REFEREN
ARY
育教僑華頁三第張七第日八初月三年子壬展夏 WAH KIU YAT PO
郭日橋
21 APR 1972
CITY HALE
五期星
555555
僑
30
1912英文中學會考試題預習專欄*
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生物科 (廿五)
Biology (25
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K-BERETEK BOŽOKE
Genetice
It is a common phenomenon that new
generations resemble their parents;
plants giving rise to plant, and animais
giving rise to animal, maize plant to maize plant, doge to doga and man to:
Herdity
There is a tendency of individuale to resemble their parents, This condi➡ tion is called Heredity.
VariationsTM
Despite the similarity between par- ents and off-apring there are signa to show variation between the individuale. The governing factors for such varia- tione may be the genetic constitution of the organism or the influence of the environment. The expression of inherited character governed by genetic constitü- tion may be strongly influenced by the environment. In other words the genetic as well as the environmental effecta on
he "offspring, are equally important.
Genetics
The phenomena of heredity and varia tion and the study of the laws governing them are called genetica.
Chromozone
Within the nucleus of the cell of every organism there is a number of thread-like structures called chromosomes. These chromosomes are carrying the gene“: tic materials which controls the charac- teristics of the body..
Chromosome number
Each cell of every organiem of a given species contains a definite num ber of chromosome. In higher plants and all animals, the chromosomes are present in pairs) ice two of each kind of chrom- osones are present, one from the male parent and the other from the female parent e.g. Human being:
Genes
Frog
Maize
-fruitflý
23 pairs
Genes are hereditary ractors lying within the chromosome and are arranged in a single series. One kind of genes differs from another in ite behaviour and controls the development of certain characteristics. Since genee are loc- ated inside the chromosomes and chrome- Bomes are existing in pairs therefore each cell must possess two of each kind of genes.
Locus of rene
The gene for each character situated at a particular point in the chromosome, that point is called the Locus for the gene. One locus can be occupied by one gone only. Homologous and non-
homologous chromosome a
homologous chromosoONES
cell
non-homologous chromosomes.
zne 2 chromosomes belonging to single pair are said to be homologous, while the chromosomes belonging to dif ferent chromosome pairs are non-homologous. Genes in gametes
In the formation of gametea. (gene- togenesis) one of each pair of chromo-. somes from the mother cell goes to the gamete therefore each game te contains only one of each kind of genes instead of 2 (such as in the formation of apern and egg).4
Law of dominance
When 2 genes for the alternate ex pressions of a character are brought to- gether in an dividual one may be shown off completely while the other not at all. The character which appears is said to be dominant, the contrasting character on the other hand is said to be recessive.
ve. The gunes involved are called respectively dominant and reces- sive genes, eg, in guinea pig, black gene dominant in maneye ka brown gene-recessiva brown – dominant
blue recessive
Symbols for genes
Usually the dominant gene is repre sented by a block letter and the reces sve gene by the corresponding small latter e.g. Guinea pig,
Black:
brown.
Certain flowers
red (d
HOMOSYKOLỡ and reterozygous conditiona
An organien with a pair of genes exactly alike is said to be homozygous or pure (AA or aa) while that with one. dominant and one recessive gene is said to be beterzygous, which is sometimes. called a hydrid (a). Thus a recessiva gene is one which will produce its effact only under homozygous condition (e.ge by for brow, in quines pig) and a dominant is one which will cruduce ita effect. whether it is homozygous (BB black) or heterozygous (Bb - also for black) Genotype and phenotype-
Genotype is a type or organis defined by its genetic constitution: (as AA6a, as). Phenotype is a type of organism defined by its observable character (such as red, white black, brown, long and short hair etc.) 0.5. guinea pig,
genotype 38
BE
phanötuna black, blacky
Symbols for different ceneration; use guinea pig, a pure black female is sed with a brown male.
Black
33
All Black
Brown
parents
gametes
generation
male P is crossed with another
Self cross,
male of÷F
3 black
gamoten
2nc
erefore the ratio of black to brown ie 3 1 in the second generation. Pparente; F. - 1st filial generations
2nd filial generations. (From Fx
Genetic ration
Genetic ratios are probability rat- ios or chance ratios, then we say 11 means 50% of one type and 50% of other; The actual number may be 421 45 * 111 or 32 313 131 x 3 - 93) which is close to 92)
Examples of genetic characters garden psa Tall plant dominant, drawf plant-
recessive, yellow seed - dominant; green geed-
recessiva," round seed dominant; wrinkled seed-
тесевві
Human being
Tongue rollers-dominan recessive.
non-rollers"
-PTC taster dominant: non-taster
recessive.
(PTC is a chemical called phenylthio- carbamide giving a bitter taste to tastus and for non-tastas, no at all).
Black eyes dominant: Dive areas recessive.
black hair-dominant; non-black hair- vercessive (yellow, red etc. Tomato plant
Tall plant-dominant: short plant
plant- recessive. Spherical fruit dominanti pear-shaped
fruit recessive,
Fruitfly
།།།་།
Brown eye-dominant; white eye-recea-
sive.
Normal large wing-dominan small wing-recessive.
(dodi #nant)
Guinea pig.
White
(recessive)
Black hair-dominant; brown-recessive, Short hair-anginantı long hair-races-
give.
【中文中學會考試題預習專欄
數學科 (廿五)
張正邦 串田次預習題解答
Za Sin2 72° - Sun 60 =
Ihop = Sin (90°-
10-18") -Si 60
- (5m 90 Cos 18 - Cos 90°5 in 18°)-
(長) (Cs 1892
= 1- Sun 18" -—--sm
5-25+1
--3-√5-2-3+ √
5 =1 = $1
右端
Nos Co2 48" - Sin 12-
***= Cos2 (30+18) -Sm2(30-18)
(Curs 30° 000 18" - Sm 30 Sun 18") = (Sun 30"
Cs 18°
Go 30°5m 18°32
= (= Coll" - +5m 18") = (opsie 4[ (√3 Cos 18" - Sun 18°) = (Gr 18-55-18)])
-(3.00 18-2√3 Coll Smil + Sm 18-C18
+ 2√3 Co. 18°5m 18" — 3.5m 18").
==(361 18-25m2 18' + Sm 18 - As2 18")
± (Cos2 18° - Sin2 18°)= =C=36"
右端
* Sun 105 215
p98 Smlos = sm (60 +45°)
=Sin 60° 6045°+ C066" Sin 45°
吾吾
4(√6 + √2)
Ex B C 68.20 G58:20 + Co81.46
* Exto = 6068 20 000820 + Sin 820
Sin 68°20
Cos (68°20-820) Cos 60°
=右端
Ed & Gos9 211
$8 22 x=18, £lsm=smg.
25mm Za lisz = Sung
0+Q (Sm2 + ep 2)2 = 1+5mx
Sun & + hot = + √itsmax
F] Suq & 69 11 12
Sus 2 + lost = √1+ Sinx — R
)-@ Su= -cs// = -√1-Smx
26002 = √1+5mx + √1-suit
as & = {(√[+smx +√1-Smix)
以不值代入得
( 1037-1/ + 1-5 + 1 / ) = = 63
= = ( √3 + JE + √5-JE)
ENE Suo 78°32′ 663 11°28' + Sm11° 28":
Cos 78°32=1
65. $35=5m 78°32′ GO11 28
+ Co. 78°32′ Sun 11°28′
Sun (78°32′ + 1)*
Sm 90° =1
簡單消去法:
-$160.
19] <BR>] 2842 + £ •
a' co +b Sind = c
a a coo+absme=ca:
aacootab Sind = ca (ab - ab') Sm B = ca~c'a Sun A = ca-
ab-ab
1904019 Sm2 0 =
同理可求
ca'-c'a
ab- Ab
-ab
@ +@ Smet Cs20 = (ca'-c'a
ab-ab
ab
(a'b-ab) = (ca-c'a) + (be' - b'c)2 例二消去下式中之日
axswa
64000
Sm20
*@19 QX Sm B +b7Q00 = 0
a x Sin3b = - by an30
az
-by Sino -(by)* (ax)3
Suo20 +60-0
/ (61)+(07)}}/(by) th
(by)
(ax)=
代入1
/ (60) 34 (ax)= [ (ax) + bys ] = a=L2
(by)
√(by)}+(ax) ((ax) + (by)] = a = b2
[(ax) + (by) $] = = a+b2
(ax)3+ (by)* = (a2_b2)**
第十五次預習題
消去下列各式各日.
ax
Sme.
∙lan O
Sec 0
Sec20+ co28==2
Che
3 x + 15 = 1
ysme
4 £600-7 Sine = co20
•Suid +
760=2