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 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.

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 = ?