1970-01-08 — Page 18

頁二張五第日一初月二十年西巴展夏 WAH KIU YAT PO

郭日僑華

四期星日八月一年〇七九一公年九十五國民潔力 育敎僑埑

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#

5

APEL

Lype Bulb

Food-storage Carbohydrates

學會考試題預習專欄

Method of reproductions

with

•

arisa

from

and

數學科(+)

14.

第九次預習題解答

僑

*#$%$#$%$# 5% 3 »» 3**#$*£#3*3 #$%&#57·9 #272#$%$#3%$#$%$#$°!% #$%$*

179英文中學會考試題預習專欄

堅道英文書院主編

生物科

BIOLOGY (10)

(-1)

the questions of last week

What do you understand by vegetative propagation"

Give a brief account of advantages and disadvantages of vegetative propagation.

1

Tegetative propagation is the development of new individual from the parent plant of a vegetative structure such as a. food-storage organ, besides providing for perennation of the plant, may also provide for its reproduction. By developing from one storage-organ several new shoots which may then become detached from the parent plant and from new individuale. This' method of reproduction without producing flowers is called vegetativa propagation."

b) advantages of vegetative propagationi

\{1) New planta develop quickly into perfect

Treplicas of their parente.

(1) The new individual_may_be_adapted_to_the

environment

Kiss) Plentiful” food is stored in thá'

Treproductive organ to Lida the plant owe

the early stage of growth.

The procesa- of reproduction ie without t (belp of gibescomplicated necban isme,

Disadvantages of vegetative propagation

Plants Fore

Jaralese 119 to be come overcrowdeu, thun; enough food supply_for_t tpe na k [restricted

((11) The healthy generatione tena "ko diminiec {{(111) There is no genetical differance desweet

the parentrand ibe now

andavad

nence. the new individusie ara lesa, adaptable to@the_change_in_ine

environment.

ind

(iv) Diseases of the parents "stwayé detune

dASUBBUD of the offspring autumatideng]

IN BIKES the type,food-storage external" restura or internall structure, and netboo of reproductian. sack of the following vegetative propagation organn

(a)}Potato)

(b) Tulip€

Crocus

Ginger

{Strawber

Dahlia

Lenticals]

(s)

(a)

(•)

The food-storage leaves triangular stem. The future leaves flowers are enveloped in the centre of food storing leaves. The annual cycle begins la Butumn during which the stored food being wee in the growth of route, leaven, sad flowers while the exces food to etored 10 tho axillary bude which gradually

Typa Corm

become the bal DE

(1)試述任意角之三角函數定義,並說各象限角之符號。 設<AOP-180日,試以日之函数表出(180°)函數值,

x<A0P=90*+¢, jx¢*£$£± (9044) z 3 h

sin20+ cont2 = 1, 04<^X. श्रु

(M) LATI★☆★ £££**, P(x,y)

VAL OP TELE P(x, y) § ̧¥#£ OP=r,

從坐標二

**

4

Foliage leaf

RJ

sen AOP

Flower bud

横坐標

cor 2A07-

半狸

Future Corms

tan LAOP

cot LAOP

横坐標

鲜花坐耀

Old Corm

•Adventitious

其中九恒為正值,而不對則依卡但氏坐權之 由上述定義,可得各象限角之符蕊如下: 第一象限角:各函數均為正值. 第二象限角: : sin, csc ZI, £4. 第三象限角.tam, cot為正;共

S

A

T C

roots

Pood-torage protein and starch. Method of reproduction:

In spring the growth of the bude from serial shoote, the terminal buda bear leaves and flowers while the lateral bude bear only Leaves. A new corm 10 formed at the base of the aerial shoot.

Type: Rhizome

·Scar of old

"aerial stem,

Terminal

bud -Axillary,

bud

•Scale leaf

·Adventitious roof.

Food-storage: Carbohydrates Method of reproduction.

The underground stem possesses distiney nodes and internodes; adventitious roots, soals leaves and axillary buds arise from nodes, During reproduction the rhizome given off a single leafy shoot bearing flower,

scale Teaf

Leat

Stalk

Runner

Termina.

hud.

ARC LAOP-

tott: sec E, #

to CR0? = 180°-6, § <POM= 0, 0 31 OP 12 <AOT=0

OP=OP. ZES PAK P'M'LOX, But. OPM 2 rt. sop′ M2. (a.a. sai I sin (180°0) =

sin

ор

MP = (+)

(+)

P

M

0

M' A

2. sin (180°-6)=sin b.

=-tans

M'P'.

cos (180°- &)—~co18, tan (18020)=

cot (180°-0)=-coto.

sec(180°-9)=—secc

coc (180° – 0) —— csc0,

3/4 AOP=90°+4, ia se

rt. 40PM crt. A QON, LE

19 sin (90°+4)==co14, 20d (90°+4)=– sin &

tan (90° +4)= =-cot & sec (90°+4)=-coco,

ant0+ cos2+=/

*

18

sen (1-8)= sin O

(2)

sin (211-8)= - sin ✪

Aum (3.77-8): = sino

M

cot (90°+4)=-tan &. csc (90°+ 0) — seco:

sin 0 = ± [1-cos2 = ± sin &

sin ( π[ + 0) = -sino,

six (21+0)= sın 8.

o=nq±o (n%*&*%) [%]

A ▲ ABC †. Z AA ₺ 4A A SE (a) a=2R sinĄ,

(b) c = bu

aim A cot B + a sin B cot A. - a cosc_

(c) C-a. cos B.

sin C

Aim B

(TE) 17 OF 111 80. 14 CD.91

¿D=180*~A (E»‡‡ «£#{* Â*a 在△ BCD中,

[補]

<BCD=90°(半园园角直角)

BC

—sim (180°-A)=sin A

=2R'sim

36 the to 191 C14 CDLAB, 90

(a) ##

•Attachment

to the parent. plant!

Net

eye

'Scale leaf, Axillary bud,

Type.

tuber

food-storage, sturch

Rethod of Reproduction.

5. spring the "Net eyea" of the

potato tuber grow into aerial shoots which develop adventitious roots in the soil. The aerial part develops and, latter in the year, the, underground axillary buds grow into short underground stems; the ends of which receive the excess food from the photosynthetic leaves and so well to produce new potato tubers while the axillary bud of the compound leaves develop into the ordinary shoots,

-Heshy leaf -Foliage leaf

·Flower bud

·Scale leaf -Axillary bud

·Triangular sta Advent trous

Foot

(f)

Advantitious Stem

roots

Type: Runner

Food-storage: No stored 1000 ans

stem.

Method of reproduction

The plant has a short underground #tem.] Toward the end of the flowering season, the bude in the axials of leaves grow outward forming long and thin stems which pos289 several scale leaves. The terminal bud of va runner turns upwards and adventitious roots are formed, thus making a new leafy plant.

yper Root tuber

·Aerial Shoot -Adventitious,

root "Root tubers'

Food-storage: carbonydratei

Method of reproduction:

During the reproduction it produces

numerous adventitious roots which, towards th end of the year, swell with stored food, Thear root tubers bear buds which, the next year, become the aerial shoots of a new plant.

Questions for this week.

What is goil composed of? Describe an experiment which will help you to detect the different

constituents present in it. Explain the occurrence and the use of nitrates in the soil.

#ABCD4.

CD= a sin &

BD = CD cot b

RADAC

-(2)

CD = bawn (180°-A) = & sin A AD=CD cot ( 180° ~ A) = - CD cot A

VÄLJA X(4)

-(3)

BD=&an A cot B. AD-a sin B cot A

·bean A cot B-(- a simt cot A) =bum A cst B + a sinВ cot A

@ C = 8D-AD==

28 14

←

b

= acos C + c cos A

C=

a cos C = C cos A ይ-

a car c C - a cor B

19❀ 2 56 £

b-acos

C- a cos B

a con 8 +

C cot A

c sin 8

sin B

bcas A c-acor B:

(3) £££: AB=100R <DAC= 56°18,

<CAB 40° 30°, «DBC=70°42′

4DBA=54*36 CD z

(解)在△DAB中

<DAB=56°18'+ 40° 30

=976°48′

• <ADB=180°-{ $b°48 ́+ 54°38'}

=28°36'

依正弦定律

AD

Ain 54° 36'

.. AD-

100

Jan. 25° 36'

100 in 54°34′

sin 28°35′

又在△ABC中,

40°38

Im beosA

54°36

NO

100

LOG 20000

Dvor 54°36'

-170.3 (2)

7.91124 19/12

1

am 28°36'1.68015-

70.3

22311

<ABC= 54°36's 70°42=125°18*

《未完国入第五張第二頁》

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