20 Basic Genetics Notes

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2 0 .1 Genetic Terms --liliC:il!l!!::..-::=--- -- !l!!!!!!l!!C!!:---.1!~~-=====-~ A . Alleles C' They are different.forms of a gene which occupy th e sa me poc;, it io n on hrm /(Jl ogou, chromosomes. Example: eye colour gene - _/ _ -~~ -l L--:~ ' brown eye _l_} t-----i : allele ' i----, other genes B. Phenotype C, IL refers to t he ap pearance of a tr ait . C. Genotype blue eye allele { It refe rs to the Kenef ic make-up or an organism. 93
"' t ,; and c xe1 ( :iSC 'J N :' ,S 8H"i< 0\ - /1·tG /iS I\ { No e, ; Rn 94 ul/c /1' .,. E. H ete ro~.ygote I . . 1 · 1 c ul ar ch aracteri s tic co nlai n 1;, two dif{ert I ,, 11 ot\'\ JCO a pa1. c ·· 111 lt 1 :-- , \\1 l1 r ~a ni~rn ,, l()\C ~ l - a l/ rlc s. F. Hy brid . . .. 1 1 . d I by crossi na two in di v idua ls w hich are ge ne ti calh \t ,, an 1nd1, 1c ua p1 o u cec . c diffe rent in o ne or more ch arac te ri s ti cs. G. Do m;nant allele 11 ,s the a ll ele whi ch c an ex p ress it se lf in h etero::ygo 11 s co ndit io n. H. Dominant character It is the phenotype w hi ch can s ho w up in hete ro:,ygo us co nd iti on; co nt ro ll ed by t he do mjnant a llel e. I. Recessive allele It is the alle le w hic h effect is mask ed in het ero::,ygo us co ndition, i. e. , it c an o nl y ex pr ess it se lf in honw :: ,ygo us co nditi on. J. Recessive chara ct er C It is th e phen oty pe w hi ch can only s how up in hom crvoo us co i 1 ct ·t· II ! ~- o 1 10 11 : co n tro ec by th e r ec e ss i ve allel e. K. Monohybrid inheritance , It refe r, to the s tud y or lh e i nheritan ce nf a single pair of contrast in g character s. L. Dihybrid inher i tance C The inh er itance nf two pai r ::, or contra-.tin g char;1c 1cr s.
Basic Genet/e-i:; 20.2 Monohybrid Inheritan ce ~~-~------- A. Exa,np le: inheritanc 1·1 e ,~ 111 111 0 11 eye c o/o11r Gen otype : h on, oz.ygous Phenotype: brown homozygous brown homozy qou 'i blue 0) This trait is controlled by a pair of alleles (two differentforms of eye colour gene): brown allele and blue allele, and therefore has two contrasting phenotypes : brown eye and blue eye. ~ Every person has two alleles ( on homologous chromosomes) to determine his/her eye colour and the two alleles are received from his/her parents, one allele from each parent. ~ If both alleles are the same, the person is said to be homozygous. C:, If both alleles are brown aJleles, the person must have brown eyes. C) If both alleles are blue alleles, the person must have blue eyes. C) If the two alleles are different (i.e. one brown and one blue), the person is said to be heterozygous. C) In this ca s e, only one of them can express itself and the other one will be ma s ked . c:, Si nce a hete ro zygous person h as brown eyes , the brown allele must have expressed it self a nd it is there fo re the dominant allele and brown eye colour is the dominant phenotype·
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---~="'"""---: Jn ~-:,~n ,v Not, ,nil r~ xucr,es • , . 11 ..,cll in the pr e1.ic n cc of the bro \':n al lele . ( '1 11<' h l1 w :d kk c ,1111wt , xpll ' " "'Q i< 96 _ 1 1 , 1 .. 1 · 1 17 ,rcl or c 1h c r eces~ i vc all ele. The blue eye cr,i, 'd h I I h, 1 11 ' , II II ', , c , , <111 " c . , ,u r " tf · . t , . 11 ,1 tl' l" occ ur s onlv in homo z yg ou , cond1t 1on. '' n·c e"i\ t \' l' p t• no ~ 'p t: ,1 ~ , - U. Men del's firs t I.o w - The Law of Segregation C ~cndcl \ hn . ·c d tn g c:<. pcrimc nt '-t on ga rd en pe a: 0 '1 r: <11 , tuJi cd : !lo we r co lo ur !two contrasting characters : r ed and v. hi teJ 0 Pur c- hr e~ d in g rcd- ll owcrcd p ea pl anb we re cr os s-po lli nat ed w ith pure-br eedin ,; wh it c- fl o wc rcd pea pl a nt s, ' R e Su lts of the breeding experiment: Pa rents: Red X ( pur e- br eeding) l c ros s-pollination all red (hybrid) se lf - pol I ination Red Phenotypic ratio: 3 b. Explanation using genetic d iagram : Defining symbols: White ( pure-breeding) White Let 'R' re pre se nt the dominant allele f d II I f or re flower and ' ' a e e or white flower r represent th · e recessive Parents: Red (homozy gote) RR t Ga mete s: R F . ) • X Rr (hete ro zyg ote) a ll red White (homozygote) rr t r
Basic Genet F,: Rr X Rr / ~ / ~ G: l{ r I< r ~r--::J F2: RR Rr Rr rr Genotypic ratio: 2 Phen otypic ratio: red 3 ~ This ratio is known as monohybrid ratio. Significance of the ratio: wh it e 0 It indicates the proportion (or %) of offspring having a part ic ul ar phenotype e.g. 3/4 or 75% of the F 2 plants will produce red flower. 0 It indicates the probability of appearance of a particular phenotype in an offspring. e.g. the chance for an F 2 plant to produce red flower is 75%. C) Mendel's first law of inheritance (in modern terms) - for a pair of alleles. o nl y ONE can be passed to the offspring through gamete . C) The Mendel's law can be applied as follows: If the two phenotypes of the offspring from a certain cross are in the 3 : l ratio , according to Mendel's law of inheritance, bo th parents must be heteroz) 1 gous. 97
98 . .J E erc i se s N otes at1• 1 x · /n ten s ve . (cr osses (. Tvpe .\ ~ 111 exa mpl e : ~ I 1 )1 '''" "" ' . l 1 1 ~ ;l I l r 11 · ' l , / 7o H' l' r I o /tJIII ' ---'.'.:l~=~~J--- "F -;; ei ;:~:-----;7 7rr, .. '...~-'~·~ ( l ', \ 11 l.'. { l l • - ro •· HK -.. Rl{ RR ·, Rr RR >- n R r' Rr Rr x, rr rr X TT D. Te st c ross all RR RR : Rr = J : I all Rr RR : R r : IT = I : 2: 1 Rr : rr = I : I alJ rr all r ed a ll red a !J red re d : white = 3 : I red : w hj te = J : l all white a. to identify the ge notype of an organism with a dominant character Example: To iden ti fy th e genotype of a red-flowered pea plant. (Given that flower colo ur is co ntro ll ed by a pair of alleles and red is dominant.) Solution: Tes t- c ro ss the plant with a white-flowered plant, if only red-flowered offspring are produc ed , then it is homozygous dominant; if some white-flowered offspring are produced, then it is heterozygou s. b. to identify the dominant allele I phenotype c:> T es t cro ss two homozygous individuals of different ph enotypes. c:> All their offspring will be heterozygous. c:> In heterozygous condition , the allele expressed I phenotype shown by the off \ prin g is dominant. E. Pedigree analysis ~ e.g. In man, th e abi lit y to roll t on gue is determined by a pair nf a ll eles.
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10 - --- 2 6, Key: D mak tonguc-rullcr D rnalc non- ro l le r Q female tongue- ro ll er Q female non-ro ll er ~~\ - - ~ ( , C:- 1 Informa ti on obta in ed from th e above pedigr ee: C:, Tongue-rolling is domina nt . C:, Both individuals 1 and 2 are hete ro zygous. F. Solving genetic proble,ns involving monohybrid inheritance a. The phenotypes and I or genotypes of the parents are gi ven, C:, deduce the phenotypes and / or genoty pe s of th e offsp ri ng and th e ir pheno l~ pie ratio and / or genotypic ratio; C:, predict the proportion of off spring th at will have a c ena in phen ol~ pe genotype; C:, determine the probability that an offspring will have a ce rt ain phenotype genotype. Example: A white-flowered pea pl a nt is se lf-c ro ssed. W ith th e help of a ge ne tic di~1gr~m. predict th e p ro portion of th e F 1 pl a nt s w hi ch w ill p rod u ce red fl cn wr ~. (Gi ve n : Red is domina nt over white.) Solution: Le t , R · rep re se nt the domina nt al le le for red fl ower and ·r· represl'n l the re cessive a ll ele for wh ite flo wer. 99 -i:::!fd
d E e1c1se s I nf :lf1SIV(I NSS B1< 1 lt 1 Q) - t;:: N otes an -x White IT it 100 P: Whit e rr t r G: rr l <' \ ' a ll w hit e . . · oducino r ed fl ower s === 0 The μro p or tt on of F, pl an ts p1 c .d d deduce the phenotypes and I (Jr . es are provt e ' b. Tire result of certam cr oss ge notypes o.f the parents. CASE 1 h t 0 ·ven that both parents show the dominant Deduce the oe notypes of t e paren s 0 1 0 ,ffi h the rec essive character. character and o ne / some of the OJJ spring s ow Example : ln mice, black fur is dominant over brown fur. Two black mice were mated gi\ ing 3 brown and 1 black mice. Assuming that the fur colour is con tr olled by a single pair of alleles, deduce th e genotypes of the par ents . Deduction: The brown offspring must be homo zy gous recessive. Therefore, th ey must haw received one recessive allele from each parent. Since both of th e parents should have a dominant allele for being black, their genotypes mu st be hetero z.ygo us. CASE2 Deduce the ge notypes (and phenotype s) of the parents o- i ven tt 1 at tl 1 t . h ::o c e wo p enotypes in the offspring are in the ratio of 3: 1. Exa mple: A mai ze plant wa s se lf-pollinated and cobs bearing both pur\Jle a 11 c1 ,, II ~ Jc ow corn grains were produce d. The number of purpl e grnins and ye llo w grains 0 11 one of the c ob~ were 53 and 18 re sp ec ti ve l y.
purple Drains ye ll ow gra in s \. , 1 ' / ••Q1t l1 eo1 0 1111 0 illi11 8 ~~ 1~1~•~••~ !si ~• 0 1110 11111,~lf' corn cob Basic Genetics ·h ., 11111i11~ rllw the () r · . I . - . . "' LI i/ i ( CJ 0 111 ,., co 111ml/ ed hy o .\III J.!. fr /Ht 11 · n/ (l//f'lf' \, r/('d11r rhf' gcnn( Y />C 0nd /Jh enm _r pe ql th e por e 11, p/cmt. De duction: The purple grains and yellow grains ar e in a 3 : I ra ti o (53 : J 8 = 3 : I J. According to Mendel's Law of inheritance, th e pare nt mu st be heterozygous and pur pl e grain is th e dominant character. There fo re th e phenoty pe of the parent mu st be purple grain (i.e. the parent plant had de Yel oped from a purple grain). CASE3 Deduce the genotypes of the parents given that the phenoty pe s of the parents an d the offspring are the same.* Example: A green pea plant was self-pollinated and all offspring produce green leaves. Ass umin g that le af colour is controlled by a pair of alleles, deduce the genotype uf rh e pea plant . Deduction: Since all the offspring have th e same phenotype as their parent. th e parent pl ant must be homozygous. c. The phenotypes of parents and the phenotypes (and phenotypic ratio) of the o jf ,;; pring are provided, deduce the dominant I recessive character for certain phenotype. CASE l Deduce the Jo m in ant / r eces~ i ve alle le of a character given th at bo th pare nt s have the same phenotype but u ff -; pri ng ar e of different ph enotyp es. ·ri . . . ' I' I - ro· 'l JC '' '] l'tri•c nu mber or orr~pr in ° . 1\ , 1s onlv true· for oq:.alll'-1n'- w ·1 1c 1 u rn P u ~ ' ' c- "' 101
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es an d E xe rc ises NSS Biology- lnt e nstve Nol , 10~ Kram pl e I: _ . " . . . . II · I by a pair ol allel e~ . Jn n b1 ced 1n g ex pe rim ent 1 _ r n 1 ._ , s co 111 , o cl . 1 , •. f ur L'U 1.) Llt 11 " · · 1 . . , 1 , g ave birth to 1hr ee gr ey and fi ve bJ, 1 k .,, I · 11 1d 11l c c 1nc1 L , c 1lf ('\ J':l l'i wer e 1n atc 1. ' , ') r, I . I ell \ ::- . . , • 1 ,- ev 0 ,- h/ <1r ·k ('(J/0111 . c XJJ mn _ \O llr an swer H · 1, ; (11 i., rhc· r<'<' c ' SS/\ '< ' ('/1(/1 u, f ti · -~ · lkduction: Rlac A 1:-- n:c~ :-- s i\' c h ec ausL ': )h ,th parents h ave gre y .fur. th ere fore e ach of th em mu 5t po ssess at i rnl/ on e gre y allele. S om1.: of the ir o ff s pri ng ha ve black.fur sh ow in g that th ey mu st ha ve recei ved w lc1as1 one a ll ele for bl ack f ur fr om e ith er of th e parent s. At le a st one of the pa ren ts is hetero zygous. in hete r ozygo us co nditi on, th e rece ss ive all ele is m as k ed. Therefo re the black allele is recessive . Example 2: ln man. the ab ilit y to roll tongue is determined by a pair of alleles. D ma le t ong ue- ro ll er D ma le non - ro ll er Q fema le tongue - ro ll er Q female non -ro ll er Wi th rc }N ef/('e I ti 1 0 U' 0 Jn 1 1 e /'Jcdi[lrec .. ,,1 11 ·(·/ • , 1 · '" • • I I,\ f I l' r/ > · or tnr,hility to ro ll 10 11 > .,,.., .- · < 111111r1111 clwracr e. . . . .~ lf t . L.rp l mn yo 11 r c 111 s 11 ·er . , • the ahtl 1( r
Basic Genetics Deduction: The abi!i(y to roll 1011 ' " ) . . I . ,I< ( ' " l 0 11111 1. 111[ h L'l' ] ll '-;t': Both p,ircnts ; 1rc tou ,. , . I , . , . ~ 11 ' ,n h ' ·' · tlwrl'll>1t · L'i 1L ·h 1 i1 · 1l1 c 111 m11 q p o<.,<., cs., u/ /co st one to11g1w-rol/i,,,:: :i lkk . Some or !heir cllildre 11 'l .. , . . , , < I L notHo/ler s , how 111 1~ 1l1 ; 1t 1l1 cy n,11 , 1 h;1vc received or ha. \/ 0 11 c alklc for llOJ _ 1 n , . . . . . 1 Ul\ , .. LI L 1oll111 g trnrn c ilh er 1)1 Ili c pcin·n h. At kast o ne of the parents is heterozygous. In hetcro zy , ious . r · _ . ~ c- ' co nuti o 11 , only th e dum1nc1nt allele will h <: n p r(' ,w' rl. ·r l 1crclorc.: the tongue-rolling allele is dominant. CASE2 Deduce th e dominant / recess iv e a. ll ele / c hara ct er given that the parents have d~fferent ph enotypes but o ff s prin g are of the same pheno(vpes.* Example: Wing length of fruit flies is controlled by a pair of alleles. In a br eeding experiment. long-winged flies were crossed with short-winged fli es . a ll the F 1 fli es , vere long- winged . Which of the par e nts shows the dominant character? Explain ru11r m1s1t ·er. Deduction: Some offsprjng must have rece iv ed a long-winged allele and a short-winged allek from their parents/ so me offspring must be heterozygous. A ll offspring have long wings. Therefore. th e long-winge<I a ll ele must be dominant over the short-winged allek. · . - - ·- . -- . - --- ~ .. . 1 lun· a Li r<'C' 111 1111h cr 11f offsprin g. l '11 , 1s nnlv 1ru i.: lu 1 11rt• . a111 o.;111-; , ... 111l Ii i.: ,11 1 P 1 ( 1 t, . ~ 103
, gy- lnlensi, e Notes and E,crcises .,. ,;J,:..--- --~- 20.3 Oihybrid Inheritance ..- ~1 '8c::= 104 ;1. Mendel' .~ . ,. 1 ,c 1111 d /aw - The / ,, aw of Ind ependent Assortment { In m odl•r n te rm s: D1 m 11 ~ ~ amc k' r, ,,ma t ion. e ith er one of a pair of , iil clcs ma y com bi ne rand " i1 h , ithrr 011e of anoth er pa ir and e nt er th e sa me gamete eel I. °"'' i OR gametes B. J1 endel's experiment on pea plant a. Re s ult s of the breedi . , ng expermzent: pu re br ee ding X homologous chromosome pair ro und and ye llow see ds pure breeding wrinkled and green seeds F1 s ee ds ar e all round . I anc ye llow (d 'h . I I ybnd) + se lf -pollination F · 31S 2 - . round a nd ye llow IOI wr i nk led a nd e ll ow , . y (new comb 1na1i on) I 08 ro unJ a nd gr ee n ( . new comb mnlion) 12 w ri 11 klcd and gr ee n
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Basic Genet/ TliC' pn 1p,1rt1t11 is. lll. J l1 - I l' llnl ,, ,, ,, .., ., 1 dih,h ri< .1 ·· •· . , 1p 1u '<. 11 11 ; 11 c d 111 ii rn 110 , ,r <J : J: 3 : I - the .I • ' 10 . h. Explanation 11 .\·inx R<' 11 eli<' dia ~ra 111 : I l'I : I } . ' IL'j)t°l.'sc nt s the I · . . L um 111ant allele fur rc ,ullll .:; ccd r n .' ]ll"l' :--. cnts I he re , , . . · 1 . · l L:SS I\ c a k le lor wrinkled see d '\ represe nt s th e domi1nnt a ll ele f . II d ~ - 0 1 ye ow see y represents th e re ce ss i ve · 1l lelc f . d c 0 1 green see P: ro und. ye ll ow seed x wrinkled. green seed G: R RYY rr yy t RY ry RrYy F, seeds are all round and yellow ( dih y brid s) t self-pollination RRYY RRYy Rr YY round & yellow round & y ell ow ro und & ye ll ow RR Yy RRy y R rYy round & ye J low rou nd & green round & ye ll ow Rr YY Rr Yy rr YY R rY y round & yell ow Rry y ro u nd & green rrY y round & yellow ro und & ye ll ow w ri nk le d & ye ll ow wrinkled & yellow RrY y Rryy IT Yy rr yy round and ye lJ ow round a nd green wri nk led & yellow wrinkl ed & gr ee n Punnett square s ho w in g all po ss ib le c omb ina ti o ns<~ { ga met es to form F 1 105 ;:_;::zd
c'e~ ,··d E~erc ses i.) ---- RE \1 \RI\.\ : J' ff crcnl ch rom osome s. Uni~ true !0 1 ge ne" ho rn c on I 20.4 I nherita nce in Humans 106 A. Multiple alleles C Th ey a n. .: a se ri es of 3 or more alleles (alternative form s) of a gene which occ ur at t he s am e locus on homologous chromo so mes. C; On ly 2 of them can be present together in a cell. C Exa mpl e: human ABO blood gro up s / 1 - domi n ant all e le r es pon s ibl e for produc in g a nti gen A on red blood ce ll s. t'1 - dom in a nt all e le r es ponsible for produ ci ng antigen B on red blood cell s. , -- recessive alIde produ c in g neith er antigen . / 1 t.1 / 11 d . d . . . b all arc co omuiant an , 1s recess iv e to oth. 0 (;e1101,·pe !"!° : s in ce th e two alleles arc equally dominant , they can express th emselves at th e sa me ti111 e in heterozyg ous co ndition. 0 Human bloot.l gro up ge not yp es : A 1:3 AB 0 Genotype \' t'rt . t 'i t11 J'i , t i r'/'1 ii A B both no ne
t;t •, 1 l1k11 ,, i : ,' II / 1 ,1 1 I /" t 11 I ( Bns1c Gonoftc., r l\lo r1n t1 r1 on Blood group A An tl qo 11 [l Blood group B Antigen A Blood group AB Blood group 0 B. Sex deten11i11ation ( ' Sex is :-1 genetically determined trait. C: 1 Membe rs of all homologous chromosome pairs are identical in s iz e and sh ape: exec-pr the sex chromosomes which are important for the deter min ation of se x. A ll other chromosomes are known as autosomes. C) In humans, the sex chromosomes in male are not alike in size and shape whil e in femaJc, the sex chromosomes are identicaJ in size and shape. C XX-XY system: in humans , 0 th e female possesses two X chromosomes while the male on e X and on e Y 0 X and Y are of different sizes; Y being shorter, contains very few genes 0 at meiosis, half of the sperms wilJ contain X and half Y Q maleness is determined by Y chromosome which controJs the differentia ti on of th e testis which in turn affects the dev elopment of the genital organs and male characteristics . 107 =-:it!
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IU I ., \C"fL f...) \ .,..,.'l vgy - ,r 11 cr 11'"' 11 C' /'ll l l /C.) ,;JI - QC U>vC , 108 1, ,, 11 «I 61 4 5 2 3 3\&l~HHU lh 6 7 l\ U AA 13 14 15 U )UC 19 20 8 9 11 Xl 21 22 10 11 12 1°C II H 16 17 18 ( t- XY )) ,i 2 31 &I 6 7 n A6 AA 13 14 15 U XJ 19 20 11 3 it u 8 9 11 XI 21 22 «i H 4 5 n ax lh 10 11 12 re ix 11 16 17 18 ,, xx Ch romosomes of male Chromosomes of female c:, The chance of a child being a boy or a girl is 50% . P: Father XY /~ G: X C. Sex-linkage in man xx 0 irl 0 y X XY boy Mother xx ! X ~ Genes carried on the sex chromosomes are said to be sex-linked. In human male. th e re is a portion of the X chromosome for which there is NO homologous region of th e y chromosome. Genes carried on th e non-homologous por ti on of rh e X ch ro mo so me there .f<Jre can expre ss th e ms e lv es even if th ey are recessive . CJ Example: c ol o ur blindn ess and ha emophilia a. Red-green colour blindness = Pers on s suffe rin g. from red -g r ee n co lour blindn ess ca nnot di s tingui sh be t, veen re d and gr ee n colo ur s.
Basic Genetic e A higher !) lT l' Cllt a μ c 01 · l lll ' I .. , ' . . . . . ~ · 1 dl t.: et il()u1hl111d 111 a pr,pula 11 on. c:, Explanation: Th C g C 11 (' ro r C () I l) ll r h 1 · I . . , ,. lll l ll C...,..., I ', a r ct.: c-,..., , vc ; 1l le lc l, 1r r1 cd r;n th e X Lh1omosnn 1 e Th ey . 1 1... 1 rnmn -, om c ()I" 111..il c c ,rric ., ncil ht r th e c () l <>U rhli ncJ gene nor it s normal al lclc . C) Exampl e: A normal \.\ ' . . . L om ...in 1s married to a colourblinrJ man Let: X rep ·e x 1 se nt s chromosome w ith th e normal a ll ele X' represents X chromo so me w ith colourblind a ll ele y represe nt s y chromosome Parental phenotype: Normal female X Parental genotype: XX Gametes: X F 1 genotype: xx c F I phenotype: normal female (carrier) F1: Normal female X (carrier) xx c /~ Gametes: X Xe F . XY CoJourblind male xcy xc XY normal male Normal male XY y /~ X y xx normal female normal normal colour-blind mal e female (carrier) male e Cris s -cross pattern of inheritance - a colourblind man transmits his colo ur blind gene through hi s daughter who is phenotypically normal but carries the colourblin cl g en e to half of her sons, i.e., the se x-linked recessive character of the g ra nd fat he r reappears in the grandson. 109 -:::J?J!
d E xe rcises ,vc Note s an -----:- N~S ;c:; S n 9, i o/ogy - /ntens oTJ hilia . d on th e X ch romosom e. B l oo u ol th I,. Haem " , . rre ne ca rn e t . , .. a r ecess i ve o . . . a nd ma y ble ed lo death . ~ T hi s ,s du e 10 ·, . . o c lot att cr inJur y . . . pe rson fa 1 Is t hac111 op l11li ac , . . . ,o n in man . 0 ;\~: till ii ,s (11L}I C c ot11n c:, Exp lanation: y allele for th e trail, a man w ith a e do es not ca rr y an Since th ey c hr omoso m . haemophili ac. h .. X c hromo s ome ts recess i ve ge ne on 15 20. 5 Variati on ---- ------------- -~ 11 0 A. Types of variation a. Discontinuous variation C) It is the variation where the character can be grouped into two or a few distin ct classes with no intermediate phenotypes. c:> Characteristics showing di sc ontinuous variation are usually controlled by one pair (o r two pair s) of alleles and are not affected by environmental factors easily. c:> Human examples: tongue-rolling, eye colour, shape of ear lobe, blood gro up , etc. b. Continuous variation c:> It is the variation of a character where there is a co n rnuou s range o intermediate phenotypes between two extremes of the character. c:> A normal distribution curve is produced. c:> Ch arac teri s tic s s howing continuous variation are produ d b . . . ce Y the combined effec ts of many genes ( many paus of alleles) and are affect db . . . e Y environmental factors more eas il y. c:> Human exa mples: hei.g ht , we ig ht , IQ, etc.
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1 200 1 100 1 000 900 800 C C'il E 700 ::i .c. 0 (/) 600 u C C'il t/) ::i 5 00 0 .c I- 400 300 200 100 I 14 ~0 ~~ 1 ~ 45 ~-: 1-:=-5 :::--- 0 -: 1~ ~-=-5 - 1 J_ 6_0 _ 1 J_ 6_5 _ 1 j_ ;- o _ 117_5 _1180 ::::::~1 :i::: 85= __, =1 :i..__ 90 Height (cm) B. Causes of genetic variation C) / ndependent assortment of chromosomes and crossing over at meiosis e to give gametes with different combinations of alleles. ~ Random fertilisation C, Mutation e a sudden, inheritable change in the genetic material (DNA) . e Cause: exposure to ionising radiation and chemicals C. Effects of environment on gene expression Ex amples : a. Effect of light on chlorophyll production c:, The all ele for c hl orophyll production (dominant) can express it self o nl y when light is present. 111 ~
d E crus es logy- lntenswo Notes ,m x _ i.1; .... ....... __,~~ ---""~- l F .riu , rim ents 0 11 twin s J. - . . . .. . , , , I 10 s wd y env ir on me nt al ef fect s on ge n (: , 0 I Jc 11t1 ca l 1w 1I1 s t11 c u scl L:X p rl'. \\ 11 l . t li cv , 11 .,, "t' J ze li c a/lu id e nt ica l. 'rt 1 1 , rn an K ' ca u sc . , -- - ,., - J "' 7 , 1 . H ·d · ii n rl in Jiffcrc nt environments wi l!1 diff er ent nut . . .., i cy arc 1 e; c , • rit1 cJn . rtn(j L 'du ca l io n. 0 Resu lt s sh ow th at 0 char ac ters sh ow in g discontinuous variation are n ot affected 0 ch ar acters sho win g continuous variation ar e aff ected to di ff crc nt cxtc nh- body we ight is significantly affected by nutrition wh il e hei oht aw' fQ · b u arc l ess affec ted. "'il,,m::: 112