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ICHHA PURAK
ICHHA PURAK

This power point presentation is designed to explain deviation of Mendelian dihybrid ratio due to interaction of genes which may be of following types 1.Two gene pairs affecting same character – 9:3:3:1 2.Epistasis, one gene hides effect of other a) Recessive Epistasis - 9:3:4 b) Dominant epistasis - 12:3:1 3.Complementary genes - 9:7 ( 2 genes responsible for production of a particular phenotype ) 4. Duplicate genes – 15:1 ( same effect given by either of two genes ) 5. Polymeric gene action - 9:6:1 6. Inhibitory gene action - 13 : 3 Each interaction is typical in itself and ratios obtained are different

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INTERACTION OF GENES MODIFICATION OF DIHYBRID RATIO EXTENSION O MENDELIAN GENETICS By Dr Ichha Purak University Professor Department of Botany Ranchi Women’s College Ranchi University,Ranchi
Mendel performed monohybrid and dihybrid crosses taking sweet pea as experimental plant In monohybrid cross only one trait and its two alternative forms were taken under consideration at a time . Two phenotypes ( Red and White flower colour) were obtained . There were only three possible genotype ( RR,Rr & rr) with Phenotypic Ratio 3:1 and Genotypic Ratio 1:2:1. RR(Homozygous ) and Rr ( Heterozygous ) are phenotypically alike and both are governed by dominant allele (R). In Dihybrid cross two traits at a time were considered ( Height of plant and colour of flower ) . Four phenotypes i.e. tall red(TR), Tall white(Tr), dwarf red(tR) and dwarf white (tr) were obtained by dihybrid cross. Phenotypic Ratio 9:3:3:1 was obtained Genotypic Ratio 1:2:2:4:1:2:1:2:1 was obtained Combinations 1,6,11 & 16 are only homozygous
DIHYBRID CROSS AND ITS ABBREVIATED RATIO Parents TTRR ttrr Parental Gametes TR tr F 1 Zygote Tt Rr tall plants Red flowers F1 Gametes TR Tr tR Tr TR TR TR 1 Tr TR 2 tR TR 3 tr TR 4 Tr TR Tr 5 Tr Tr 6 tR Tr 7 tr Tr 8 tR TR tR 9 Tr tR 10 tR tR 11 tr tR 12 tr TR tr 13 Tr tr 14 tR tr 15 tr tr 16 F2 Generation 16 Possible Combinations (P) TR (Tall Red) : 1,2,3,4, 5,7, 9,10,13 = 9 ( R ) Tr (Tall White) : 6,8.14 = 3 ( R ) tR ( Dwarf Red) ; 11,12.15 =3 ( P ) tr ( Dwarf White ) : 16 =1
By seeing Mendel’s di hybrid Ratio, it becomes clear that different characters are inherited independently . Because of that four combinations are produced in F2 generation Mendel concluded on the basis of results of monohybrid and di hybrid crosses performed on sweet pea that a) One trait (character) is controlled by a gene (factor) existing in two alternative forms (alleles) . The alleles are responsible for two alternative allelomorphs. b) That different traits assorted independently ( segregation of two genes is independent of each other)
After Mendel’s rediscovery by Hugo Devries, Carl Correns and EV Tschermak, many geneticists repeated the same pattern of work using other organisms and different phenotypes but the results were not always according to Mendel’s laws of inheritance, fit with the principles as law of dominance and law of independent assortment. It became apparent that it is not necessary that a single character (trait ) is controlled by one gene only and it is also not true that a gene has only two alleles, may have many alleles and the phenotype may be result of interaction of two genes and their alleles or a character may be controlled by more than one gene and its alleles. In such a condition 2 or more genes may interact, the resulting phenotype may be result of interaction.
Suppose Gene A is responsible for phenotype A (Red) and Gene B is responsible for phenotype B (white ) ,when both A and B are present together may give rise to a new phenotype C (Pink) , but these genes still obey Independent Assortment pattern. If A and b are two genes, both dominant over their respective recessive alleles a & b ,then interaction will depend on presence of both dominant alleles A & B, Absence of A or B or absence of Both A & B. Mendel was fortunate to get 3:1 and 9:3:3:1 ratio uniformly because a) The traits he considered were governed by only one pair of alleles of a gene. b) The genes and their alleles for different characters were located on different homologous pairs of chromosomes or even if on the same homologous pair the genes for different trait were present at a distance sufficient for cross over
INTERACTION OF GENES Deviation of Di-hybrid Ratio
Genes usually function or express themselves singly or individually. But, many cases are known where two genes of the same allelic pair or genes of two or more different allelic pairs influence one another. This is called gene interaction Non-allelic gene Interactions These are interactions between genes located on the same chromosome or on different but non-homologous chromosomes controlling a single phenotype to produce a different expression. Each interaction is typical in itself and ratios obtained are different from those of the Mendelian dihybrid ratios. Some of these interactions of genes are explained here which fall under this category and deviate from Mendel's ratios.
The interaction of genes may be of following types 1) Two gene pairs affecting same character – 9:3:3:1 2) Epistasis, one gene hides effect of other a) Recessive Epistasis - 9:3:4 b) Dominant epistasis - 12:3:1 3) Complementary genes - 9:7 ( 2 genes responsible for production of a particular phenotype ) 4) Duplicate genes – 15:1 ( same effect given by either of two genes ) 5) Polymeric gene action - 9:6:1 6) Inhibitory gene action - 13 : 3
Gene Interaction Inheritance Pattern &Example A-/B- 9/16 A-/bb 3/16 aa/B- 3/16 aabb 1/16 RATIO Additive gene effect Comb shape Chicken 9 Walnut 3 Rose 3 Pea 1 Single 9:3:3:1 W:R:P:S Recessive Epistasis Coat Colour Mouse 9 Agouti 3 Albino 3 Black 1 Albino 9:3:4 Ag:Bl:Al Dominant Epistasis Fruit colour squash 9 white 3 white 3 yellow 1 green 12:3:1 W:Y:G Complement ary genes Flower Colour Lathyrus 9 Purple 3 White 3 White 1 white 9:7 P:W Duplicate genes Fruit Shape capsella 9 3 3 1 15:1 Tri: Top Polymeric Gene action Fruit Shape Squash 9 Disc 3 Circular 3 circular 1 Long 9:6:1 D:C:L Inhibitory gene action Maize Aleurone colour 9 White 3 Red 3 white 1 White 13:3 W:R
Two gene pairs affecting the same character : comb shape in chicken ( 9:3:3:1-for single character) Gene R- rose comb -Rp Gene P –Pea Comb-rP The dominant alleles of each of the two genes produce separate forms of phenotype when they are alone ( heterozygous) Both R and P when brought together form a new phenotype walnut Allele R dominant over r Allele P dominant over p rr and pp produce single comb ADDITIVE GENE EFFECT
F M RP Rp rP rp RP RP RP 1 Rp RP 2 rP RP 3 rp RP 4 Rp RP Rp 5 Rp Rp 6 rP Rp 7 rp Rp 8 rP RP rP 9 Rp rP 10 rP rP 11 rp rP 12 rp RP rp 13 Rp rp 14 rP rp 15 rp rp 16 F1 gametes RP-Walnut 1,2,3,4.5,7,9,10,13= 9 Rp –Rose 6,8,14=3 rP- pea 11,12,15=3 rp- Single 16=1 so the ratio is 9:3:3:1 just like normal dihybid ratio for two traits but it is for single trait i.e. comb shape having 4 different forms Parents RRpp x rrPP (Rose) (Pea) P. gametes Rp rP F1 RrPp walnut GENETIC EXPRESSION F2 Generation 9:3:3:1
A Cross Involving a Two-Gene Interaction : 9:3:3:1 ratio for single trait RRpp rrPP RrPp rrpp Inheritance of comb morphology in chicken described by William Bateson and Reginald Punnett in 1906 : First example of gene interaction Four different comb morphologies
F2 Generation F2 generation consisted of chickens with four types of combs 9 walnut : 3 rose : 3 pea : 1 single Bateson and Punnett reasoned that comb morphology is determined by two different genes
(Additive Gene Action) There are 4 phenotypes in F2,so are governed by more than one gene If it was governed by one gene and its two alleles F2 would have shown only 3 phenotypes in 1:2:1 ratio Where both A and B are present add colour so new phenotype brown is produced F2 phenotype classes • Dominance Relationships: – Tan is dominant to green – Gray is dominant to green – Brown is dominant to gray, green and tan. – Tan and Gray are incompletely dominant, giving rise to brown. • Genotypic classes: – Brown: A_B_ – Tan: A_bb – Gray: aaB_ – Green: aabb 2nd Example Lentil Seed colour 9:3:3:1
EPISTASIS One gene hides the effect of other gene. It is different from Mendel’s Dominance which is meant for intragenic alleles (alleles of a gene) but here dominance works at intergenic level (alleles of different genes ). One gene masks the expression of another non-allelic gene. A gene which masks (hides) the action of another gene (non allelic) is termed as epistatic gene. The gene whose effects are masked is called hypostatic gene For example if two gene A and C with their alleles a and c take part in a cross then epistasis can be of following types 1) Recessive Epistasis -Recessive allele ( c ) of one gene may hide the effect of dominant allele (A ) of other gene 2) Dominant Epistasis Dominant allele ( A ) of one gene may hide the effect of dominant allele ( C ) of other gene.
a) Recessive epistasis: Here the recessive allele masks the effect of dominant allele of other gene. In mice the wild body colour is known as agouti(greyish) and is controlled by a gene A which is hypostatic to recessive allele c. The dominant allele C in the presence of 'a' gives coloured mice. In the presence of dominant allele C, A gives rise to agouti. So, CCaa will be coloured and ccAA will be albino. When coloured mice (CCaa) are crossed with albino (ccAA), agouti mice (CcAa) appear in F1. cc masks the effect of AA and is therefore epistatic. Consequently, cc AA is albino. The ratio 9 : 3 : 3 : 1 is modified to 9 : 3 : 4. The combination ccaa is also albino due to the absence of both the dominant alleles.
Recessive Epistasis Example : Coat colour of Mouse Coat colour is controlled by Gene A , Allele A is hypostatic to recessive allele (c ) The dominant allele C in absence of A gives coloured mice When both C and A are present colour is Agouti ( wild type most common ) due to banded hair : Near skin Grey yellow Black Two other colours are Albino and solid black AACC (agouti) x aacc (albino) AaCc (all agouti) AaCc x AaCc A-C- Agouti 9/16 A-cc Albino 3/16 aaC Black 3/16 aacc Albino 1/16 The c locus is epistatic to the A locus. 9 (Agouti ) :3 ( Black ) :4 ( Albino) F2 ratio Genotype Phenotype
Parents CCaa X ccAA Black Albino p. gametes Ca cA F1 CcAa ------------------------------------Agouti F1 Gametes→ ↓ CA Ca cA ca CA CA CA 1 Agouti Ca CA 2 Agouti cA CA 3 Agouti ca CA 4 agouti Ca CA Ca 5 Agouti Ca Ca 6 Coloured cA Ca 7 Agouti ca Ca 8 coloured cA CA cA 9 Agouti Ca cA 10 Agouti cA cA 11 Albino ca cA 12 albino ca CA Ca 13 Agouti Ca Ca 14 Coloured cA Ca 15 albino Ca Ca 16 Albino CA –Agouti- 1,2,3,4,5,7,9,10,13=9 Ca- Black- 6,8,14=3 cA- Albino- 11,12,15=3 (c masks the effect of A ) ca- albino -16=1 So the ratio is 9:3:4 ( Agouti:Black:Albino ) F2 Generation Genetic Expression Cross between Black and Albino : Genetic Expression
Interaction of genes for slide share
F2 Phenotypic ratio : 9 (agouti) : 3 (Coloured) : 4 (Albino)
(b) Dominant epistasis: In summer squash or Cucurbita pepo, there are three types of fruit colour - yellow, green and white. White colour is dominant over other colours, while yellow is dominant over green. Gene for white colour (W) masks the effects of yellow colour gene (Y). So yellow colour is formed only when the dominant epistatic gene is represented by its recessive allele (w). When the hypostatic gene is also recessive (y), the colour of the fruit is green. White Fruit - WY , Wy Yellow Fruit - wY Green Fruit – wwyy A cross between a pure breeding white summer squash, (WWYY) with a pure breeding green summer squash, (wwyy) yields white fruits in the F1 generation. Upon selfing of F1 the F2 generation comes to have 12 white fruit : 3 yellow fruit : 1 green fruit.
Parents WWYY X wwyy White Green P gametes WY wy F1 Generation Hybrid WwYy --------------------------------------------White F1 Gametes Female Male WY Wy wY wy WY WY WY 1 white Wy WY2 white wY WY 3 white wy WY 4 white Wy WY Wy 5 white Wy Wy 6 white wY Wy 7 white wy Wy 8 white wY WY wY9 white Wy wY 10 white wY wY 11 Yellow wy wY 12 Yellow wy WY Wy 13 white Wy Wy 14 white wY wy 15 Yellow wy wy 16 green WY-White- 1,2,3,4,5,7,9,10,13=9 Wy-White- 6,8,14=3 wY-yellow- 11,12,15=3 wy-Green-16=1 F2 generation 12 white : 3Yellow :1 green Genetic Expression
F1- AaBb X AaBb 9 A B –white 3Aa-bb –Yellow 3 aa-B- White 1 aabb –Green A causes Yellow but in presence of B can not express ,produce white FRUIT COLOUR SQUASH : 12 : 3 : 1 13:3 12:3:1
Two genes are responsible for a particular phenotype. Production of one phenotype requires dominant alleles of both the genes controlling the character. The complementary genes are two genes present on separate gene loci that interact together to produce dominant phenotypic character, neither of them if present alone, can expresses itself. It means that these genes are complementary to each other Example is flower colour of Lathyrus odoratus( Keshari/pea grass) .The colour of flower is either purple or white. Purple colour is produced only when dominant A is complemented by Dominant Allele B COMPLEMENTARY GENES
Bateson and Punnet have demonstrated that in sweet pea (Lathyrus odoratus) purple colour of flowers develop as a result of interaction of two dominant genes C and P. In the absence of dominant gene C or P or both, the flowers are white. It is believed that gene C produces an enzyme that catalyzes the formation of necessary raw material for the synthesis of pigment anthocyanin and gene P produces an enzyme which transforms the raw material into the pigment. It means the pigment anthocyanin is the product of two biochemical reactions, the end product of one reaction forms the substrate for the other. Product of Gene C Product of Gene P Substrate A Substrate B Anthocyani (Purple) Therefore, if a plant has ccPP, ccPp, CCpp or Ccpp genotypes, it bears only white flowers. Purple flowers are formed in plants having genotype CCPP or CCPp or CcPP or CcPp. From checker board, it is clear that 9 : 7 ratio between purple and white is a modification of 9 : 3 : 3 : 1 ratio.
Genetic Expression Parents : AAbb X aaBB White White P.gametes Ab aB F1 AaBb --------------------------------- Purple F1 Gametes Female→ AB Ab aB ab AB AB AB 1 AB Ab 2 AB aB 3 AB ab 4 Ab Ab AB 5 Ab Ab 6 Ab aB 7 Ab Ab 8 aB aB AB 9 aB Ab 10 aB aB 11 aB ab 12 ab ab AB 13 ab Ab 14 ab aB 15 ab ab 16 So the ratio becomes 9:7 ( Purple : White ) AB- Purple-1,2,3,4.5.7.9,10,13= 9 Where both (A ) and ( B ) are present flowers are coloured ( when A is complemented by B then only flower is coloured ) Where either (A ) or ( B ) is present colour is white Ab- White- 6,8,14 = 3 aB- White – 11,12,15 = 3 ab- White -16 = 1 Total white- 3+3+1=7 F2 Generation Male COMPLEMENTAY GENES KESHARI
A Cross Producing a 9:7 ratio 9 C_P_ : 3 C_pp :3 ccP_ : 1 ccpp purple white
Complementary gene action - interactions arise because the two genes encode proteins that participate in sequence in a biochemical pathway Enzyme C and enzyme P cooperate to make a product, therefore they complement one another Enzyme C Purple pigment Colourless intermediate Colourless precursor
F2 phenotypic ratio - 9 (Purple) : 7 (White) Results of an experiment showing inheritance of flower colour in Lathyrus odoratus controlled by complementary genes
Interaction of genes for slide share
Duplicate genes If the dominant alleles of two gene loci produce the same phenotype, whether inherit together or separately, the 9 : 3 : 3 : 1 ratio is modified into a 15 : 1 ratio. The capsules of shepherd's purse (Capsella) occur in two different shapes, i.e., triangular and top-shaped. When a plant with triangular capsule is crossed with one having top-shaped capsule, in F1 only triangular character appears. The F1 offspring by self crossing produced the F2 generation with the triangular and top-shaped capsules in the ratio of 15 : 1. Two independently segregating dominant genes (A and B) have been found to influence the shape of capsule in the same way. All genotypes having dominant alleles of both or either of these genes (A and B) would produce plants with triangular-shaped capsules. Only those with the genotype aabb would produce plants with top - shaped capsules.
Same effect is given by either of two genes (A ) or (B) Example is Capsella bursa -pastoris (Shepherd’s Purse ) Fruit shape of two types a) Triangular (Heart shape ) capsule b) Top shaped (Narrow ) capsule Genetic Expression Parents AABB X aabb Triangular Top shaped P gametes AB ab F1 (Hybrid ) AaBb ------------- triangular F1 gametes AB Ab aB ab AB AB AB 1 AB Ab 2 AB aB 3 AB ab 4 Ab Ab AB 5 Ab Ab 6 Ab aB 7 Ab Ab 8 aB aB AB 9 aB Ab 10 aB aB 11 aB ab 12 ab ab AB 13 ab Ab 14 ab aB 15 ab ab 16 AB- 1,2,3,4,5, 7,9,10,13=9 (Triangular) Ab-6,8,14=3(Triangular ) aB-11,12,15 =3 (Tringular ) ab-16 =1 (top shaped ) So the ratio is 15:1 This is example of gene interaction,two genes involved in same pathway.It is based on the idea that some genes may be present more than once in the genome DUPLICATE GENES F2 generation 15 (Triangular): 1 (Top)
A_ or B_ = heart shape aa and bb = narrow shape Fruit shape in Shepherd’s purse Duplicate Dominant Epistasis
In a cross between two lines differing in fruit shape (Heart shape vs narrow ) F1 generation shows all heart shaped like intragenic dominance but F2 generation shows a ratio of 15:1 and not 3:1 as that of monohybrid cross. It means that trait fruit shape is controlled by two genes (A and B ) and their Alleles a and b. 15 :1 ratio is therefore modification of dihybrid ratio 9:3:3:1 in which 9,3 and 3 are grouped.The triangular shape results by the presence of at least one dominant allele of either gene. The two gene appear to be identical in function and is in contrast with complemenatary genes or 9:7 ratio where both dominant alleles are required for a phenotype and they complement each other. Duplicate genes provide alternative genetic determination of a specific phenotype
Duplicate Gene Action Epistasis TV TV Tv Tv tV tV tv tv TTVV TTVv TtVV TtVv TTVv TTvv TtVv Ttvv TtVV TtVv ttVV ttVv TtVv Ttvv ttVv ttvv TTVV Triangular ttvv Ovate TtVv All triangular F1 (TtVv) x F1 (TtVv) x F1 generation 15:1 ratio results Shepherd’s Purse (Capsella ) Fruit Shape Triangular / Top (Ovate) shape
F2 phenotypic ratio : 15 (triangular ) : 1 (Top shaped )
P AABB aabb AB ab AaBb F1 AaBb X AaBb F2 15/16 Red : 1/16 white Whenever a dominant gene is present, the trait is expressed. One allele is sufficient to produce the pigment. 2nd Example : Petal colour in snapdragon Redundancy : Duplicate Genes
Polymeric gene action 9:6:1 Two completely dominant genes controlling a character produce same phenotype, when their dominant alleles are alone, But when dominant alleles are together , the phenotypic effect is enhanced and become cumulative or additive effect Example -1 Awn length on Barley fruit AABB x aabb A /B- Median Awn A & B – long Awn ab- Awnless Example- 2 Fruit shape Squash (Disc/Circular/Long) 9:6:1 AaBb X AaBb 9 A B – Disc 3 A-bb – Circular 3 aa-B- Circular 1 aabb – Long
Genetic Expression Parents AABB( Long awn ) X aabb(Awnless) P gametes AB ab F1 AaBb Long awn F1 gametes AB Ab aB ab AB AB AB 1 AB Ab 2 AB aB 3 AB ab 4 Ab Ab AB 5 Ab Ab 6 Ab aB 7 Ab Ab 8 aB aB AB 9 aB Ab 10 aB aB 11 aB ab 12 ab ab AB 13 ab Ab 14 ab aB 15 ab ab 16 AB – Long Awn -1,2,3,4,5,7,9,10,13=9 Ab- Medium Awn -6,8,14=3 aB- medium Awn-11,12,15=3 ab-Awnless-16=1 F2 Generation : Phenotypic ratio 9(Long) : 6(Medium) :1(awnless) Example awn length on Barley fruit AB- Median Awn A & B – long Awn ab- Awnless
INHIBITORY GENE ACTION Inhibitory gene action Example Maize Aleurone colour 13:3 One dominant gene produces concerned phenotype and its recessive allele produces contrasting phenotype. The second gene (dominant) has no effect on concerned phenotype but stops expression of dominant allele of first gene, so when both dominant alleles are present ,phenotype as that of recessive homozygote is produced . Genetic Expression Parents RRII( White aleurone ) X rrii (White aleurone ) P gametes RI ri F1 hybrid ----------------RrIi ------------------------White F1 gametes
Genetic Expression Parents RRII( White aleurone ) X rrii (White aleurone ) P gametes RI ri F1 hybrid ----------------RrIi ------------------------White F1 gametes RI Ri rI ri RI RI RI Ri RI rI RI ri RI Ri RI Ri Ri Ri Red rI Ri ri Ri Red rI RI rI Ri rI rI rI rirI ri RI ri Ri ri Red rI ri ri ri RI-White-1,2,3,4,5,7.9.10,13=9 Ri- Red- 6,8,14=3 rI- white -11,12,15=3 ri-white-16=1 So phenotypic Ratio becomes 13 (White ) : 3 (Red) Inhibitory gene action Example Maize Aleurone colour 13:3
Second Example 13 :3 Inhibitory gene ( one gene inhibits expression of the other ) 13: 3 Example –Feather colour in Fowl ( White/Coloured ) Epistasis AaBb X AaBb 9 A B –white 3 A-bb –white 3 aa-B- coloured 1 aabb –white B is responsible for colour but in presence of A cannot express
Masking gene action (12:3:1) Dominant alleles of two genes affecting a character produce distinct phenotypes when they are alone,but when dominant alleles of both genes are present together , expression of dominant allele of one gene masks the expression of other and when both genes are present in recessive state , a different phenotype is produced Example Barley seed colour –Black/Yellow/White Parents BByy (Black) X bbYY (Yellow ) P gametes By bY F1 (Hybrid ) BbYy Black F1 gametes
BY By bY by BY BY BY By BY bY BY by BY By By BY By By bY By by By bY bY BY bY By bY bY by bY by by BY by By by bY by by Parents BByy (Black) X bbYY (Yellow ) P gametes By bY F1 (Hybrid ) BbYy Black F1 gametes BY – Black - 1,2,3,4,5,7,9,10,13=9 By –Black- 6,8,14=3 bY- Yellow-11,12,15=3 by-white-16=1 F2 Generation ---- 12 (Black) : 3 (Yellow ) : 1 (White ) When both B and Y are present both express but Black colour is so intensive that yellow colour produced by Y is not detected B gives Black colour,Y gives Yellow colour and b and y donot produce colour
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Interaction of genes for slide share

  • 1. INTERACTION OF GENES MODIFICATION OF DIHYBRID RATIO EXTENSION O MENDELIAN GENETICS By Dr Ichha Purak University Professor Department of Botany Ranchi Women’s College Ranchi University,Ranchi
  • 2. Mendel performed monohybrid and dihybrid crosses taking sweet pea as experimental plant In monohybrid cross only one trait and its two alternative forms were taken under consideration at a time . Two phenotypes ( Red and White flower colour) were obtained . There were only three possible genotype ( RR,Rr & rr) with Phenotypic Ratio 3:1 and Genotypic Ratio 1:2:1. RR(Homozygous ) and Rr ( Heterozygous ) are phenotypically alike and both are governed by dominant allele (R). In Dihybrid cross two traits at a time were considered ( Height of plant and colour of flower ) . Four phenotypes i.e. tall red(TR), Tall white(Tr), dwarf red(tR) and dwarf white (tr) were obtained by dihybrid cross. Phenotypic Ratio 9:3:3:1 was obtained Genotypic Ratio 1:2:2:4:1:2:1:2:1 was obtained Combinations 1,6,11 & 16 are only homozygous
  • 3. DIHYBRID CROSS AND ITS ABBREVIATED RATIO Parents TTRR ttrr Parental Gametes TR tr F 1 Zygote Tt Rr tall plants Red flowers F1 Gametes TR Tr tR Tr TR TR TR 1 Tr TR 2 tR TR 3 tr TR 4 Tr TR Tr 5 Tr Tr 6 tR Tr 7 tr Tr 8 tR TR tR 9 Tr tR 10 tR tR 11 tr tR 12 tr TR tr 13 Tr tr 14 tR tr 15 tr tr 16 F2 Generation 16 Possible Combinations (P) TR (Tall Red) : 1,2,3,4, 5,7, 9,10,13 = 9 ( R ) Tr (Tall White) : 6,8.14 = 3 ( R ) tR ( Dwarf Red) ; 11,12.15 =3 ( P ) tr ( Dwarf White ) : 16 =1
  • 4. By seeing Mendel’s di hybrid Ratio, it becomes clear that different characters are inherited independently . Because of that four combinations are produced in F2 generation Mendel concluded on the basis of results of monohybrid and di hybrid crosses performed on sweet pea that a) One trait (character) is controlled by a gene (factor) existing in two alternative forms (alleles) . The alleles are responsible for two alternative allelomorphs. b) That different traits assorted independently ( segregation of two genes is independent of each other)
  • 5. After Mendel’s rediscovery by Hugo Devries, Carl Correns and EV Tschermak, many geneticists repeated the same pattern of work using other organisms and different phenotypes but the results were not always according to Mendel’s laws of inheritance, fit with the principles as law of dominance and law of independent assortment. It became apparent that it is not necessary that a single character (trait ) is controlled by one gene only and it is also not true that a gene has only two alleles, may have many alleles and the phenotype may be result of interaction of two genes and their alleles or a character may be controlled by more than one gene and its alleles. In such a condition 2 or more genes may interact, the resulting phenotype may be result of interaction.
  • 6. Suppose Gene A is responsible for phenotype A (Red) and Gene B is responsible for phenotype B (white ) ,when both A and B are present together may give rise to a new phenotype C (Pink) , but these genes still obey Independent Assortment pattern. If A and b are two genes, both dominant over their respective recessive alleles a & b ,then interaction will depend on presence of both dominant alleles A & B, Absence of A or B or absence of Both A & B. Mendel was fortunate to get 3:1 and 9:3:3:1 ratio uniformly because a) The traits he considered were governed by only one pair of alleles of a gene. b) The genes and their alleles for different characters were located on different homologous pairs of chromosomes or even if on the same homologous pair the genes for different trait were present at a distance sufficient for cross over
  • 7. INTERACTION OF GENES Deviation of Di-hybrid Ratio
  • 8. Genes usually function or express themselves singly or individually. But, many cases are known where two genes of the same allelic pair or genes of two or more different allelic pairs influence one another. This is called gene interaction Non-allelic gene Interactions These are interactions between genes located on the same chromosome or on different but non-homologous chromosomes controlling a single phenotype to produce a different expression. Each interaction is typical in itself and ratios obtained are different from those of the Mendelian dihybrid ratios. Some of these interactions of genes are explained here which fall under this category and deviate from Mendel's ratios.
  • 9. The interaction of genes may be of following types 1) Two gene pairs affecting same character – 9:3:3:1 2) Epistasis, one gene hides effect of other a) Recessive Epistasis - 9:3:4 b) Dominant epistasis - 12:3:1 3) Complementary genes - 9:7 ( 2 genes responsible for production of a particular phenotype ) 4) Duplicate genes – 15:1 ( same effect given by either of two genes ) 5) Polymeric gene action - 9:6:1 6) Inhibitory gene action - 13 : 3
  • 10. Gene Interaction Inheritance Pattern &Example A-/B- 9/16 A-/bb 3/16 aa/B- 3/16 aabb 1/16 RATIO Additive gene effect Comb shape Chicken 9 Walnut 3 Rose 3 Pea 1 Single 9:3:3:1 W:R:P:S Recessive Epistasis Coat Colour Mouse 9 Agouti 3 Albino 3 Black 1 Albino 9:3:4 Ag:Bl:Al Dominant Epistasis Fruit colour squash 9 white 3 white 3 yellow 1 green 12:3:1 W:Y:G Complement ary genes Flower Colour Lathyrus 9 Purple 3 White 3 White 1 white 9:7 P:W Duplicate genes Fruit Shape capsella 9 3 3 1 15:1 Tri: Top Polymeric Gene action Fruit Shape Squash 9 Disc 3 Circular 3 circular 1 Long 9:6:1 D:C:L Inhibitory gene action Maize Aleurone colour 9 White 3 Red 3 white 1 White 13:3 W:R
  • 11. Two gene pairs affecting the same character : comb shape in chicken ( 9:3:3:1-for single character) Gene R- rose comb -Rp Gene P –Pea Comb-rP The dominant alleles of each of the two genes produce separate forms of phenotype when they are alone ( heterozygous) Both R and P when brought together form a new phenotype walnut Allele R dominant over r Allele P dominant over p rr and pp produce single comb ADDITIVE GENE EFFECT
  • 12. F M RP Rp rP rp RP RP RP 1 Rp RP 2 rP RP 3 rp RP 4 Rp RP Rp 5 Rp Rp 6 rP Rp 7 rp Rp 8 rP RP rP 9 Rp rP 10 rP rP 11 rp rP 12 rp RP rp 13 Rp rp 14 rP rp 15 rp rp 16 F1 gametes RP-Walnut 1,2,3,4.5,7,9,10,13= 9 Rp –Rose 6,8,14=3 rP- pea 11,12,15=3 rp- Single 16=1 so the ratio is 9:3:3:1 just like normal dihybid ratio for two traits but it is for single trait i.e. comb shape having 4 different forms Parents RRpp x rrPP (Rose) (Pea) P. gametes Rp rP F1 RrPp walnut GENETIC EXPRESSION F2 Generation 9:3:3:1
  • 13. A Cross Involving a Two-Gene Interaction : 9:3:3:1 ratio for single trait RRpp rrPP RrPp rrpp Inheritance of comb morphology in chicken described by William Bateson and Reginald Punnett in 1906 : First example of gene interaction Four different comb morphologies
  • 14. F2 Generation F2 generation consisted of chickens with four types of combs 9 walnut : 3 rose : 3 pea : 1 single Bateson and Punnett reasoned that comb morphology is determined by two different genes
  • 15. (Additive Gene Action) There are 4 phenotypes in F2,so are governed by more than one gene If it was governed by one gene and its two alleles F2 would have shown only 3 phenotypes in 1:2:1 ratio Where both A and B are present add colour so new phenotype brown is produced F2 phenotype classes • Dominance Relationships: – Tan is dominant to green – Gray is dominant to green – Brown is dominant to gray, green and tan. – Tan and Gray are incompletely dominant, giving rise to brown. • Genotypic classes: – Brown: A_B_ – Tan: A_bb – Gray: aaB_ – Green: aabb 2nd Example Lentil Seed colour 9:3:3:1
  • 16. EPISTASIS One gene hides the effect of other gene. It is different from Mendel’s Dominance which is meant for intragenic alleles (alleles of a gene) but here dominance works at intergenic level (alleles of different genes ). One gene masks the expression of another non-allelic gene. A gene which masks (hides) the action of another gene (non allelic) is termed as epistatic gene. The gene whose effects are masked is called hypostatic gene For example if two gene A and C with their alleles a and c take part in a cross then epistasis can be of following types 1) Recessive Epistasis -Recessive allele ( c ) of one gene may hide the effect of dominant allele (A ) of other gene 2) Dominant Epistasis Dominant allele ( A ) of one gene may hide the effect of dominant allele ( C ) of other gene.
  • 17. a) Recessive epistasis: Here the recessive allele masks the effect of dominant allele of other gene. In mice the wild body colour is known as agouti(greyish) and is controlled by a gene A which is hypostatic to recessive allele c. The dominant allele C in the presence of 'a' gives coloured mice. In the presence of dominant allele C, A gives rise to agouti. So, CCaa will be coloured and ccAA will be albino. When coloured mice (CCaa) are crossed with albino (ccAA), agouti mice (CcAa) appear in F1. cc masks the effect of AA and is therefore epistatic. Consequently, cc AA is albino. The ratio 9 : 3 : 3 : 1 is modified to 9 : 3 : 4. The combination ccaa is also albino due to the absence of both the dominant alleles.
  • 18. Recessive Epistasis Example : Coat colour of Mouse Coat colour is controlled by Gene A , Allele A is hypostatic to recessive allele (c ) The dominant allele C in absence of A gives coloured mice When both C and A are present colour is Agouti ( wild type most common ) due to banded hair : Near skin Grey yellow Black Two other colours are Albino and solid black AACC (agouti) x aacc (albino) AaCc (all agouti) AaCc x AaCc A-C- Agouti 9/16 A-cc Albino 3/16 aaC Black 3/16 aacc Albino 1/16 The c locus is epistatic to the A locus. 9 (Agouti ) :3 ( Black ) :4 ( Albino) F2 ratio Genotype Phenotype
  • 19. Parents CCaa X ccAA Black Albino p. gametes Ca cA F1 CcAa ------------------------------------Agouti F1 Gametes→ ↓ CA Ca cA ca CA CA CA 1 Agouti Ca CA 2 Agouti cA CA 3 Agouti ca CA 4 agouti Ca CA Ca 5 Agouti Ca Ca 6 Coloured cA Ca 7 Agouti ca Ca 8 coloured cA CA cA 9 Agouti Ca cA 10 Agouti cA cA 11 Albino ca cA 12 albino ca CA Ca 13 Agouti Ca Ca 14 Coloured cA Ca 15 albino Ca Ca 16 Albino CA –Agouti- 1,2,3,4,5,7,9,10,13=9 Ca- Black- 6,8,14=3 cA- Albino- 11,12,15=3 (c masks the effect of A ) ca- albino -16=1 So the ratio is 9:3:4 ( Agouti:Black:Albino ) F2 Generation Genetic Expression Cross between Black and Albino : Genetic Expression
  • 21. F2 Phenotypic ratio : 9 (agouti) : 3 (Coloured) : 4 (Albino)
  • 22. (b) Dominant epistasis: In summer squash or Cucurbita pepo, there are three types of fruit colour - yellow, green and white. White colour is dominant over other colours, while yellow is dominant over green. Gene for white colour (W) masks the effects of yellow colour gene (Y). So yellow colour is formed only when the dominant epistatic gene is represented by its recessive allele (w). When the hypostatic gene is also recessive (y), the colour of the fruit is green. White Fruit - WY , Wy Yellow Fruit - wY Green Fruit – wwyy A cross between a pure breeding white summer squash, (WWYY) with a pure breeding green summer squash, (wwyy) yields white fruits in the F1 generation. Upon selfing of F1 the F2 generation comes to have 12 white fruit : 3 yellow fruit : 1 green fruit.
  • 23. Parents WWYY X wwyy White Green P gametes WY wy F1 Generation Hybrid WwYy --------------------------------------------White F1 Gametes Female Male WY Wy wY wy WY WY WY 1 white Wy WY2 white wY WY 3 white wy WY 4 white Wy WY Wy 5 white Wy Wy 6 white wY Wy 7 white wy Wy 8 white wY WY wY9 white Wy wY 10 white wY wY 11 Yellow wy wY 12 Yellow wy WY Wy 13 white Wy Wy 14 white wY wy 15 Yellow wy wy 16 green WY-White- 1,2,3,4,5,7,9,10,13=9 Wy-White- 6,8,14=3 wY-yellow- 11,12,15=3 wy-Green-16=1 F2 generation 12 white : 3Yellow :1 green Genetic Expression
  • 24. F1- AaBb X AaBb 9 A B –white 3Aa-bb –Yellow 3 aa-B- White 1 aabb –Green A causes Yellow but in presence of B can not express ,produce white FRUIT COLOUR SQUASH : 12 : 3 : 1 13:3 12:3:1
  • 25. Two genes are responsible for a particular phenotype. Production of one phenotype requires dominant alleles of both the genes controlling the character. The complementary genes are two genes present on separate gene loci that interact together to produce dominant phenotypic character, neither of them if present alone, can expresses itself. It means that these genes are complementary to each other Example is flower colour of Lathyrus odoratus( Keshari/pea grass) .The colour of flower is either purple or white. Purple colour is produced only when dominant A is complemented by Dominant Allele B COMPLEMENTARY GENES
  • 26. Bateson and Punnet have demonstrated that in sweet pea (Lathyrus odoratus) purple colour of flowers develop as a result of interaction of two dominant genes C and P. In the absence of dominant gene C or P or both, the flowers are white. It is believed that gene C produces an enzyme that catalyzes the formation of necessary raw material for the synthesis of pigment anthocyanin and gene P produces an enzyme which transforms the raw material into the pigment. It means the pigment anthocyanin is the product of two biochemical reactions, the end product of one reaction forms the substrate for the other. Product of Gene C Product of Gene P Substrate A Substrate B Anthocyani (Purple) Therefore, if a plant has ccPP, ccPp, CCpp or Ccpp genotypes, it bears only white flowers. Purple flowers are formed in plants having genotype CCPP or CCPp or CcPP or CcPp. From checker board, it is clear that 9 : 7 ratio between purple and white is a modification of 9 : 3 : 3 : 1 ratio.
  • 27. Genetic Expression Parents : AAbb X aaBB White White P.gametes Ab aB F1 AaBb --------------------------------- Purple F1 Gametes Female→ AB Ab aB ab AB AB AB 1 AB Ab 2 AB aB 3 AB ab 4 Ab Ab AB 5 Ab Ab 6 Ab aB 7 Ab Ab 8 aB aB AB 9 aB Ab 10 aB aB 11 aB ab 12 ab ab AB 13 ab Ab 14 ab aB 15 ab ab 16 So the ratio becomes 9:7 ( Purple : White ) AB- Purple-1,2,3,4.5.7.9,10,13= 9 Where both (A ) and ( B ) are present flowers are coloured ( when A is complemented by B then only flower is coloured ) Where either (A ) or ( B ) is present colour is white Ab- White- 6,8,14 = 3 aB- White – 11,12,15 = 3 ab- White -16 = 1 Total white- 3+3+1=7 F2 Generation Male COMPLEMENTAY GENES KESHARI
  • 28. A Cross Producing a 9:7 ratio 9 C_P_ : 3 C_pp :3 ccP_ : 1 ccpp purple white
  • 29. Complementary gene action - interactions arise because the two genes encode proteins that participate in sequence in a biochemical pathway Enzyme C and enzyme P cooperate to make a product, therefore they complement one another Enzyme C Purple pigment Colourless intermediate Colourless precursor
  • 30. F2 phenotypic ratio - 9 (Purple) : 7 (White) Results of an experiment showing inheritance of flower colour in Lathyrus odoratus controlled by complementary genes
  • 32. Duplicate genes If the dominant alleles of two gene loci produce the same phenotype, whether inherit together or separately, the 9 : 3 : 3 : 1 ratio is modified into a 15 : 1 ratio. The capsules of shepherd's purse (Capsella) occur in two different shapes, i.e., triangular and top-shaped. When a plant with triangular capsule is crossed with one having top-shaped capsule, in F1 only triangular character appears. The F1 offspring by self crossing produced the F2 generation with the triangular and top-shaped capsules in the ratio of 15 : 1. Two independently segregating dominant genes (A and B) have been found to influence the shape of capsule in the same way. All genotypes having dominant alleles of both or either of these genes (A and B) would produce plants with triangular-shaped capsules. Only those with the genotype aabb would produce plants with top - shaped capsules.
  • 33. Same effect is given by either of two genes (A ) or (B) Example is Capsella bursa -pastoris (Shepherd’s Purse ) Fruit shape of two types a) Triangular (Heart shape ) capsule b) Top shaped (Narrow ) capsule Genetic Expression Parents AABB X aabb Triangular Top shaped P gametes AB ab F1 (Hybrid ) AaBb ------------- triangular F1 gametes AB Ab aB ab AB AB AB 1 AB Ab 2 AB aB 3 AB ab 4 Ab Ab AB 5 Ab Ab 6 Ab aB 7 Ab Ab 8 aB aB AB 9 aB Ab 10 aB aB 11 aB ab 12 ab ab AB 13 ab Ab 14 ab aB 15 ab ab 16 AB- 1,2,3,4,5, 7,9,10,13=9 (Triangular) Ab-6,8,14=3(Triangular ) aB-11,12,15 =3 (Tringular ) ab-16 =1 (top shaped ) So the ratio is 15:1 This is example of gene interaction,two genes involved in same pathway.It is based on the idea that some genes may be present more than once in the genome DUPLICATE GENES F2 generation 15 (Triangular): 1 (Top)
  • 34. A_ or B_ = heart shape aa and bb = narrow shape Fruit shape in Shepherd’s purse Duplicate Dominant Epistasis
  • 35. In a cross between two lines differing in fruit shape (Heart shape vs narrow ) F1 generation shows all heart shaped like intragenic dominance but F2 generation shows a ratio of 15:1 and not 3:1 as that of monohybrid cross. It means that trait fruit shape is controlled by two genes (A and B ) and their Alleles a and b. 15 :1 ratio is therefore modification of dihybrid ratio 9:3:3:1 in which 9,3 and 3 are grouped.The triangular shape results by the presence of at least one dominant allele of either gene. The two gene appear to be identical in function and is in contrast with complemenatary genes or 9:7 ratio where both dominant alleles are required for a phenotype and they complement each other. Duplicate genes provide alternative genetic determination of a specific phenotype
  • 36. Duplicate Gene Action Epistasis TV TV Tv Tv tV tV tv tv TTVV TTVv TtVV TtVv TTVv TTvv TtVv Ttvv TtVV TtVv ttVV ttVv TtVv Ttvv ttVv ttvv TTVV Triangular ttvv Ovate TtVv All triangular F1 (TtVv) x F1 (TtVv) x F1 generation 15:1 ratio results Shepherd’s Purse (Capsella ) Fruit Shape Triangular / Top (Ovate) shape
  • 37. F2 phenotypic ratio : 15 (triangular ) : 1 (Top shaped )
  • 38. P AABB aabb AB ab AaBb F1 AaBb X AaBb F2 15/16 Red : 1/16 white Whenever a dominant gene is present, the trait is expressed. One allele is sufficient to produce the pigment. 2nd Example : Petal colour in snapdragon Redundancy : Duplicate Genes
  • 39. Polymeric gene action 9:6:1 Two completely dominant genes controlling a character produce same phenotype, when their dominant alleles are alone, But when dominant alleles are together , the phenotypic effect is enhanced and become cumulative or additive effect Example -1 Awn length on Barley fruit AABB x aabb A /B- Median Awn A & B – long Awn ab- Awnless Example- 2 Fruit shape Squash (Disc/Circular/Long) 9:6:1 AaBb X AaBb 9 A B – Disc 3 A-bb – Circular 3 aa-B- Circular 1 aabb – Long
  • 40. Genetic Expression Parents AABB( Long awn ) X aabb(Awnless) P gametes AB ab F1 AaBb Long awn F1 gametes AB Ab aB ab AB AB AB 1 AB Ab 2 AB aB 3 AB ab 4 Ab Ab AB 5 Ab Ab 6 Ab aB 7 Ab Ab 8 aB aB AB 9 aB Ab 10 aB aB 11 aB ab 12 ab ab AB 13 ab Ab 14 ab aB 15 ab ab 16 AB – Long Awn -1,2,3,4,5,7,9,10,13=9 Ab- Medium Awn -6,8,14=3 aB- medium Awn-11,12,15=3 ab-Awnless-16=1 F2 Generation : Phenotypic ratio 9(Long) : 6(Medium) :1(awnless) Example awn length on Barley fruit AB- Median Awn A & B – long Awn ab- Awnless
  • 41. INHIBITORY GENE ACTION Inhibitory gene action Example Maize Aleurone colour 13:3 One dominant gene produces concerned phenotype and its recessive allele produces contrasting phenotype. The second gene (dominant) has no effect on concerned phenotype but stops expression of dominant allele of first gene, so when both dominant alleles are present ,phenotype as that of recessive homozygote is produced . Genetic Expression Parents RRII( White aleurone ) X rrii (White aleurone ) P gametes RI ri F1 hybrid ----------------RrIi ------------------------White F1 gametes
  • 42. Genetic Expression Parents RRII( White aleurone ) X rrii (White aleurone ) P gametes RI ri F1 hybrid ----------------RrIi ------------------------White F1 gametes RI Ri rI ri RI RI RI Ri RI rI RI ri RI Ri RI Ri Ri Ri Red rI Ri ri Ri Red rI RI rI Ri rI rI rI rirI ri RI ri Ri ri Red rI ri ri ri RI-White-1,2,3,4,5,7.9.10,13=9 Ri- Red- 6,8,14=3 rI- white -11,12,15=3 ri-white-16=1 So phenotypic Ratio becomes 13 (White ) : 3 (Red) Inhibitory gene action Example Maize Aleurone colour 13:3
  • 43. Second Example 13 :3 Inhibitory gene ( one gene inhibits expression of the other ) 13: 3 Example –Feather colour in Fowl ( White/Coloured ) Epistasis AaBb X AaBb 9 A B –white 3 A-bb –white 3 aa-B- coloured 1 aabb –white B is responsible for colour but in presence of A cannot express
  • 44. Masking gene action (12:3:1) Dominant alleles of two genes affecting a character produce distinct phenotypes when they are alone,but when dominant alleles of both genes are present together , expression of dominant allele of one gene masks the expression of other and when both genes are present in recessive state , a different phenotype is produced Example Barley seed colour –Black/Yellow/White Parents BByy (Black) X bbYY (Yellow ) P gametes By bY F1 (Hybrid ) BbYy Black F1 gametes
  • 45. BY By bY by BY BY BY By BY bY BY by BY By By BY By By bY By by By bY bY BY bY By bY bY by bY by by BY by By by bY by by Parents BByy (Black) X bbYY (Yellow ) P gametes By bY F1 (Hybrid ) BbYy Black F1 gametes BY – Black - 1,2,3,4,5,7,9,10,13=9 By –Black- 6,8,14=3 bY- Yellow-11,12,15=3 by-white-16=1 F2 Generation ---- 12 (Black) : 3 (Yellow ) : 1 (White ) When both B and Y are present both express but Black colour is so intensive that yellow colour produced by Y is not detected B gives Black colour,Y gives Yellow colour and b and y donot produce colour