REFERANCE LIBRARY
-2 JAN 1970
育教僑華 頁三第張 第
WAH KIU YAT PO
報日僑翠
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170英文中學會考試題預習專欄
堅道英文書院主編
生物科
(JL)
BIOLOGY (9)
Inswer to the questions of last week
1. Why is it necessary that plants should scatter
And.
their seeds? Describe the chief methods of seede or fruits dispersal, ar1 give one example of each.
է
The important reason for the dispersal of seeds or fruits is to give the seeds a better chance of successful life. By being dispersed they grow away from the competition which would face then,. if they fell to the ground immediately around the parent. The process also serves to spread the species over a wide area. The chief methods of seeds or fruits dispersal are outlined as follows. (a) By animals – Animals assist in the distribution
of many fruits and seeds, usually by the develooment on the seeds or fruita of processes (e.g. hooks or stiff hairs) which cause them to adhere to the bodies or the coats of the animale. Example is provided by goose-grass The collection of nuts by squirrels also Tesulta in disoersal of seede,
Kany fruits are fleshy so that animals would like to eat them. In this case, the seeds ara either rejected by the animals or they may pass out undigested with the animal's faeces. e.g. orango.
(b) By wina - In this case, the pericarp showe
modifications which will result in the fall of the fruit to the ground being delayed in order to give the fruit time to be blown away from the parent plant. Wings or plumes formed extensions of the pericarp are the most favoured. In cases where the seeds are dispersed after leaving the fruit, the testa may be extended into wings or plumes. e.2. seeds of cotton, and fruits of Dandelion.
(c) By water - The fruit or seed is provided tough
and fibrous pericarp or testa which protects it from damage in water, and enables it to float on water for a long distance. *.g. coconut.
(d) By self-dispersal - The seeds of many dry
fruita may be distributed by other means. In many cases, the seeds are shot out of the fruit, with: some violence as a result of the strain set up if the wall of the fruit, which eventually bursta. The pea is a good example of this type of dispersal.
2. (a) State the conditions necessary for seed
germination and the further conditiona required for healthy growth of seedlings to mature plante.
(b) Describe epigeal germination as shown by a
named saad.
(c) State clearly the functions of cotyledons in
the different stages of germination of this seed.
Ans...
(a) There are four conditions necessary for seed
germination:
(1) Water (moisture);
(i) is to saturate the contents of the
seed in order to aotivate the dorman ^embryo.
(11) is to dissolve the materials stored in
the cotyledon or endosperm being absorbed for the growth of the aubryog
(111) is to distend the tenta for easy
emergence of the fadiole and plumule
(2) Uzygen is tó release energy for embryonic
sruption during respiration.
(3) A suitable temperature will just stimulate the protoplasm to have vivid activity.
At a high temperature the protoplasm
in the seed in destroyed while at a low temperature the protoplasm in the seed ir inactivated.
-
(4) Sunlight Among the sun rays red and blue. lights are essential for plant growth. However for seed germination. light is unnecessary.
(2) Temperature
i just to keep normal aotivity of the protopiasm in the growing plants.
(3) Oxygen – is required for respiration in
order to maintain a good supply of energy : to the growing plante after séod
germination.
(4) Wafer is to maintain the process of photosynthesis of the plant and to
dissolve the mineral salts for the growing plants.
(5) Mineral salts - such as potassium, sodium,
calcium, magnesium, iron, phosphorus, chlorine, Bulphur and silica in the form of nitrates, sulphates, and phosphates are needed for the normal growth of mature planta.
(b) The disgrams below showing epigeal
germination of a castor oil seeds
soil level
-Hypocotyl
·Radicle
Tegmen
•Split testa
-Radicte
(A) Emergence of the
Radicle
Rost hairs
Root cap
(B) Elongation of the
hypocotyl
-Cotyledon
of the hypocotyl
Hypocotyl arch
Hypocotyl
Secondary Roots
(c) Further elongation
Cotyledon's stalk
(D) Liberation, and
withdrawal of the cotyledons
•plumule
(E) Liberation of the plumile
and the formation of the seedling... Secondary Roots
On germination radicle which first emerges downwards, and then hypocotyl straightens and lifts seed out of soil surface. The testa drops off and the endosperm
can be seen on outside of the two cotyledons, Enzymes from the cotyledons digest the
andosperm into soluble substances which are passed to the growing shoot and root. When the endosperm is all used the ootyledone open out and form the first pair of green leaves.
•
(a) The cotyledons, after exposure at the surface
of the ground, grow apart, set free themselves, expand and become green. When the stored food in them is consumed completely they are able to
make food for the growing plant by photosynthesis for a certain period of time.
Simultaneously they also can secrete
certain enzymes to change insoluble food into soluble which is absorbed by the growing plumule and the radicle.
The cotyledons of this seed also act as protective organs for the young plumule which lies between them at the seed stage and during the early stage of germination.
Questions for this week
1. (a) What do you understand by vegetative
propagation?
(b) Give a brief account of advantages and
disadvantages of vegetative propagation.
四期星日一月一年〇七九一曆公年九十五國民華中
(按第張第二路)
光珠鐵中文中學會考試題預習專欄
(九) ·喬仲强-
數學科 (九)
3. AZ1S, AƑŽ¥IT
作团.
4.以日為心,BF為半程作团
BC&D.
5.以C為心,CD為半/
摆作园、
6. A, B)C = 4
行求
(証明)機
Q. E. F.
(5)已ᅀABC,求以A,B,C各国心作三团,僅BC外
切,且怕典A园内切,
(EX2) A ABC.
(求作)以A, B, C备為因心作三园,使B,C两国外切,且婚 典A园内切
(分析)設DEF分别高BC两国,AB两园A
及A,C两园的切真.. AB-AF= AQ☀12, BE=BD= BE¥42, CD-CF=C & $1.
*AE÷AB+BE=AB+BD,
AFAC+ CF= AC+ CD
* AE+AFAB+BD+ ACT CD
AB+AC+ BC.
D
·· AE=AF-(AB+AC+BC)=$ (= A**MF)
X BD-BE-AE¬AB÷S-C, CD=S-b. D, E, F 别离 AABC对A角旁加园的
(141) 1.14 28, ×CA
平分线使交Ia.
2.
I. 44 IaD LBC.
3. 以B為心,為半挂 作國交AB延线於E.
4.以C為心, CD為半程
149.
5. VAŽIC, AESTI
448.
6、A,B,C三因為好求。 (证明)拨备
Q.E. F
(6)内摆柃园而外切於园之四边形,其边和奌之联 线互相密直、
(已知) ABCD為园之內接四边形,
1 ABCD 外切於EFGH园.
(*) EGLFH··
(-) 1, BK GH, ME
•DG=DH (W 2.441=42 (FRAX1) 3. EADGH
2D+41 +22=180° (Am) *K*) 4. 4D=180*~241 (HAW) 5.联GF,則41=43 (弦切角)
6. LD=180°-223 (¥1****)
7. M+7QJ£ 18=180-224 (16)
8. 但<D+∠B=180°(园内接四边形对角互補)
9.
JO.
II.
(180°-243)+(180-214)=180° (##)
180°= 243+244 (等量公理)
123+24= 90°
AGKF 4, 4GKF+43+24=180° (=A+{@Õ¤I)
* 4GKF=90°
12, EGLFH (ÍHALZH‡1)
Q.E.D.
(2-) 1. <D=2(GFEH-HG) = 1⁄2 (GF+FE+EN-HQ)
< B ÷ / (EH + H G +GF-FE).
两切线之交角,以其野夹之孤之差之半度之
2. <D+18+1 (2GF+2EH) = GF+EN (#ahe}
3. <D+B= =180°(园内接四边形对角相補)
4
GF+ÉH = 180° (4)
<GKF (GF+ EH)==× 180°— 90a
(两相交张昕成之角,以其弧及其对頂对弧之和之半度之)
6. EGLFH (11)
第九次預習題
Q.E.D.
(1)試述任意角之三角函數定義,並說出各象限角之符號
CADP180°-日,試以日文函數
表出(180-6)之函數值
X$E <AOP=90°++, £ •×¢ż £ €
表出904中之函数值
說Ain8+0861.求與中關係式(日,中任意
(b) c=
(2)鈍角△ABC中,A角為鈍角,試証
a= 2R sin A
(a)
(c) b-a cosc
C-a cof B
(3)如右图:AB=100尺
<DAC=56°18 4CAB=40′30′′,
– bainĄ oot Bra sinß cot A,
For Dealiny growth of seedlings to the mature plants, the following conditions are supplementedi
(1) Sunlight - 18 essential because after a long period of absorption of the stored food in the cotyledon or endosperm during seed germination the process of photo- ayathesis follows to make food for the growing seedlings to mature plants. Light is utilized to help for the process.
2. State the type, food-storage, external ́reatura
(or internal structure), and method of reproduction of each of the following vegetative propagation organs
potato
tulip
aroque
ginger strawberry dahlia
4DBC=70°42′, «DBA=54°36′.
#co.
(4)設為△ABC的內切园半
橘」の為其面積 試
(a) s
b) taw &
(C) sin £ = [(5-1) (5-0), (d) cor£ = $(5-2)
(5)三园两两相外切,其半挂各8.8.10吋旅 圍成图形之(0) 周界
(B)面積。
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