Skip to content
Dabour HubLEARN · PRACTICE · EXPLORESign inStudent

16.2.516 · Inherited Change

Predicting Inheritance- Chi-squared Test — Cambridge AS & A Level Biology 9700 Revision Notes & Topical Questions

5 min read 8 parts Lesson 12 of 18 1 quiz

🩺
100% Curriculum Specification Match Lesson Doctor Verified
The only platform with 100% curriculum specification match and verified interactive question banks.
0%read
Lesson summary

16.2.5 Predicting Inheritance: Chi-squared Test — CAMBRIDGE AS & A LEVEL BIOLOGY 9700 — REVISION NOTES notes: Use the chi-squared test to decide whether observed results differ significantly from expected results.Observed and expected — differences may be significant or due to ch…

  • Significant difference
  • Comparison
  • Dependence
  • Probability level
  • Degrees of freedom
  • Greater than or equal to the critical value

Section 1 of 8Purpose of the test

Observed and expected
  • Observed and expected differences may be significant or due to chance.
Significant difference
  • Significant difference suggests something else is happening, such as linkage between genes.
Test
  • Test the chi-squared test finds whether the difference between observed and expected results is significant.
Data
  • Data used with categorical data, which can be grouped.

Section 2 of 8Calculating chi-squared

1
Obtain the expected (E) and observed (O) results.
2
Calculate the difference O − E for each class.
3
Square each difference, so the sign does not matter.
4
Divide each squared difference by the expected value.
5
Add the values to give the chi-squared value.

Section 3 of 8Critical value

Comparison
  • Comparison compare the chi-squared value with a critical value from a table.
Dependence
  • Dependence the critical value depends on the probability level and the degrees of freedom.
Probability level
  • Probability level biologists use 0.05 (5 %).
Degrees of freedom
  • Degrees of freedom number of classes − 1.
Degrees of freedomp = 0.1p = 0.05p = 0.01p = 0.001
12.713.846.6410.83
24.605.999.2113.82
36.257.8211.3416.27
47.789.4913.2818.46

Section 4 of 8Drawing conclusions

Greater than or equal to the critical value
  • Greater than or equal to the critical value a significant difference;
    • a factor other than chance causes it;
    • the null hypothesis is rejected.
Smaller than the critical value
  • Smaller than the critical value no significant difference;
    • differences are due to chance;
    • the null hypothesis is accepted.
Between two probability levels
  • Between two probability levels the probability of chance lies between those levels.
Above the 0.001 value
  • Above the 0.001 value less than 0.1 % probability that chance explains the difference.

Section 5 of 8Using results

1Reasons for a significant result
  • Reasons for a significant result linkage between genes, epistasis or other factors.
2Table
  • Table setting out O, E, O − E, (O − E)2 and (O − E)2/E helps avoid errors.

Section 6 of 8More from the specification

More from the specification
Autosomal linkage
  • Genes on the same chromosome inherited together.
Punnett square
  • A grid predicting offspring genotypes.
Genetic diagram
  • Layout showing parents, gametes and offspring of a cross.
Allele
  • A version of a gene.
Genotype
  • The alleles an organism has.
Recessive
  • Expressed only when no dominant allele is present.
Dominant
  • Expressed when present in the genotype.
Chromosome
  • A thread-like structure of DNA carrying genes.
Homozygous
  • Having two identical alleles.
Disease
  • A condition in which part of the body does not function properly.
Haemophilia
  • A sex-linked recessive blood-clotting disorder.
Expected ratio
  • Expected ratio a dihybrid cross between two heterozygotes is expected to give phenotypes in a 9 : 3 : 3 : 1 ratio.

Section 7 of 8Key terms

TermMeaning
Chi-squared testTest comparing observed and expected results.
Null hypothesisStatement that there is no significant difference.
Degrees of freedomNumber of classes minus one.
Critical valueValue from the table at a chosen probability.
Observed valueResult actually counted.
Expected valueResult predicted by theory.
Categorical dataData that fit into groups.

Section 8 of 8Exam technique

1
Using degrees of freedom equal to the number of classes — it is classes − 1.
2
Saying a high chi-squared supports the null hypothesis — it leads to rejecting it.
3
Using percentages instead of counts — use actual numbers.
4
Leaving out the null hypothesis — it must be stated.

Last updated: