Home > Knowledge Base > Assignment Samples > Assignment Sample: Mendelian Genetics Problem Set With Chi-Square Testing

Assignment Sample: Mendelian Genetics Problem Set With Chi-Square Testing

Published by at July 30th, 2026 , Revised On July 30, 2026

Type: Assignment  |  Subject: Biology  |  Level: Undergraduate  |  Word Count: ~2000 words

This model assignment was produced by an Essays UK specialist as reference material for learning purposes only. For support in this field, see our biology assignment specialists.

The Brief

You are a first-year Biology undergraduate. Complete the attached genetics problem set, applying Mendel’s laws and chi-square goodness-of-fit testing to two data sets from dihybrid cross experiments. Show all working, state your hypotheses clearly, and interpret the statistical significance of your results in relation to independent assortment and genetic linkage.

Model Answer

Introduction

Mendelian genetics provides the theoretical basis for predicting how traits are inherited across generations, yet real experimental data rarely match theoretical ratios with perfect precision. Random sampling variation, classification error and biological factors such as genetic linkage can all cause observed phenotype counts to diverge from the ratios predicted by Mendel’s laws of segregation and independent assortment (Griffiths et al., 2015). The chi-square (χ²) goodness-of-fit test is the standard statistical procedure used to determine whether an observed distribution of phenotypes deviates significantly from an expected theoretical ratio, or whether the deviation is small enough to be attributed to chance alone (Sokal and Rohlf, 2012).

This problem set applies the chi-square test to two experimental data sets drawn from classic dihybrid cross designs. Problem 1 examines an F2 generation expected to segregate in the 9:3:3:1 ratio characteristic of two independently assorting gene pairs, following the general design used by Mendel (1866) in his original pea-plant experiments. Problem 2 examines a dihybrid testcross expected to produce a 1:1:1:1 ratio under independent assortment, using a design analogous to that employed in early Drosophila linkage studies (Morgan, 1911). For each data set, the aim is to state a null and alternative hypothesis, calculate the chi-square statistic step by step, compare the calculated value against the appropriate critical value, and interpret the biological meaning of the result, including the possibility of genetic linkage where independent assortment is rejected. The two problems are deliberately paired so that the same statistical procedure can be shown to support two very different biological conclusions, depending entirely on what the data reveal.

Approach and Method

Both problems follow the same statistical procedure. For each phenotypic category, the expected count (E) is calculated by multiplying the total sample size by the theoretical proportion predicted under Mendelian inheritance. The chi-square statistic is then calculated using the standard formula χ² = Σ [(O − E)² ÷ E], where O is the observed count and E is the expected count for each category, summed across all phenotypic categories (Pierce, 2020). The degrees of freedom (df) for a goodness-of-fit test are equal to the number of phenotypic categories minus one, since the total sample size is fixed once the category counts are known.

At the conventional significance level of α = 0.05, the calculated χ² value is compared against the critical value taken from the chi-square distribution table for the appropriate degrees of freedom (Zar, 2010). If the calculated χ² is less than the critical value, the null hypothesis — that the observed data fit the expected Mendelian ratio — is retained. If the calculated χ² exceeds the critical value, the null hypothesis is rejected, and the deviation is considered statistically significant at the 5% level (Snedecor and Cochran, 1989). Before applying the test, both data sets were checked against the standard assumptions of the chi-square goodness-of-fit test: observations must be independent of one another, categories must be mutually exclusive and exhaustive, and expected counts should generally exceed five in each category to ensure the chi-square approximation to the underlying sampling distribution remains valid (Hartl and Ruvolo, 2012). Both problems below comfortably satisfy these conditions, with minimum expected counts of 40 and 250 respectively, well above the conventional threshold.

Worked Solution: Problem 1 – Dihybrid F2 Cross

A cross was set up between a homozygous pea plant with round, yellow seeds (RRYY) and a homozygous plant with wrinkled, green seeds (rryy). The F1 generation, all heterozygous (RrYy) and phenotypically round and yellow, was self-pollinated to produce an F2 generation. Under independent assortment, the two gene pairs — seed shape and seed colour — are expected to segregate in the classic 9:3:3:1 phenotypic ratio: round yellow, round green, wrinkled yellow, and wrinkled green (Mendel, 1866). A total of 640 F2 offspring were scored by phenotype.

Null hypothesis (H0): the observed phenotype counts fit the expected 9:3:3:1 ratio.
Alternative hypothesis (H1): the observed phenotype counts do not fit the expected 9:3:3:1 ratio.

The expected count for each category is found by multiplying the total of 640 by the relevant fraction: 640 × 9/16 = 360; 640 × 3/16 = 120; 640 × 3/16 = 120; and 640 × 1/16 = 40. These expected counts, together with the observed counts and the individual chi-square contributions, are set out in Table 1.

Phenotype Observed (O) Expected (E) O − E (O − E)² (O − E)² ÷ E
Round, yellow 355 360 −5 25 0.069
Round, green 123 120 3 9 0.075
Wrinkled, yellow 116 120 −4 16 0.133
Wrinkled, green 46 40 6 36 0.900
Total 640 640 χ² = 1.18

Summing the final column gives χ² = 0.069 + 0.075 + 0.133 + 0.900 = 1.18 (to two decimal places). With four phenotypic categories, degrees of freedom = 4 − 1 = 3. The critical value of χ² at α = 0.05 for df = 3 is 7.815 (Zar, 2010). Because the calculated χ² of 1.18 is well below the critical value of 7.815, the null hypothesis is retained: there is no statistically significant difference between the observed and expected counts, and the data are consistent with seed shape and seed colour assorting independently in a 9:3:3:1 ratio, exactly as Mendelian theory predicts for two gene pairs located on different chromosome pairs.

Worked Solution: Problem 2 – Dihybrid Testcross

A second experiment used a testcross design in the fruit fly Drosophila melanogaster, crossing a heterozygous grey-bodied, normal-winged fly (GgNn) with a homozygous recessive black-bodied, vestigial-winged fly (ggnn). If the genes controlling body colour and wing shape assort independently, this testcross is expected to produce four phenotypic classes — grey normal, black vestigial, grey vestigial, and black normal — in an equal 1:1:1:1 ratio (Morgan, 1911). A total of 1,000 offspring were scored.

Null hypothesis (H0): the two genes assort independently, so offspring fall into the four phenotypic classes in a 1:1:1:1 ratio.
Alternative hypothesis (H1): the two genes do not assort independently.

Under H0, each of the four categories has an expected count of 1,000 ÷ 4 = 250. Table 2 sets out the observed counts alongside the expected counts and chi-square contributions.

Phenotype Observed (O) Expected (E) O − E (O − E)² (O − E)² ÷ E
Grey, normal wing (parental) 412 250 162 26,244 104.98
Black, vestigial wing (parental) 408 250 158 24,964 99.86
Grey, vestigial wing (recombinant) 92 250 −158 24,964 99.86
Black, normal wing (recombinant) 88 250 −162 26,244 104.98
Total 1,000 1,000 χ² = 409.66

Summing the final column gives χ² = 104.98 + 99.86 + 99.86 + 104.98 = 409.66 (to two decimal places). With four categories, degrees of freedom = 3 again, and the critical value at α = 0.05 remains 7.815 (Zar, 2010). Because the calculated χ² of 409.66 is very much larger than the critical value, the null hypothesis of independent assortment is firmly rejected: p is far below 0.001. The parental phenotypic combinations — grey normal and black vestigial — occur far more frequently than the recombinant combinations — grey vestigial and black normal — which is the classic statistical signature of genetic linkage rather than independent assortment (Hartl and Ruvolo, 2012). The recombination frequency can be estimated directly as the proportion of recombinant offspring: (92 + 88) ÷ 1,000 = 0.18, or 18%. Using the standard convention that a recombination frequency of 1% corresponds approximately to one map unit, or centimorgan, the two loci are estimated to lie roughly 18 map units apart on the same chromosome.

Evaluation

The contrast between the two problems illustrates both the usefulness and the limits of the chi-square goodness-of-fit test. In Problem 1, a low χ² value close to the value expected under H0 supported the conclusion that seed shape and seed colour assort independently, consistent with these two genes lying on different chromosomes, or far enough apart on the same chromosome that crossing over effectively randomises their combination at each generation (Griffiths et al., 2015). In Problem 2, a very large χ² value led to firm rejection of independent assortment, and the specific pattern of results — an excess of parental-type offspring and a corresponding deficit of recombinant-type offspring — pointed directly to genetic linkage as the underlying biological cause (Klug et al., 2019).

It is important to note that the chi-square test itself does not identify the cause of a significant deviation; it only indicates that the deviation is unlikely to have arisen through chance sampling alone. A significant χ² value could, in principle, arise from causes other than linkage, such as reduced viability of a particular genotype combination, non-random mating, systematic scoring or classification errors, or a genuine departure from the assumed genetic model (Sokal and Rohlf, 2012). In this case, however, the specific pattern of parental-type excess and recombinant-type deficit, combined with the biological plausibility of the two loci lying on the same chromosome, makes linkage by far the most parsimonious explanation for the deviation observed in Problem 2.

A further methodological point concerns sample size and the validity of the chi-square approximation itself. Both problems used expected counts comfortably above the conventional minimum of five per category, so the chi-square approximation to the true underlying sampling distribution is expected to be reliable in both cases (Hartl and Ruvolo, 2012). Had multiple independent chi-square tests been performed across several gene pairs within the same wider study, a correction for multiple comparisons would also need to be considered in order to avoid inflating the overall false-positive rate across the set of tests (Rice, 1989); this was not required here, since only two independent tests were carried out and each addressed a clearly distinct research question. Overall, the analysis demonstrates that statistical hypothesis testing is an essential complement to Mendelian ratio prediction, allowing a researcher to move from a purely qualitative expectation to a quantitative, statistically defensible conclusion about the underlying genetic architecture of a cross.

Conclusion

This problem set has applied the chi-square goodness-of-fit test to two Mendelian genetics data sets, using the same procedure to reach two contrasting conclusions. Problem 1 showed that seed shape and seed colour in the F2 generation of a dihybrid cross fit the expected 9:3:3:1 ratio closely (χ² = 1.18, df = 3, p > 0.05), supporting independent assortment of the two gene pairs. Problem 2 showed that body colour and wing shape in a Drosophila testcross deviated very significantly from the expected 1:1:1:1 ratio (χ² = 409.66, df = 3, p < 0.001), providing strong statistical evidence of genetic linkage with an estimated recombination frequency of 18%, corresponding to roughly 18 map units between the two loci. Together, these results demonstrate how a single statistical formula, applied consistently and with clearly stated hypotheses, allows a genetics student to distinguish true independent assortment from linked inheritance using nothing more than carefully tabulated cross data and a standard chi-square calculation.

More broadly, this exercise highlights why chi-square testing has remained central to introductory genetics teaching for over a century. Mendel himself worked without formal significance testing, relying instead on the closeness of his ratios to theoretical expectations to argue for particulate inheritance (Mendel, 1866); it was only with the later development of the chi-square distribution that geneticists gained a rigorous, quantitative way of asking exactly how close is close enough. Modern genetic mapping techniques, including the construction of linkage maps from recombination frequencies of the kind calculated in Problem 2, still rest on precisely this logic: a statistically significant deviation from independent assortment is the starting signal that two loci may be physically linked, prompting further investigation such as three-point testcrosses or, in contemporary research, molecular mapping using DNA markers (Klug et al., 2019). Students first encountering chi-square testing through problems such as these therefore gain more than a mechanical calculation skill; they gain an appreciation of how nineteenth-century plant-breeding observations and twentieth-century statistical theory combined to give rise to the quantitative discipline of genetic mapping that underlies modern genomics.

References

  • Griffiths, A.J.F., Wessler, S.R., Carroll, S.B. and Doebley, J. (2015) Introduction to Genetic Analysis. 11th edn. New York: W.H. Freeman.
  • Hartl, D.L. and Ruvolo, M. (2012) Genetics: Analysis of Genes and Genomes. 8th edn. Burlington, MA: Jones & Bartlett Learning.
  • Klug, W.S., Cummings, M.R., Spencer, C.A. and Palladino, M.A. (2019) Concepts of Genetics. 12th edn. Harlow: Pearson.
  • Mendel, G. (1866) ‘Versuche über Pflanzen-Hybriden’, Verhandlungen des naturforschenden Vereines in Brünn, 4, pp. 3–47.
  • Morgan, T.H. (1911) ‘Random segregation versus coupling in Mendelian inheritance’, Science, 34(873), p. 384.
  • Pierce, B.A. (2020) Genetics: A Conceptual Approach. 7th edn. New York: W.H. Freeman.
  • Rice, W.R. (1989) ‘Analyzing tables of statistical tests’, Evolution, 43(1), pp. 223–225.
  • Snedecor, G.W. and Cochran, W.G. (1989) Statistical Methods. 8th edn. Ames, IA: Iowa State University Press.
  • Sokal, R.R. and Rohlf, F.J. (2012) Biometry: The Principles and Practice of Statistics in Biological Research. 4th edn. New York: W.H. Freeman.
  • Zar, J.H. (2010) Biostatistical Analysis. 5th edn. Upper Saddle River, NJ: Pearson Prentice Hall.

Need a Model Assignment Written to Your Exact Brief?

Our 350+ UK-qualified writers deliver referenced model documents from £15 per 250 words, with free plagiarism and AI-detection reports.

Order Your Model Assignment

Frequently Asked Questions

About Jesse Pinkman

Avatar for Jesse PinkmanJessie Pinkman has been writing since childhood when her mother gave her a book where she could write her stories. Since then Jessie has always loved to write about the topics she loves. She graduated from Birmingham University in 2012, worked as a teaching assistant, and then turned to full-time writing in 2016.

You May Also Like

WhatsApp Live Chat