| Date | May 2022 | Marks available | 1 | Reference code | 22M.2.HL.TZ2.3 |
| Level | Higher level | Paper | Paper 2 | Time zone | Time zone 2 |
| Command term | Calculate | Question number | 3 | Adapted from | N/A |
Question
A group of students used quadrat sampling and the chi-squared test to find out whether the distributions of two plant species were associated with each other or not. These two species were found in the ground vegetation in a woodland ecosystem.
[Source: (left) Pixabay.
(right) Topic, J., n.d. Veronica montana 2. [image online] Available at: http://www.freenatureimages.eu/Plants/
Flora%20S-Z/Veronica%20montana/#Veronica%2520montana%25202%252C%2520Bosereprijs%252C%2520Saxifr
aga-Jasenka%2520Topic.jpg [Accessed 3 December 2019].
The numbers of quadrats with one, both or neither species present were counted and recorded. The observed frequencies from 150 quadrats are shown in the following contingency table.
State the alternative hypothesis for this study.
To calculate chi-squared, expected values must first be calculated. Assuming that there is no association between the two species, calculate the expected number of quadrats in which both species would be present, showing your working.
State the number of degrees of freedom for this test to determine the critical value of chi-squared.
When the data in the table were used to calculate chi-squared, the calculated value was 0.056. The critical value is 3.84. Explain the conclusion that can be drawn from the calculated and critical values for chi-squared.
Markscheme
there is a positive / negative association between the two species;
they tend to grow together / they tend to grow apart;
OWTTE
- ;
- 25.7;
Award [1] for proper values chosen/equation or [1] for answer.
1 (df)
OR
(r-1) (c-1);
- (when the calculated value is smaller than the critical value) there is no significant association between the two species / H0/null hypothesis accepted;
- it is random chance if both species are either present or absent in most quadrats;
Examiners report
There were many correct answers, but a number confused the alternate and null hypotheses.
Calculating the degrees of freedom proved problematic for many.
Syllabus sections
- 22M.1.HL.TZ1.36: An individual is heterozygous for two linked genes . To investigate the frequency of...
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22M.2.HL.TZ2.3c:
State the number of degrees of freedom for this test to determine the critical value of chi-squared.
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21M.2.HL.TZ2.5a:
Autosomal genes are located in chromosomes that are not sex chromosomes. The inheritance of autosomal genes is affected by whether the genes are linked or unlinked. Explain the two types of inheritance, using the example of parents that are heterozygous for two genes A and B.
- 19N.2.HL.TZ0.2b: Identify the recombinants.
- 16N.1.SL.TZ0.16: The diagram shows a pedigree chart. What does it reveal about the inheritance of the blood...
- 20N.1.HL.TZ0.36: Many commercially produced bananas are triploid instead of diploid. The nucleus of a triploid...
- 18N.1.HL.TZ0.36: The image shows variation in height of adult humans. What can explain the variation? A. ...
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19M.1.HL.TZ2.20:
William Bateson and Reginald Punnett used the sweet pea (Lathyrus odoratus) in genetics studies in the early 20th century. Pure-breeding plants that produced purple flowers and long pollen grains were crossed with pure-breeding plants that produced red flowers and round pollen grains. The resulting offspring all produced purple flowers and long pollen grains. Two of the F1 generation plants were crossed. The table shows the ratio of phenotypes in the F2 generation.
What is an explanation for these experimental results?
A. Purple flowers and long pollen grains are dominant and the alleles have assorted independently.
B. The genes for flower colour and pollen shape are linked and all plants producing long pollen grains are recombinants.
C. The genes for flower colour and pollen shape are linked and all plants producing red flowers are recombinants.
D. Plants producing purple flowers and round pollen grains arose through crossing over.
- 17M.1.SL.TZ1.1: Which structure found in eukaryotes has a single membrane? A. Nucleus B. Lysosome C....
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22M.2.HL.TZ1.8c:
Explain the reasons for variation in human height.
- 16N.1.HL.TZ0.35: Some of the ratios that Morgan investigated in genetic crosses did not correspond with...
- 19N.2.HL.TZ0.2c: The chi-squared value was calculated as shown. Deduce, with reasons, whether the observed...
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21M.1.HL.TZ1.36:
In fruit flies (Drosophila melanogaster), grey bodies (b+) are dominant to black bodies (b) and normal wings (vg+) are dominant to vestigial wings (vg). Homozygous vestigial winged, black bodied flies were crossed with individuals that were heterozygous for both traits. 2300 individuals were counted and the phenotypes observed were recorded as shown.
965 normal wings, grey bodies
944 vestigial wings, black bodies
206 vestigial wings, grey bodies
185 normal wings, black bodiesWhich statement is valid?
A. The predicted phenotypic ratio was 9:3:3: 1.
B. There is independent assortment of wings but not body colour.
C. The expected number of vestigial winged, grey bodied flies was 575.
D. The traits are on different chromosomes.
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21N.2.HL.TZ0.2c:
Explain gene linkage and its effects on inheritance.
- 21N.1.HL.TZ1.35: A dihybrid cross was carried out between two plants to determine whether the genes for seed...
- 19M.2.HL.TZ1.2d.i: A fly that is homozygous dominant for both body colour and wing size mates with a fly that is...
- 21N.2.HL.TZ0.6c: Distinguish between continuous and discrete variation, using examples.
- 20N.2.HL.TZ0.5b: Using a Punnett square, determine how a smooth-haired puppy could be produced in the offspring.
- 20N.2.HL.TZ0.5a: Identify the genotype of the male wire-haired dog.
- 17N.1.HL.TZ0.36: In a plant, dark leaves are dominant to pale leaves and yellow seeds are dominant to white...
- 17M.1.HL.TZ2.19: The genetic determination of dogs’ coats can be quite complex, with many different genes...
- 22M.2.HL.TZ2.3a: State the alternative hypothesis for this study.
- 19M.2.HL.TZ1.2d.ii: The offspring, which were all heterozygous for grey body and normal wings, were crossed with...
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22M.2.HL.TZ2.3d:
When the data in the table were used to calculate chi-squared, the calculated value was 0.056. The critical value is 3.84. Explain the conclusion that can be drawn from the calculated and critical values for chi-squared.
