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六期星
日九十月四年二七九一曆公年一十六國民奘十
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1972英文中學會考試題預習專欄:
each chromosome
splits to two
-membrane
1972
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and move to opposite sides.
disappeared
生物科(十六)
中文中學會考試題預習專欄 C
世界歷史科 (廿六)
歐陽者放
蟹鐵
化學科
1912英文中學會考試題預習專欄
(六)
Biology (26)
Ca11 Cell Division and Gametogenesis
A cell is the basic unit of life-
from which living organisms are built up. Each cell containa a mucleus with cytoplasm around. The living semiperme¬ able boundary membrane on the outside. is called plasma mendrana (-cell membrane cytoplasmic membrane). In plant cell an additional non-living rigid cell wall is found enclosing this protoplasmic struc- ture.
plasma membrane.
Lactoplasm
groups of chrosones formed
Animal Cell
(clear)
endoplasm granular
2 daughty zuclei
2 daughter cells produced also diploid as mother cell
cytoplasm divides
chromo
ytoplasm
The proceÈS
Tat divis-
nucleus
cell wall
nucleus
Plant
plasma membrane-
call
vacuole (with sap)
qucleus + cytoplasm - protoplas
are
Diploid and haploid cells
With the nucleus of a cell there thread-like structured called chromo- somes (chron-colour; some-body) which become distinct during cell division owing to their ability to take up basic dyes sasily after staining,
Bomes
There is a number of chromosoma inside the nucleas which can be disting. uished into different types. Chromosomes are often present in paire i.e. the cell possesses two of each tind of chromos-> It is said to be in the diploid or (21) state." But in certain cella such as the gametes. Only one of each kind of chromosomes is present and is said to be in the haploid" or (n) condition.
nucleus
Diploid cell
(20)
chromosomes:
nucleus
Haploid cell' ~(n)
Different organisms may have diff? ferent number of chromosomes; human being
23 pairs, frog-7 pairs, fruit-fly JennA PRÁTOL maize-10 pairs.
Cell Division
Definition's Cell division is the the
omatic splitting of the live ing cell body into two or more parte,
There are two main types of cell division namely, niste. osis or ordinary cell divie-
Fion and Meiosis in the for!
mation of gametes'.
Meiosis
diploid nother
20
ist divison
diploid
to ther cell
cell
20
Vitosis
The process.
chromsenes,
"diploid mother
cell..
nucleus
2nd divi
Вод
duplication
of chromosome;"
spindies
chr
4: hapI- oid
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duplication chromosome
"A (1) (2)
B (3)(4):
Bpindles
similar chromo-
separation chromosome pair
omes pair toga.
thar
haploid nuglei Cormed
Splitting of
2 haploid cells completed (2nd 23x3"- aploid nuclei
chromosomes to a and move
to opposite sides
(as in mitosis)
haploid cells produced,
Compare mitosis & merusaS
mitosis
The process. One other cell produces 2 daugh ter cells Daughter cella having same num- ber of chromosomes. as the mother cel
2n) (2n
Occurrence
3. In different parts of the. multicellular
organism.
Ригорве
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Ons mother celi produces 4 daugh- ter cells.
2. Halving the num ber of chromos- Oree in daughter cells.
4. Provide more cells4.
for growth and re- -pair of worm away
tissues,
(3)
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of male gametes in animal is called Spermatogenesis, while the process of eza formation is called daxenesit. Sperm and ovum are produced as a result of maiosië..
Spermatogenesis
in + testis
diploid Sperm mother call (primary
Spermatocyte)
ondary/ spermato n cyte
habloid apera
In sex organs:
genes
Formation of gametes for. sexual reproduc- tion
In unicellular org-5. Allow mixing anise this is a of genetic Method products from tion binary fission, both parents as in Amoeba and Para- during fortii-
zation. Such mecium.
that the off spring poss68 pen. certain characteristics of bothe
Gametogenesis
Gametogenesis is the process of gamete formation in the sex organs, sucл ae in the testis and ovary of animale The formation and flowers or plants.
3 small polar
bodies. (useless and
die away
Ovary
diploid egg
mother cell (primary Oocyte)
secondary:oocyte,
Bingle large haploid övün
One sperm mother cell will give rise to four small eperm, while one egg mother cell is producing only One single large. ovum containing food storage which is used for the early stage of developmen of the embryo after fertilization.
Chemistry 26
(2) Manufacture of ammonia by Haber
process. M
The essence of this industrial
process is the catalytic combinatio of nitrogen and hydrogen
(A) Hydrogen can be obtain form water
gas. The latter is a mixture of carbon monoxide and hydrogen, prod- uced by passing steam over red hot coke
C+
CO water gar
9) Nitrogen can be obtained either from
fractional distillation of liquid
air or from producer gas. The lat ter ie made by passing air over. heated cokes:
air. C.
producer gas
(C) In the reaction chamber one vpiURE I
of nitrogen combines with three volumes of hydrogen in the presence: of iron as catalyst, aluminium oxide as promotor. This reaction takes place at about 500 C and a pressure of 250 atmospheres.
3H
500°C
2 NH
250 ata.
(D) Carbon monoxide can be converted into more hydrogen and carbon dioride by reaction with steam under specific conditions. CO2 is removed in solution under pressure and hydr- ogen 18 channelled to the reactio chamber for further reaction.
80 + CO LÀ HA
(E) The ammonia gas is condersed and
purified. The unreacted gases: are recirculated for further proces- sing.
VOLUMETRIC ANALYSIS.
In the following paragrapna, certain Stud- useful experiments are considered. ents should familiarise themselves with problems which are set on experimental basis in volumetric analysis.. Experiment (1)
To determine the equivalent weight of a civen acid, e.ɛ. tartaric acid or citric acid, using a standard solution of sodium hydroxide.
Step (1) Prepare a standard solution of the given acid, f.e dissolves
known weight of the acid in known volume of solution, Steipii) Titrate 25 al. of the acid solution against the given standard sodium hydroïide solution; note the volume of: or alkali required to change the solution just pink. Phen-
used as the
Step (iii
indicator in a weak acid,. strong base titration
By making use of the equation NAV the normality of
the acid solution be found
top (iv) Equivalen weight, E.W.
Concentration of acid solution în normality of acid solution
Experiment (2).
To determine the number of molecules of water of crystallization in, for exam ples, Na, 00 • xH o
Step (1) Dissolve a known weight, say a
, of Na C. XHO crystals in a known volume of solution. Step (ii) The weight of anhydrous sodium carbonate, bg, in these cry- stals is determined by titra.... tion against a standard acid solution.
the
Step (111) Hence, a gn crystals contain.
bg anhydrous saitidez weight of water-
of water
(a:
of anhydrous sodium carbonate
(iv) The molecular wt. of water
18
The molecular wte of annydr-. Que Bodium carbonate 106 If x- the number of molecules of water of crystallization
then
182
106
From this equation, where a and ara round from experiment, I can be calculated.”