trihybrid cross , test cross chi squares

trihybrid cross
test cross
back cross
probability
chi square test

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 F 1  heterozygote is determined by the formula 2 n , where n represents the number of characters. Thus in a tri-hybrid cross 2 3  = 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. Back Cross T he 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

9. 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.

10. 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: R esults 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)

11. Types of Events 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.

12. 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 1 st time) = 4/10 Probabilty of 2 nd event depends on result of event A. if first ball drawn was red probability of 2 nd ball drawn being red will be P (Drawing red ball 2 nd time) =3/9 If in event A 1 st ball drawn was not red then probability of 2 nd 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.

13. 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.

14. 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)

15. 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.

16. 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

17. 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

18. 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 = ?