11tc 0 1111 of the invcsligauon LI man,;)
.. lt.. L 1 ... tl.1. 3-'· "-6./~, 1!'!.! ••• It _.1,,tJ ... ~~ ...... t:.{..(J, .. ~ .. ~.J .. k.:f.;~.-··'· ~.b.""• ~ I
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h\) i\l.'l1Hml lc,r the Jiffcr;:m:c• !11 rhc rate ol photo~ynthc:sis in the two plnnlS. (3 rruub)
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(d) Account f<,r thcditlcrcnc~ in the rate ofphotogynthcsis in the two plants bctWeCn the(= L) - I following hl,hl intcnsiticM: )
(1)
(i1)
1-1 units (2 marks) ~r,~ ~L!-:p,:f. ..... l.~~ .... ~~ ... !.~ ... 1. ..... 14. ... '!:!:.li. ..... !Jf. .. -f;.b.fi.IJ,_1~.t~.HS .. Ni.R .... , , ,. J..
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(c) ( i) Predict the rate of photosynthesis at light intensity of 16 units. ( \ mark)
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........................................................... ·-····-·········· ................. ...................................... -.. (ii) Give a reason for your answer in (e) (i) above. ( I martc)
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Ke,rya Cmificate of Secondary Education, 1017 231/2
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SI ,h., ont inCcrn11I nri.l o ne cxlcrn,1 facu,r ll111 could be lil1llt1ng tn the investigation (2 matb)
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I "Jlla1111hc onpmtancc: ofprutccling the forest ecosystem with reference to the following: · (20 marks)
(•) l'l111111tc , hM1;W
llJ I biocJ1, crsity
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