Mendelian Genetics: Singapore Junior Biology Olympiad Training PDF

Dunman High School SJBO Mendelian Genetics...

Dunman High School SJBO Mendelian Genetics DUNMAN HIGH SCHOOL Singapore Junior Biology Olympiad Training MENDELIAN GENETICS 1. IMPORTANT DEFINITIONS.............................................................................................. 2 2. MENDEL’S LAWS............................................................................................................ 4 2.1 MENDEL'S FIRST LAW: LAW OF SEGREGATION..................................................... 4 2.2 MENDEL’S SECOND LAW: LAW OF INDEPENDENT ASSORTMENT....................... 5 3. MONOHYBRID INHERITANCE........................................................................................ 7 3.1 THE TEST CROSS...................................................................................................... 8 3.2 MODIFICATION OF THE CLASSICAL 3:1 RATIO....................................................... 9 3.2.1 Incomplete Dominance.......................................................................................... 9 3.2.2 Co-dominance.......................................................................................................11 3.2.3 LETHAL GENES...................................................................................................12 3.2.4 MULTIPLE ALLELES............................................................................................14 3.2.5 SEX-LINKED GENES...........................................................................................15 4. DIHYBRID INHERITANCE...............................................................................................21 4.1 CHROMOSOMAL BEHAVIOUR AND DIHYBRID INHERITANCE..............................23 4.2 DIHYBRID TEST CROSS...........................................................................................24 4.3 MODIFICATION OF THE CLASSICAL DIHYBRID RATIO..........................................25 4.3.1 Epistasis...............................................................................................................25 4.3.2 Linked genes.........................................................................................................29 5. CHI-SQUARED (c 2) TEST...............................................................................................34 1 Dunman High School SJBO Mendelian Genetics 1. IMPORTANT DEFINITIONS 1. Gene: It is a heritable sequence of nucleotides along a DNA molecule, which codes for a polypeptide, and hence may have a phenotypic effect / specific function. A gene can be regarded as basic unit of inheritance. 2. Locus: Within each species, each gene occupies a fixed position on a particular chromosome. This place on a chromosome is called a locus. 3. Allele: It is an alternative form or version of a gene that can be present on either one or both of a pair of homologous chromosomes. Alleles of a gene occupy the same locus on a pair of homologous chromosomes. Usually, they occur in pairs in a diploid cell although only one of the pair is represented in a gamete. There may also be cases where there are three or four alleles for one gene. Alleles of a gene impart a particular characteristic. For example, each human has a gene that controls eye color, but there are variations among these genes in accordance with the specific color the gene "codes" for. 4. Genotype: Genotype is the term for the genetic constitution of an organism that produces a phenotype. 5. Phenotype: This refers to the observable physical and chemical characteristics of an individual. The phenotype depends on the interaction between the genotype of an individual and the environment in which development occurs, e.g. axial (side) or terminal (vertical) position of flowers. 6. Gene pair: This refers to the two alleles of a particular gene present in a diploid cell. 7. Homozygous: If the alleles of a gene pair (at a given locus) are identical, e.g. TT or tt, the pair of alleles is homozygous. An individual with homozygous alleles is referred to as a homozygote. 8. Heterozygous: If the alleles of a gene pair (at a given locus) are different, e.g. Tt, the pair of alleles is heterozygous. An individual with heterozygous alleles is referred to as a heterozygote. 9. Dominant Allele: A dominant allele can express itself whether it occurs in the homozygous or heterozygous condition, e.g. TT or Tt. A capital letter is always used to represent a dominant allele. An individual homozygous for a dominant allele is described as homozygous dominant. 2 Dunman High School SJBO Mendelian Genetics 10. Recessive Allele: A recessive allele is one whose effect is expressed in the phenotype of a diploid individual only in the presence of another identical allele, e.g. tt. A small letter is always used to represent a recessive allele. An individual homozygous for a recessive allele is said to be homozygous recessive. Note: It is incorrect to describe an allele as recessive without reference to the allele it is recessive to. It should be, for example, the allele for white eye is recessive to the allele for red eye. 11. Cross: This is the deliberate mating of two individuals in genetic analysis. 12. Selfing: Selfing refers to the fertilisation of a female gamete by a male gamete from the same individual. Selfing over many generations can produce a pure line. Members of a pure line are homozygous at the gene locus in question and are said to breed true or are true-breeding or pure breeding. They are either homozygous dominant or homozygous recessive but never heterozygous. Used for plants, but not animals. 13. Sibling Mating: This refers to mating between siblings. As with selfing, sibling mating may be dangerous due to inbreeding. Offspring have a high chance of inheriting any deleterious alleles in the family gene pool. 14. Monohybrid Monohybrid inheritance is the inheritance of a single characteristic 15. Dihybrid Dihybrid inheritance is the inheritance of two characteristics 16. Test Cross: In a test cross, the individual of unknown genotype is crossed with one who is homozygous recessive e.g. tt, at the locus in question. 17. Backcross This refers to the crossing of a heterozygous offspring with its parent. (Parent can be homozygous recessive e.g. Tt x tt or homozygous dominant e.g. Tt x TT) 18. F1 generation F1 stands for Filial 1, the first filial generation seeds/plants or animal offspring resulting from a cross mating of distinctly different parental types. 19. F2 generation F2 stands for Filial 2, the second filial generation consists of the progeny (seeds/plants or animal offspring) resulting from the selfing / sibling mating of the F1 generation. 3 Dunman High School SJBO Mendelian Genetics 2. MENDEL’S LAWS 2.1 MENDEL'S FIRST LAW: LAW OF SEGREGATION Genes occur in pairs and that each gamete receives only one allele of each pair (in other words, the alleles segregate during meiosis). A parent who has two alleles (a copy of the allele A1 and a copy of the allele A2) passes on allele A1 to 50% of the gametes and allele A2 to the other 50% of the gametes. The two alleles of the organism are separated into different gametes, thus ensuring variation. For each trait, an organism inherits two alleles, one from each parent. This means that when somatic cells are produced from the fusion of two gametes, one allele comes from the mother, one from the father. These alleles may be the same (true-breeding organisms, e.g. TT and tt), or different (hybrids, e.g. Tt). This law is related to the separation of homologous chromosomes during anaphase I of meiosis. Anaphase I Fig. Separation of homologous chromosomes in anaphase I of meiosis results in segregation of alleles. 4 Dunman High School SJBO Mendelian Genetics 2.2 MENDEL’S SECOND LAW: LAW OF INDEPENDENT ASSORTMENT The Law of segregation was based on Mendel’s work involving monohybrid crosses. The concept of allele segregation refers to a single genetic locus only. The Law of independent assortment was based on work involving dihybrid crosses and refers to 2 or more genetic loci. Mendel's law of independent assortment states that two different genes on different pairs of homologous chromosome segregate independently: any one of a pair of characteristics may combine with either one of another pair. The Law of independent assortment applies only to genes that are unlinked. Two genes are said to be unlinked when they are situated on different pairs of homologous chromosomes. Two genes are said to be linked when they are on the same pair of homologous chromosomes. (refer to figure below) Genes on the same pair of homologous chromosomes are affected by linkage and crossing over. Fig. (a) The genes Aa and Bb are on the same pair of homologous chromosomes, i.e. they are linked. (b) The genes Aa and Bb are on different pairs of homologous chromosomes, i.e. they are unlinked This law is related to the independent assortment and random segregation of homologous chromosomes during metaphase I and anaphase I, of meiosis respectively. 5 Dunman High School SJBO Mendelian Genetics With reference to the figure above: The orientation of homologous chromosomes shown in (a) is equally as likely as that in (b) because the arrangement of one homologous pair is independent of the other. As a result of independent assortment (Metaphase I) and random segregation (Anaphase I) of homologous chromosomes, the genotype AaBb gives rise to four possible combinations of alleles (AB, ab, Ab and aB), which occur with equal frequency in the gametes. With reference to Figure on left: The diploid parent has two genes where the two alleles of each gene separate during gamete formation. As the two genes are unlinked, there are four possible combinations of gametes (SY, sy, Sy, and sY) where all of which will occur in equal proportions. In other words, there is a 1:1:1:1 ratio of the types of gametes that are formed. 6 Dunman High School SJBO Mendelian Genetics 3. MONOHYBRID INHERITANCE A monohybrid cross is a breeding experiment between parental generation (P generation) organisms that differ in one trait. Mendel conducted crosses between pure-breeding (i.e. homozygous for a particular trait) plants which had different characters. An example of Genetic cross in Garden pea (Pisum sativum) where the allele for tallness is dominant to allele for shortness is shown below. Symbols Let T represents the allele for tall plant Let t represents the allele for short plant Parental phenotypes : Tall X Short Parental genotypes : TT X tt Gametes formed : T X t F1 Genotype : Tt F1 Phenotype : all tall Selfing of F1 Selfing F1 generation : Tt X Tt Gametes formed : T t X T t Using a Punnett square to show the fusion of gametes produced by F1 generation T t T TT Tt t Tt tt F2 genotype : TT Tt Tt tt F2 phenotype : tall short F2 phenotypic ratio: 3 : 1 Results: In F2 generation, ratio of tall plants to short plants is 3:1. This 3:1 ratio is the monohybrid ratio. 7 Dunman High School SJBO Mendelian Genetics 3.1 THE TEST CROSS An individual that shows the effect of a dominant allele can have two possible genotypes. For example, a tall pea plant could either be homozygous dominant (TT) or heterozygous (Tt) for the gene. The phenotype is identical in both cases. A test cross with a homozygous recessive individual can be carried out to determine the genotype accurately. Why is a homozygous recessive individual used? A homozygous recessive individual is used because all its gametes must contain the recessive allele of the gene, which will have no effect on the phenotype of the offspring. One can then deduce the unknown genotype (either TT or Tt) from the phenotypes of the offspring. A. A test cross on a tall pea plant whose genotype is TT will produce only tall offsprings. Parental genotype : TT X tt Gametes formed : T X t F1 Genotype : Tt F1 Phenotype : all tall B. A test cross on a tall pea plant whose genotype is Tt will produce tall and short offsprings in the ratio of 1:1. Parental genotype : Tt X tt Gametes formed : T t X t F1 Genotype : Tt tt F1 Phenotype : tall short F1 Phenotypic ratio : 1 : 1 8 Dunman High School SJBO Mendelian Genetics When is a test cross conducted? When you want to determine the genotype of an organism that exhibits a dominant trait. e.g. purple flowers are the dominate trait. Plant with unknown genotype has purple flowers. Possible genotypes: PP (homozygous dominant) or Pp (heterozygous). Cross the organism in doubt (unknown genotype) with an individual expressing the homozygous recessive trait (pp), e.g. white-flowered parent that is homozygous, and determine the ratio of the offspring. If the offspring are all purple flowers, then the unknown organism is homozygous dominant (PP), if half of the offspring are purple flowers and the other half white flowers, then the unknown organism is a heterozygote (Pp). 3.2 MODIFICATION OF THE CLASSICAL 3:1 RATIO The breeding of many other organisms produced results different from that of Mendel’s experiments on monohybrid inheritance. This is due to the interaction of genes. Modifications of the classical 3:1 ratio for the F2 generation for monohybrid inheritance may be due to the following reasons: (i) Incomplete Dominance (ii) Codominance (iii) Lethal Genes (iv) Multiple Alleles (v) Sex-linked genes 3.2.1 Incomplete Dominance In incomplete dominance and co-dominance, the locus controlling a particular characteristic also has two alleles. However, there are three variations (phenotypes) for the characteristic. Incomplete dominance is the situation in which one allele for a specific trait is not completely dominant over the other allele, resulting in the heterozygote exhibiting a combined phenotype which is an intermediate between the two homozygous forms. The alleles are described as semi-dominant to each other. An example of incomplete dominance occurs in snapdragon (Antirrhinum). Flower colour in snapdragon is controlled by two alleles of a flower-colour gene. Individuals that are homozygous for the red-flower allele have red flowers while those that are homozygous for the white-flower allele have white flowers. Heterozygotes have pink flowers; a flower colour which is intermediate between red and white. 9 Dunman High School SJBO Mendelian Genetics The letter “C” with a superscript represents an allele for flower colour. Let CR represents the allele for red flower. CW represents the allele for white flower. Table on flower colour in snapdragon Genotypes Phenotypes CRCR Red flower CRCW Pink flower CWCW White flower A cross is made between two pure-breeding strains of snapdragon; one red and one white. All individuals of the F1 generation have pink flowers. Selfing the F1 generation gives a ratio of 1 red : 2 pink : 1 white in the F2 generation. Parental phenotype : Red X White flower flower Parental genotype : CRCR X CWCW Gametes formed : CR X CW F1 genotype and phenotype : CRCW (All have pink flowers.) Selfing F1 generation : CRCW X CRCW Gametes formed : CR CW X CR CW Punnett square to show the fusion of gametes produced by the F1 generation: CR CW CR CRCR CRCW CW CRCW CWCW F2 genotype : CRCR 2 CRCW CWCW F2 phenotype : red flower pink flower white flower F2 phenotypic ratio : 1 2 1 10 Dunman High School SJBO Mendelian Genetics 3.2.2 Co-dominance Co-dominance is the situation in which both alleles are equally expressed in the phenotype of the heterozygote. An example of co-dominance is provided by flower colour in some plant species, which is controlled by a single gene (Table below). Since neither allele (for both red flowers and white flowers) is recessive, the capital- and small-letter system of representing dominant and recessive alleles is inappropriate. Instead, capital letters (in superscript) are used for both alleles. Use "F" for the flower color allele. Let FR represents the allele for red flowers. FW represents the allele for white flowers. Table on flower colour in some plant species Genotypes Phenotypes FRFR Red flowers FWFW White flowers FRFW Red & white spotted flowers Genetic basis for co-dominance Parental phenotypes : Red flowers x White flowers Parental genotypes : FRFR FWFW Gametes formed : FR FW F1 Genotype : FRFW F1 Phenotype : All red & white spotted flowers 11 Dunman High School SJBO Mendelian Genetics 3.2.3 LETHAL GENES Lethal Gene - a gene that leads to the death of an individual; these can be either dominant or recessive in nature. Modification of the 3:1 ratio occurs when the bearer of certain genotypes is killed by drastic phenotypic effects. Most lethal genes are recessive. Example: Chlorophyll in maize plant Let: G = allele for the ability to produce chlorophyll (dominant) g = allele for the inability to produce chlorophyll (recessive) Parental phenotype Gg Gg Parents (P) Parental genotype (Green) (Green) Gametes formed by G g G g Meiosis F1 genotype F GG Gg Gg gg 1 Albino - dies Offspring that are homozygous for the recessive allele gg are albino and die when the seedling has exhausted the food reserves. Phenotypic ratio was initially a 3 green : 1 albino but becomes a 3 green : 0 albino within a fortnight. Such lethal genes are said to produce their effects post-embryonically. Dominant lethal genes are also known. For the genotype A_ (i.e. AA or Aa) to be lethal and yet exist, they must produce their effects after reproduction has occurred if the gene is to be transmitted to later generations. E.g. Huntington’s chorea, a rare progressive degenerative disease of the human nervous system that usually begins to show after the age of about forty. Most lethal genes prevent the embryo of the bearer from developing or initially permit it to develop prior to abortion. E.g. Haemophilic genes of the human female, yellow fur in mice. 12 Dunman High School SJBO Mendelian Genetics Coat Color in Mice In 1904, a cross was made between a yellow-coated mouse and a mouse with a gray coat. The gray- coated mouse was extensively inbred and therefore was considered to be pure bred. Using the following gene symbols: Let y represents the allele for gray coat Let Y represents the allele for yellow coat The genotype of the gray mouse must be yy as yellow coat is dominant to gray coat. If the parental mice were homozygous dominant or YY, we would not see any gray mice from the cross. Therefore the genotype of the yellow parental mouse must be heterozygous or Yy. A cross made between two yellow F1 mice (Yy x Yy) should give a ratio of 3 yellow : 1 gray. Expected Punnett Square Female Gametes Y y Male Y YY Yy Gametes (yellow) (yellow) y Yy yy (yellow) (gray) However, a ratio of 2 yellow to 1 gray mouse was observed. This is because the YY genotype is lethal. The 2:1 ratio is the typical ratio for a lethal gene 13 Dunman High School SJBO Mendelian Genetics 3.2.4 MULTIPLE ALLELES If a gene controlling a characteristic has three or more alleles, the alleles are called multiple alleles. In any one individual, at most two alleles can be present. An example of multiple alleles is provided by the alleles of the gene controlling the ABO blood group system in humans. Genetic basis for human ABO blood group system The gene for the ABO blood group system is conventionally represented by the symbol I. The gene has three alleles. IA represents the allele for the production of type A antigen. IB represents the allele for the production of type B antigen. Io represents the allele that produces neither antigen. The alleles IA and IB show equal dominance with respect to each other, i.e. they are co- dominant, but both are dominant to Io. Only two of the three alleles can be present in an individual. The table below shows the possible genotypes one can have. For blood group AB, both type A and B antigens are present on the red blood cell membrane. Table on ABO blood group system in humans Genotypes Phenotypes IAIA Blood group A IAIo Blood group A I BI B Blood group B I I B o Blood group B IAIB Blood group AB IoIo Blood group O Other examples of multiple alleles 1. The colours of the rabbit fur (agouti, chinchilla, Himalayan and albino) are controlled by four alleles (C+, CCh, Ch, C) in the same locus. 2. The flower colours of the snapdragon (from red to ivory) are controlled by 9 alleles. 14 Dunman High School SJBO Mendelian Genetics 3.2.5 SEX-LINKED GENES Human cells have 23 pairs of chromosomes; the first 22 pairs are known as autosomes or autosomal chromosomes and the 23rd pair as sex chromosomes. X,Y system is used for sex determination by many animal and plant species Sex determination in some organisms: Organism Male Female 1. Drosophila XY XX (same as mammals) 2. Grasshoppers, cockroaches XO XX 3. Birds (including poultry) ZZ ZW Sex determination in human: Females have 2 identical sex chromosomes – XX (homogametic) – all eggs carry the X chromosome. The X chromosome is large and it encodes many genes. In female mammals, one of the two X chromosomes is effectively ‘shut down’ very early in embryonic development – appears as a tightly coiled darkly staining body attached to the inside of the nuclear membrane – Barr body (inactive X chromosome in female somatic cell). Males have 1 X and 1 Y chromosome – XY (heterogametic) – 50% of sperms carry X chromosome and 50% carry Y chromosome. The Y chromosome is small with few genes (not homologous to X in the traditional sense but has the same pairing region for synapsis). Function of Y chromosome varies with different organism. In human, it controls the differentiation of testes from undifferentiated gonads of developing embryo. The Y chromosome carries very few genes – genetically inert. Y chromosome carries some genes that X chromosome does not (and vice versa). These unique genes that are carried on the sex chromosomes are called sex-linked genes. Since the X chromosome is typically longer, it bears a lot of genes not found on the Y chromosome, and thus most sex-linked genes are named X-linked genes. Examples of traits controlled by X chromosome-linked genes are hemophilia and Red- green color blindness. 15 Dunman High School SJBO Mendelian Genetics EXAMPLE 1: HEMOPHILIA IN HUMANS – sex-linked, recessive condition. Hemophilia is a hereditary disease where there is a reduced ability of blood to clot, due to deficiency of one of the blood clotting factors. This results in continuous bleeding (haemorrhage) from even a small cut. Haemophilia is caused by a recessive sex-linked gene on the X chromosome. Males are hemizygous – receiving their only X chromosome from their mother. Females are homozygous – inheriting X chromosomes from both parents. If a male has a mutant X and a normal Y chromosome, he will be affected by the X-linked disease. If a female has a defective gene on one of her two X chromosomes, she will be protected from its effects by the normal gene on her second X chromosome. Genetic basis for hemophilia Genotypes Phenotypes XH XH Normal female XHXh Normal female (carrier) XhXh Hemophilia female XHY Normal male XhY Hemophilia male Let H represents the normal allele for blood clotting (dominant) h represents the allele for haemophilia (recessive) X represents female chromosome Y represents male chromosome Scenario 1: Unaffected mother and affected father x x F1 genotype F1 phenotype A son, whose mother has two normal alleles, will not be affected by hemophilia even if the father has the defective gene. A daughter of the same parents will be a heterozygous carrier. 16 Dunman High School SJBO Mendelian Genetics Scenario 2: Carrier mother and unaffected father x x F1 genotype F1 phenotype A heterozygous carrier mother and an unaffected father pass the gene for hemophilia on to half of their children. Half the daughters will be carriers and half the sons will be hemophilic. The rest of the siblings will be unaffected. Daughters can only be carriers (normal gene on the second X chromosome masks the defect) as long as one parent is genotypically normal. If a son receives the defective gene from his mother, he will be hemophilic because the Y chromosome cannot counteract the defective gene located on his X chromosome. Scenario 3: Affected mother and unaffected father 17 Dunman High School SJBO Mendelian Genetics Why is hemophilia more common in males than in females? Males only need one copy of the defective allele to suffer from hemophilia whereas female require two copies of the defective allele to be a sufferer. EXAMPLE 2: RED-GREEN COLOUR BLINDNESS Red-green colour blindness is the inability to distinguish between red and green. Genetic basis for colour blindness The allele for normal vision is dominant to that for colour blindness. Let C represent the normal allele for normal vision (dominant) c represent the allele for colour blindness (recessive) X represents female chromosome Y represents male chromosome Parental phenotypes : normal female x colour blind male Parental genotypes : XCXC x XcY Gametes formed : XC x Xc Y F1 Genotypes : XCXc X CY F1 Phenotypes : normal carrier normal male female Crossing F1 generation : XCXc x X CY Gametes formed : XC Xc x XC Y Punnett square to show the fusion of gametes produced by the F1 generation: XC Y XC XCXC X CY Xc XCXc XcY F2 Genotypes : XCXC XCXc X CY XcY F2 Phenotypes : normal normal normal colour female carrier male blind female male 18 Dunman High School SJBO Mendelian Genetics The recessive allele causing colour blindness is passed from one sex to the other at each generation. The father passes it to his daughters, who become carriers. The daughters may in turn pass it to their sons, who will be colour blind. As the male is XY, his Y chromosome is inherited from his father, so the X chromosome with the colour blind allele must be inherited from the mother. Colour blind females can only arise from a cross between a carrier female and a colour blind male. Red-green colour blindness is more common in males than in females. Males always inherit the colour blind allele from their mothers. Males cannot pass on colour blindness to their sons since the Y-allele does not have any of the colour blindness alleles. SUMMARY FOR SEX-LINKED GENES Sex-linked characteristics can be detected by their unique pattern of inheritance. - The male offspring usually obtain these features from their mothers. - Daughters get their sex-linked features from both parents. - Usually only one of the genders is more affected than the other. - If the gene is Y-linked, only the gender that receives the Y chromosome will inherit the gene. 19 Dunman High School SJBO Mendelian Genetics Try this: In the following human pedigree, circles indicate females, square males. Filled-in circles and squares indicated red-green colour-blind individuals; open circles and squares indicate normal vision. Table on Red-green colour blindness in humans Genotypes Phenotypes XCXC Normal female XCXc Normal carrier female XcXc Colour blind female X CY Normal male XcY Colour blind male State, as far as possible, the genotype of each individual in the pedigree. 20 Dunman High School SJBO Mendelian Genetics 4. DIHYBRID INHERITANCE A dihybrid cross is one that is carried out between parents that differ in two characteristics (traits), which are controlled by genes at two different loci. Assuming the genes are unlinked (on different pairs of homologous chromosomes), the dihybrid ratio is 9 : 3 : 3 : 1. Method of Mendel’s experiment Pure-bred pea plants grown from round seeds with yellow cotyledons were crossed with pure-bred pea plants grown from wrinkled seeds with green cotyledons. The above cross gave rise to the F1 generation. The F1 plants were then selfed to produce the F2 generation. Results of Mendel’s experiment All individuals of the F1 generation produced round seeds with yellow cotyledons. The F2 generation showed the following phenotypes: 9 produced round seeds with yellow cotyledons (parental phenotype) 3 produced round seeds with green cotyledons (non-parental phenotype) 3 produced wrinkled seeds with yellow cotyledons (non-parental phenotype) 1 produced wrinkled seeds with green cotyledons (parental phenotype) Genetic Basis of Mendel’s experiment The phenotypes apparent in the F1 generation are dominant while the other phenotypes, which disappear in the F1 generation but reappear in the F2 generation, are recessive. Therefore, the alleles for round seed and yellow cotyledon are dominant to those for wrinkled seed and green cotyledon, respectively. Let R represent the allele for round seed r represent the allele for wrinkled seed Y represent the allele for yellow cotyledon y represent the allele for green cotyledon 21 Dunman High School SJBO Mendelian Genetics round seed, wrinkled seed, x Parental phenotype : : yellow cotyledon green cotyledon Parental genotype : : RRYY x rryy : RY x ry Gametes formed : : RrYy F1 genotype and phenotype : (All round seed and yellow cotyledon) : F1 x F1 : RrYy x RrYy Gametes formed : RY Ry rY ry RY Ry rY ry x : F2 genotype : RY Ry rY ry RY RRYY RRYy RrYY RrYy Ry RRYy RRyy RrYy Rryy rY RrYY RrYy rrYY rrYy ry RrYy Rryy rrYy rryy 1 RRYY 2 Rryy 1 rrYY 1 rryy 2 RRYy 1 RRyy 2 rrYy 2 RrYY 4 RrYy F2 phenotype : round seed, round seed, wrinkled seed, wrinkled seed, yellow cotyledon green cotyledon yellow cotyledon green cotyledon F2 phenotypic ratio : 9 : 3 : 3 : 1 22 Dunman High School SJBO Mendelian Genetics 4.1 CHROMOSOMAL BEHAVIOUR AND DIHYBRID INHERITANCE The alleles which determine the two pairs of contrasting characteristics are located on different pairs of homologous autosomes. Because the chromosomes of one pair separate independently of the other pair, the alleles segregate independently. 23 Dunman High School SJBO Mendelian Genetics 4.2 DIHYBRID TEST CROSS Pea plants with genotypes RRYY, RrYY, RRYy and RrYy (sometimes written as R_Y_) will all show the effect of dominant alleles. In order to differentiate RrYy from the other genotypes, RRYY, RrYY and RRYy, a test cross can be carried out. This involves crossing the plant with unknown genotype (dominant phenotype) with a homozygous recessive for both traits, namely a pea plant with wrinkled seed and green cotyledon (rryy). If the genotype of the unknown was RrYy, the test cross will produced offsprings of the following phenotypes in the ratio of 1 : 1 : 1 : 1. Phenotype of offsprings: Round seed with yellow cotyledon Round seed with green cotyledon Wrinkled seed with yellow cotyledon Wrinkled seed with green cotyledon Test cross on F1 generation Parental phenotype : round seed, x wrinkled seed, yellow cotyledon green cotyledon Parental genotype : RrYy x rryy Gametes formed : RY Ry rY ry ry x F1 genotype : RrYy Rryy rrYy rryy F1 phenotype : round, round, wrinkled, wrinkled, yellow green yellow green F1 phenotypic ratio : 1 : 1 : 1 : 1 Table of summary on phenotypic ratios on monohybrid and dihybrid crosses Phenotypic ratios in Types of Inheritance F2 Test cross on heterozygote F1 generation generation in F1 generation Monohybrid All with same phenotype as 3:1 1:1 cross parent with dominant trait All with same phenotype as Dihybrid cross 9:3:3:1 1:1:1:1 parent with dominant traits 24 Dunman High School SJBO Mendelian Genetics 4.3 MODIFICATION OF THE CLASSICAL DIHYBRID RATIO 4.3.1 Epistasis Different genes can interact to control the phenotypic expression of a single trait. Many phenotypes are controlled by more than one gene and hence depend upon the combined effect of a large number of other genes The expression of a gene can be modified by alleles at one or more other gene locus because of interactions between their products at the biochemical or cellular level. This can result in the production of an altered phenotypic ratio, such as the case of epistasis. Epistasis is defined as an interaction between two different genes such that one gene (epistatic gene – sometimes also called ‘inhibiting genes’) interferes with or even inhibits / suppresses the phenotypic expression of the other gene (hypostatic gene). Example 1 – Hair colour in Labrador retrievers (recessive epistasis) In recessive epistasis, the presence of two recessive copies of an allele of the epistatic gene at a locus will inhibit the expression of the hypostatic gene. Black fur Brown fur Yellow fur The different colours arise from variations in the amount and distribution of melanin. One gene is involved in melanin production. Allele B (black) of this gene is dominant to allele b (brown). A second gene influences the deposition of melanin pigment in individual hairs. Allele E of this gene promotes melanin deposition, but if the dog has two recessive alleles (ee), deposition is reduced and fur will be yellow. Thus, _ _ e e will have yellow fur. Hence the gene for pigment deposition locus alters the phenotypic expression of the gene at the black/brown locus. The gene for pigment deposition is epistatic to the gene that codes of black/brown pigment. 25 Dunman High School SJBO Mendelian Genetics Symbols: Let B represent the dominant allele for black fur b represent the recessive allele for brown fur E represent the dominant allele for pigment deposition e represent the recessive allele for blocked/reduced pigment deposition Parental phenotype: Black fur X Yellow fur Parental genotype: BBEE X bbee Gametes formed: BE X be F1 genotype: BbEe F1 phenotype: All black fur F1 x F1 : BbEe X BbEe Gametes formed: BE Be bE be X BE Be bE be Punnett square (to show the fusion of gametes produced by the F1 generation): BE Be bE be BE BBEE BBEe BbEE BbEe Be BBEe BBee BbEe Bbee bE BbEE BbEe bbEE bbEe be BbEe Bbee bbEe bbee F2 genotype: 1 BBEE 1 bbEE 1 BBee 2 BBEe 2 bbEe 2 Bbee 2 BbEE 1 bbee 4 BbEe F2 phenotype: Black fur Brown fur Yellow fur F2 phenotypic ratio: 9 : 3 : 4 The typical phenotypic ratio 9 : 3 : 3 : 1 becomes modified into 9 : 3 : 4 ratio. 26 Dunman High School SJBO Mendelian Genetics Example 2 – Coat colour in cats (dominant epistasis) In dominant epistasis, the presence of one dominant copy of an allele at a locus is sufficient to inhibit the expression of the hypostatic gene. When pure-bred brown cats are mated with pure-bred white cats, all the F1 offspring are white. Sibling mating of F1 generation produced 118 white, 32 black and 10 brown kittens. The allele B for black is dominant to the allele b for brown. The allele, W is epistatic to the alleles of the colour gene (B and b). Parental phenotype : brown X white Parental genotype : wwbb X WWBB Gametes formed : wb X WB F1 Genotype and : WwBb phenotype White F1 x F1 : WwBb X WwBb Gametes formed : WB Wb wB wb WB Wb wB wb Punnett square: WB Wb wB wb WB WWBB WWBb WwBB WwBb Wb WWBb WWbb WwBb Wwbb wB WwBB WwBb wwBB wwBb wb WwBb Wwbb wwBb wwbb F2 genotype: 1WWBB, 2WWBb, 1WWbb, 1wwBB, 1wwbb 2WwBB, 4WwBb, 2Wwbb 2wwBb F2 phenotype: white black brown F2 phenotypic ratio: 12 : 3 : 1 Presence of one copy of the allele W results in white coat colour regardless the genotype of gene B. Hence this results in a ratio of 12 white (W_ _ _) : 3 black (wwB_): 1 brown (wwbb) 27 Dunman High School SJBO Mendelian Genetics Example 3 – Feather colour in chicken (dominant epistasis) The allele, I, that is epistatic to the colour gene C. Individuals carrying the dominant allele I will have white plumage even if they are carrying the dominant allele C for colour. For example, an individual with the genotype IiCC will be white. Note that individuals which are homozygous recessive for the colour gene (cc) will also be white. Hence, in this case, the typical phenotypic ratio 9 : 3 : 3 : 1 becomes modified into 13 : 3 ratio. Table of summary on some modified phenotypic ratios produced by gene interactions Normal dihybrid cross 9 3 3 1 9:3:3:1 AB Ab aB ab Recessive epistasis 9 3 4 9:3:4 of bb acting on A and a alleles AB a b Dominant epistasis 12 3 1 12:3:1 of A acting on B and b alleles A aB ab 28 Dunman High School SJBO Mendelian Genetics 4.3.2 Linked genes Alleles of different genes only assort independently if they are located on different chromosomes [UNLINKED GENES] separated by a great enough distance on the same chromosome that recombination occurs at least half of the time Alleles of different genes does not assort independently if they are located on the same chromosome [LINKED GENES] Linked genes do not show the 9:3:3:1 ratio for dihybrid inheritance, instead a variety of ratios are produced. A1. DIHYBRID INHERITANCE OF LINKED LOCI WITHOUT CROSSING OVER The two genes are located so close to each other physically on the same chromosome that they always nearly end up inside the same gamete (no crossing over). This is known as complete linkage. 1 Experiments and Results (a) When pure-breeding Drosophila with long wings and broad abdomen are crossed with pure-breeding Drosophila with vestigial wings and narrow abdomen, all the F1 offspring have long wings and broad abdomen. (b) When the F1 flies are mated, the F2 generation fails to give the 9:3:3:1 ratio. Instead, about 75% of the F2 offspring have long wings and broad abdomen, and the remaining 25% have vestigial wings and narrow abdomen. Note that these are the parental phenotypes. 2 Genetic basis for F2 generation (a) The explanation for this is that the genes determining the length of wings and the width of abdomen are located very near each other on the same chromosome. In other words, the two genes are tightly linked. They are usually inherited as a single unit or ‘linkage group’. (b) When two genes are tightly linked, a chiasma will not form between them and they will not be separated by crossing over. The two genes tend to be inherited together. (c) The closer the two alleles are together, the lower the chance that a cross-over event may occur between them, possibly separating the alleles. (d) Two tightly linked genes can be thought of as only one gene and not two, and so the F2 generation gives a phenotypic ratio of 3:1. Note that this ratio is the same as that for monohybrid inheritance. 29 Dunman High School SJBO Mendelian Genetics Let L represents the dominant allele for long wing l represents the recessive allele for vestigial wing B represents the dominant allele for broad abdomen b represents the recessive allele for narrow abdomen The alleles for long wing and broad abdomen are dominant to those for vestigial wing and narrow abdomen, respectively. Parental : Long wing, x Vestigial wing, phenotypes broad abdomen narrow abdomen Parental : _L B_ x _l b_ genotypes _______ ______ L B l b Gametes : formed _L B_ x _l b_ F1 genotype : _L B_ and phenotype ______ l b All have long-wings and broad abdomen. F1 x F1 : _L B_ x _L B_ _______ _______ l b l b Gametes : L B l b x L B l b formed F2 genotype : _L B_ _L B_ _l b_ _l b_ ______ ______ ______ ______ L B l b L B l b F2 phenotype : Long wings, Long wings, Long wings, Vestigial broad broad broad wings, abdomen abdomen abdomen narrow abdomen F2 phenotypic : ratio 3 : 1 30 Dunman High School SJBO Mendelian Genetics Genetic basis for test cross on F1 generation A test cross carried out on the F1 generation gives a phenotypic ratio of 1:1. Note that this ratio is the same as that for monohybrid inheritance. Test cross on : _L B_ x _l b_ F1 generation ______ ______ l b l b Gametes : _L B_ _l b_ x _l b_ formed Genotypes of : _L B_ _l b_ offspring ______ ______ l b l b Phenotypes of : Long wings, broad abdomen Vestigial wings, narrow abdomen offspring Phenotypic : 1 : 1 ratio A2. DIHYBRID INHERITANCE OF LINKED LOCI WITH CROSSING OVER The probability that crossing over between sister chromatids (during prophase I of meiosis) disrupts linkage between 2 genes is proportional to the distance that separates the 2 loci. In other words, linkage is more vulnerable to crossover when the distance between the 2 genes is greater. This is known as partial linkage. 31 Dunman High School SJBO Mendelian Genetics 1 Experiments and Results (a) In fruit fly Drosophila, the genes for body colour and wing length are linked. When pure-breeding Drosophila with grey body and normal wings were crossed with pure-breeding Drosophila with black body and vestigial wings, all the F1 offspring showed grey body and normal wings (b) Sibling mating of F1 offspring failed to give a phenotypic ratio of 9:3:3:1 or 3:1 in the F2 generation. 2 Genetic basis for experiments Let b+ represents the dominant allele for grey body b represents the recessive allele for black body vg+ represents the dominant allele for normal wings vg represents the recessive allele for vestigial wings Parental : Grey body and x Black body and phenotypes normal wings vestigial wings Parental : _b+ vg+_ x _b vg_ genotypes _________ _________ b+ vg+ b vg Gametes : formed b+ vg+_ x _b vg_ F1 genotype : _b+ vg+_ and _________ phenotype b vg All with grey body and normal wings Test cross on : _b+ vg+_ x _b vg_ F1 generation _________ _________ b vg b vg (No crossing over) (Crossing over) Gametes b+ vg+ b vg b+ vg b vg+ x b vg formed : Genotypes _b+ vg+_ _b vg_ _b+ vg_ _b vg+_ of : _________ ________ ________ ________ offspring b vg b vg b vg b vg Phenotypes Grey body, Black body, Grey body, Black body, of offspring : normal vestigial vestigial normal wings wings wings wings Number of offspring : 965 944 206 185 32 Dunman High School SJBO Mendelian Genetics What is the reason for this apparent ‘discrepancy’? If these two genes (body colour and wing length) were on different chromosomes, the alleles from the F1 dihybrid would segregate independently into gametes, and we would see the expected 1:1:1:1 phenotypic ratio in the test cross. If these two genes were on the same chromosome, we would expect each allele combination, b+ vg+ and b vg, to stay together as gametes are formed. Thus we would expect that only offspring with parental phenotypes would be produced. Since most offspring have a parental genotype, we can conclude that the genes for body colour and wing size are located on the same chromosome. However, the production of a small number of offspring with non-parental phenotypes indicated that a crossover occasionally breaks the linkage between genes on the same chromosome. Because crossing over between the b and vg loci occurs in some but not all ovum- producing cells, more ova with parental-type chromosomes than with recombinant ones are produced in the mating females (refer to the figure below). RECOMBINATION OF LINKED GENES 33 Dunman High School SJBO Mendelian Genetics 5. CHI-SQUARED (c 2) TEST We have learnt several Mendelian ratios and each of these ratios has a significant meaning. But how can we decide if our data really fits the Mendelian ratios? Could it be only a coincidence that our data seems fit the Mendelian ratios? A statistical test that can be applied here is the Chi-Squared (or Goodness-of-Fit) test. The Chi-Squared test is a statistical method which helps to determine if any deviations (differences) from the expected values (ratios) are purely due to chance (thus data is valid) or due to other factors (other than chance, e.g. epistasis or multiple gene effects), which create a bias towards a particular value. The chi-squared (χ2) test allows the evaluation of the results of breeding experiments and ecological sampling. The chi-squared (χ2) is used to test if the observed numbers are significantly different from the expected numbers in a genetic experiment. Why are observed results different from expected results? Sample size is too small Biased sampling Chance variation / variation is statistically insignificant Differential survival of sperm / ova and/or zygotes with particular genotypes (refer to lethal genes) Wrong genetic model used to predict expected ratios A. THE NULL HYPOTHESIS (Ho) It is stated in the form: “There IS NO significant difference between the observed number and the expected number of two samples”. It assumes that any differences are due to chance alone. The null hypothesis is accepted if there is no significant difference / discrepancy between the expected and the observed of two samples. B. THE ALTERNATIVE HYPOTHESIS (H1) It is stated in the form: “There IS a significant difference between the observed number and the expected number of two samples.” States that the difference is not due to chance alone. Other factors must be involved to affect the results. The alternative hypothesis is accepted if there is significant difference / discrepancy between the expected and the observed of two samples. C. LEVEL OF SIGNIFICANCE (degree of confidence or critical value) (p) To decide whether to accept or to reject a hypothesis, we need to evaluate the size of the discrepancy (difference) between the observed number and the expected number. Statisticians like to be at least 95% certain (95% confidence level) before drawing any conclusions. i.e. the level of rejection in biological investigations is normally taken as p = 0.05 or the 5% critical level This means that by chance alone, the χ2 statistic will exceed this value only 5% of the time. 34 Dunman High School SJBO Mendelian Genetics What is the p value? It is the probability that chance alone could produce the difference. If the P value is 0.05, that means that there is a 5% chance of observing a difference as large as you observed even if the two population means are identical. Random sampling from identical populations would lead to a difference smaller than you observed in 95% of experiments and larger than you observed in only 5% of experiments. If we set the critical value as p = 0.05, Then at p < 0.05, these is less than 5% probability that the difference between observed and expected numbers is due to chance alone. If we set the critical value as p = 0.001, Then at p < 0.001, there is less than a 0.1% probability that the difference between observed and expected numbers is due to chance alone. D. DEGREES OF FREEDOM (df or v) This concept arises because we wish to regard the total number of observed individuals as a fixed or given quantity. Let’s say there are two classes, e.g. tall and dwarf plants. If we have determined the number of tall plants, we will automatically know the number of dwarf plants. So one class is variable / free, while the second is dependent on the first. The number of degrees of freedom is one less than the number of classes / phenotypes: o df or v = n - 1, where n is the total number of classes E. CHI-SQUARED (χ2) CALCULATION Formula for χ2 calculation χ2 = Σ (O − E ) 2 v=c–1 E where Σ = ‘sum of…’ O = observed ‘value’ v = degrees of freedom E = expected ‘value’ c = number of classes How is this calculated χ2 value going to help you to accept or reject the null hypothesis? calculated χ2 value < critical (tabulated) χ2 value Accept null hypothesis. at the correct degree of freedom calculated χ2 value > critical χ2 (tabulated) value Reject null hypothesis. at the correct degree of freedom 35 Dunman High School SJBO Mendelian Genetics F. THE CHI-SQUARED STATISTICAL TABLE 36 Dunman High School SJBO Mendelian Genetics EXAMPLE Assuming you have a cob of corn that theoretically has purple and yellow kernels in the ratio 3:1 respectively. You counted 400 kernels of which 322 are purple and 78 are yellow. Use a c 2 test to determine is the difference between the observed number and the expected number is significant. Null hypothesis (Ho): The kernel colour was controlled by a single pair of genes (monohybrid inheritance). Allele for purple kernel is dominant to allele for yellow kernel. Thus, the corn was produced as a result of a cross between two heterozygotes. The purple and yellow kernels in the offspring are hence in the ratio 3:1 respectively. Alternative hypothesis (H1): The corn was not produced as a result of a cross between two heterozygotes. The observed ratio is not 3:1. Phenotype / Observed Expected Expected Deviations (O-E)2 (O-E)2 class (O) ratio (E) (O-E) E Purple 322 3 300 22 484 1.613 Yellow 78 1 100 -22 484 4.840 Total 400 4 400 ∑ (O-E)2 E = 6.453 The calculated c 2 value is 6.453. Degree of freedom = no. of classes – 1 = 2 – 1 =1 Reject H1 Reject H0 With reference to the χ2 table, the critical χ2 value = 3.841. Since the calculated χ2 value 6.453 > critical χ2 value 3.841 at p = 0.05, null hypothesis is rejected. At 1 degree of freedom, the calculated value of χ2 (6.453) lies between 5.024 and 6.635. The corresponding probability is 0.01