### trihybrid cross , test cross chi squares

• 1. Trihybrid Cross Test Cross Back Cross Probability Chi-Square
• 2. Trihybrid Cross • Cross between the two individuals of a species for studying inheritance of three pairs of factors or alleles belonging to three different genes. • The phenotypic ratio of the trihybrid cross is 27:9:9:9:3:3:3:1 among F2 generation • In real organisms, thousands of genes are segregating at each meiotic event. As long as two genes are located on different chromosomes, they will segregate independently from one another.
• 3. • Suppose true breed varieties of plant. (Three traits: Seed Shape, Color of Cotyledons, and Color of flower) • F1 hybrid produces 8 types of gametes. These have equal chances to combine with any of the 8 types of gametes produced by the other parent resulting in 64 different combinations. • In each case number of gametes formed by F1 heterozygote is determined by the formula 2n, where n represents the number of characters. Thus in a tri- hybrid cross 23 = 8 gametes result.
• 4. Test Cross • A test cross is used to find out the genotype of any plant with dominant expression when it is not known whether it is homozygous (pure) or heterozygous for that trait. • The parent is always homozygous recessive for all the genes • The purpose of the test cross is to determine the genetic makeup of the dominant organism. • When an organism shows a dominant character, it could be homozygous or heterozygous for that character. Using the homozygous recessive organism, the genotype of the organism can be tested. • Homozygous dominant individuals produce only one gamete, while heterozygous produce 2 kinds of gametes with equal frequency
• 5. • Let’s assume a tall pea plant with no knowledge of its parentage. Since tallness is a dominant feature in peas, your plant may be homozygous (TT) or heterozygous (Tt), but you'd have no idea. In this case, a test cross can be used to establish its genotype. • If the plant were homozygous (TT), a test cross would produce all tall progeny (TT* tt: all Tt); if the plant were heterozygous (Tt), the test cross would produce half tall progeny and half short progeny (Tt* tt: Tt and tt). • In monohybrid testcross with a homozygote (TT) gives all one phenotype • In heterozygote (Tt), it gives 1:1 phenotypic ratio, indicating one pair is segregating
• 6. Dihybrid cross • A mating experiment between two organisms that are identically hybrid for two traits. • A hybrid organism is one that is heterozygous, which means that it carries two different alleles at a particular genetic position, or locus. • The Dihybrid test cross-ratio is 1:1:1:1. • One parent from F1 generation having heterozygous condition gets crossed with a parent which is double homozygous and recessive in nature. They are allowed to fertilize with each other. • It is used to test the linkage between the genes.
• 7. The following chart shows how to calculate the results of test cross. (Note that wrinkled seeds should have the r allele). • The image describes using the FOIL method of determining all the possible outcomes. • In this case, all of the offspring are going to be RrYy. • Using the FOIL method, you arrive at 4 possible gametes from the heterozygous parent: RY, Ry, rY, and ry. • Get 4 possible genetic combinations RrYy, Rryy, rrYy, and rryy) with the single gamete type produced by the test cross parent. • The ratio on the bottom would be 1:1:1:1.
• 8.
• 9. Back Cross • The mating of a hybrid organism (offspring of genetically unlike parents) with one of its parents or with an organism genetically similar to the parent. • It is useful in genetics studies for isolating (separating out) certain characteristics in a related group of animals or plants. • The purpose of a backcross is to recover elite genotypes and to produce offsprings that are genetically similar or closer to parents. • Every testcross is a backcross, but every backcross is not a testcross
• 10. Probability • Genetics is the study of inheritance, but it is also a study of probability most eukaryotic organisms are diploid, each cell contains two copies of every chromosome, so there are two copies of each gene that controls a trait (alleles). • In sexual reproduction, these two copies of each chromosome separate, and are randomly sorted into the reproductive cells (gametes). • When gametes from two different parents combine in fertilization, new combinations of alleles are created. • Plant and animal cells contain many thousands of different genes and have two copies of every gene. • The two copies of the gene may or may not be identical, and one may be dominant in determining the phenotype while the other is recessive. • Inheritance of characteristics is the result of random chance • The genes that an individual organism inherits depends on the “luck of the draw,” and the luck of the draw is dependent on the laws of probability.
• 11. • The proportion of total number of equally likely, equally probable, mutual exclusive outcomes which satisfy the events. • Its value ranges from 0-100 or 0-1. • 0 means no event occurs • 1 means certainty that it will occur. • Expressed as % or proportion. • Outcome: Results obtained on an experiment account is called outcome. Tossing is an experiment and head, or tail is outcome. • Sample space: Total of all possible outcomes of an experiment. Outcome is a point in sample. • Event: A part of sample space. It consists of no. of outcomes of experiment. • Mutually Exclusive: When we consider two or more events in such a way that one event makes occurrence of other events impossible. For example, when we toss a coin appearance of head makes the occurrence of tail impossible and vice versa. Out of two possible outcomes, only one will occur at a time. Similarly a gamete from diploid heterozygote (Aa) can either have allele “A” or allele “a” but never both (under normal condition)
• 12. Types of Events 1. Independent Event • Two events A and B are said to be independent when occurrence of one does not influence probability of occurrence of other. • For example Tossing of two different coins simultaneously. Appearance of head or tail of one coin is not influenced by appearance of head or tail on second coin. • Birth of a child in a family is independent of previous or future births.
• 13. Dependent Event • Two events A and B are said to be dependent when occurrence of one event influences probability of occurrence of second event or other event. • For example there are 4 red and 6 white balls in a bag. If event A is drawing first ball without replacement and event B is drawing the second ball. • P (Drawing any ball) = 1/10 • P (drawing a red ball 1st time) = 4/10 • Probabilty of 2nd event depends on result of event A. if first ball drawn was red probability of 2nd ball drawn being red will be • P (Drawing red ball 2nd time) =3/9 • If in event A 1st ball drawn was not red then probability of 2nd ball drawn being red will be 4/9. • Conditional Probability: in such cases where probability of event B depends upon information of event A is called conditional probability. And represented as P(B/A) • Conditional probability of B given that A has already occurred.
• 14. Laws of Probability • The Rule of Multiplication: The chance that two or more independent events will occur together is equal to the product of the probabilities of each individual event. • What are the chances of drawing a red nine from a standard sets of cards? • Answer: 1/26 (1 chance in 26), because there is 1/2 chance of drawing a red card and 1 chance in 13 of drawing a nine. Therefore, 1/2 x 1/13 = 1/26 or 1 chance in 26 of drawing a red nine.
• 15. • Multiplication Rule for Independent Events: A & B are two independent events the probability occurrence of both events simultaneously equal to product of their separate probabilities. • Written as Follows: P(A and B) = P(A). P(B) • Past events have no influence on future events. • If a coin is tossed 5 times, and each time a head appears, then what is the chance that the next toss will be heads? • Answer: 1/2 (1 chance in 2), because coins have 2 side • Multiplication Rule for dependent Events: A & B are two independent events the probability occurrence of A and B events occur simultaneously is equal to product of event A probability and conditional probability of event B. • Represented as follows: • P(A and B) = P(A). P(B/A) • P(B and A) = P(B). P(A/B)
• 16. • The Rule of Addition: The chance of an event occurring when that event can occur in two or more different ways is equal to the sum of the probabilities of each individual event. • Represented as follows: • P (either A or B) = P(A) + P(B) • If 2 coins are tossed, what is the chance that the toss will yield 2 unmatched coins (1 head & 1 tail)? • Answer: 1/2 (1 chance in 2) because the combination of 2 unmatched coins can come about in 2 ways: Result A (coin #1 heads, coin #2 tails) as well as Result B(coin #1 tails, coin #2 heads). Therefore (1/2 x 1/2) + (1/2 x 1/2) = 1/2, or the chance of Result A plus the chance of Result B.
• 17. Chi-Square Test • An important question to answer in any genetic experiment is how can we decide if our data fits any of the Mendelian ratios we have discussed. • A statistical test that can test out ratios is the Chi-Square or Goodness of Fit test. • We need a test to evaluate a genetic hypothesis that converts deviation from expected values into probability occurring unequal chances • In this we consider sample size and no. of variables. • Symbolized by χ2
• 18. • It is a statistical procedure that enables investigators to determine how closely an experiment-obtained set of values fits in a given theoretical expectation • Developed by karl Pearson in 1990 • Its value is zero, when expected and observed values are equal O=E • The greater the discrepancy between O and E, greater the χ2 value • Large value of χ2 determines poor fit and small value indicates good fit • The minimum value of χ2 is said to be “best” • Chi-square formula is: • χ2= Σ = (O-E)2 E
• 19. • If χ2 calculated is less than χ2 tabulated Ho will be accepted and concluded that the differences between O and E values are by chance and not a real difference • Example: We have observed four phenotypic class in F2 315:108:101:32=556. While expected ratio is 9:3:3:1. Are the expected values according to our expectations? • Ho : D = 0 or O-E = 0 • Ha : D ≠ 0 or O-E ≠ 0 • α : 5% or 0.05 • χ2 = ?
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