To verify Mendel’s Monohybrid cross

Exercise: To verify Mendel’s Monohybrid cross

NOTE: Student have to work in pairs to perform this experiment and record the data in the observation and record note book with the help of the teacher.

Need not consider this Monohybrid cross experiment for Board Practical Examination.

Aim:

To verify Mendel’s Monohybrid cross.

Principle:

When two pure lines with contrastingtraits of a particular character (phenotype) are crossed to produce the next generation (F₁generation), all the members of the progeny are of only one phenotype, i.e. of one of the two parents. The phenotype that appears is called dominant and the one that doesnot appear is called recessive. When the F₁ plants are selfed, the progeny i.e. the F₂ generation, is in the ratio of 3 dominant : 1 recessive (¾ : ¼ of 75% : 25%). This reappearance of the recessive phenotype in F₂ generation, verifies Mendel’s Monohybrid cross.

Requirements:

64 yellow and 64 green plastic beads, all of exactly same shape and size (when beads are not available, pea seeds may be painted and used). Plastic beakers, petri dish and a napkin / hand towel.

Procedure

Make the student to work in pairs to perform the experiment. Follow the steps in given sequence.

·        Put 64 yellow beads in one beaker and 64 green beads in the other to represent male and female gametes respectively. Let the yellow bead be indicated by ‘Y’ and the green bead by ‘y’

·        Take a bead from each container and place them together (it represents fertilization) on the hand towel spread before you on the table.

·        Just like the previous step, continue to pick beads and arrange them in pairs. Thus 64 pairs of beads are obtained representing the 64 heterozygous F₁ progeny.

·        Put 32 F₁ progeny in one petridish and the remaining 32 in another petridish (representing the F₁ males and females).

·        To obtain the F₂ generation, the student should withdraw one bead from one beaker labelled male and one from the other beaker labelled female keeping his / her eyes closed (to ensure randomness) and put them together on the hand towel spread over the table. Continue this process till all the beads are paired. Thus 64 offsprings of F₂ progeny are obtained.

·        Note the genotype (YY or Yy or yy) of each pair and their possible phenotype.

·        Pool all the data and calculate the genotypic and phenotypic ratios.

Observation:

Record the result in the following table:

Phenotypic ratio : in F1 ____________________

                          : in F2  ____________________                          

Genotypic ratio   : in F1 ____________________

                          : in F2 ____________________

Inference:

The results are so because when the F₁ individuals are crossed together to raise the F₂ generation, each F1 individual produces two types of gametes: 50% having dominant allele and the remaining 50% having recessive allele. These gametes undergo random fusion during fertilization to produce the F2 generation. According to simple probability of mixing of opposite sex gametes, offsprings of three genotypes are likely to appear as follows:   Among these, proportion of dominant phenotype would be YY + Yy = yellow and recessive phenotype yy = green, which occur in 3 : 1 or 75% : 25% ratio.

This ratio of 3 :1 in the F₂ suggests that the hybrids or heterozygotes of F₁ generation have two contrasting factors or alleles of dominant and recessive type. These factors, though remain together for a long time, do not contaminate or mix with each other. They separate or segregate at the time of gamete formation so that a gamete carries only one factor, either dominat or recessive.

Precautions:

·        Take a sufficiently large number of seeds for analysis to minimise the error.

·        Observe the contrasting form of trait carefully.