Emily Bullas Inheritance Assignment
Investigating Monohybrid Inheritance
Method: For this experiment, fruit flies are crossed together and offspring are sedated using a Fly
Nap kit so that they can be counted more easily, using a magnifying glass and sorting brush. The
gene which was investigated is responsible for eye colour – the homozygous recessive phenotype is
white eyes, while the homozygous dominant phenotype is red eyes. The first cross is known as the
parental cross and uses 15 female homozygous dominant (red eyed) flies and 30 male heterozygous
recessive (white eyed) flies. This produced the F1 generation, which were all heterozygous, with red
eyes. Next, 15 female heterozygous flies were crossed with 15 male heterozygous flies to produce
the F2 generation, which were a mixture of red and white eyed phenotypes. These F2 generation
flies were then sedated and counted, and the results recorded in a table, seen in table 1 below.
Results:
Table 1: Practical Results for Monohybrid Cross
Phenotype Number of Flies: Number of Flies: Number of Flies: Number of Flies:
Repeat 1 Repeat 2 Repeat 3 Repeat 4
Red Eyes 79 88 43 64
White Eyes 18 11 1 15
Table 1 above shows the results for the practical investigating monohybrid inheritance. There were
four repeats to make the results more reliable.
The expected results can be predicted using a punnet square, as seen in figure 1 below.
Figure 1: Punnet Square for Monohybrid Cross
This punnet square shows that the expected results of the F1 cross is 50% homozygous dominant,
25% homozygous recessive and 25% heterozygous, represented by the red, blue and green colours
respectively. The predicted phenotypic ratio for red eyes to white eyes in 3:1, or 75% red eyes and
25% white eyes.
The chi-squared test will be used on the results to test whether this applies to the observed data.
The null hypothesis is that the observed data is no different to the expected data. Whereas the
alternative hypothesis is that the observed data is different to the expected data. The expected data
for this cross is a Mendelian ratio of 3:1. The calculations for this can be seen below.
, Table 2: Chi-Squared Calculation for Repeat 1
Phenotype Observed Expected O-E (O-E)² (O-E)² / E
Red 79 72.75 6.25 39.1 0.49
White 18 24.25 -6.25 39.1 2.17
X² = 2.66
Degrees of freedom = n-1 = 2-1 = 1. Critical value (CV) = 3.841
X² = 2.66 < 3.841 = CV, so accept null hypothesis.
Table 3: Chi-Squared Calculation for Repeat 2
Phenotype Observed Expected O-E (O-E)² (O-E)² / E
Red 88 74.75 13.75 189.1 2.55
White 11 24.75 -13.75 189.1 7.64
X² = 10.19
X² = 10.19 > 3.841 = CV, so reject null hypothesis for alternative hypothesis.
Table 4: Chi-Squared Calculation for Repeat 3
Phenotype Observed Expected O-E (O-E)² (O-E)² / E
Red 43 33 10 100 3.03
White 1 11 -10 100 9.09
X² = 12.12
X² = 12.12 > 3.841 = CV, so reject null hypothesis for alternative hypothesis.
Table 5: Chi-Squared Calculation for Repeat 4
Phenotype Observed Expected O-E (O-E)² (O-E)² / E
Red 64 59.25 4.75 22.56 0.38
White 15 19.75 -4.75 22.56 1.14
X² = 1.52
X² = 1.52 < 3.841 = CV, so accept null hypothesis.
As seen in tables 2-5, repeats 1 and 4 passed the chi-squared test, so the null hypothesis is accepted,
meaning that the observed results obey the Mendelian rules of inheritance. Repeats 2 and 3,
however, do not pass the chi-squared test and so the null hypothesis is rejected, and the alternative
hypothesis accepted, meaning that the observed do not obey the Mendelian rules of inheritance.
This does not necessarily mean that the genes do not follow Mendelian inheritance, as there are
many reasons why the results may not represent the true phenotypic ratios. Firstly, the sample size
is too small to make accurate judgements about the phenotypic ratios produced. Furthermore,
fertilisation of gametes is not an ideal process – the gametes which are fertilised are random, some
gametes may not be fertilised and once fertilised, some may not develop. There may also be
