The roles of genes in determining the phenotype (3)
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1.
Question 1
A botanist is studying a population of flowering plants in a particular habitat. They want to investigate whether there is a significant difference in the proportion of plants with red flowers compared to those with white flowers. The botanist randomly selects 100 plants and records the flower colour. The results are shown below:
| Red Flowers | White Flowers |
| 45 | 55 |
(a) State a suitable hypothesis for this experiment.
(b) Calculate the expected numbers of red and white flowered plants, assuming there is no difference between the proportions of red and white flowers.
(c) Calculate the chi-squared statistic for this experiment. (You may use the formula provided in the Mathematical requirements.)
(d) Determine the degrees of freedom for this test.
(e) State the critical value for a chi-squared test with 1 degree of freedom at a 5% level of significance.
(f) Based on your calculations, what conclusion can you draw about the difference in the proportion of red and white flowered plants?
Answer 1
(a) Null hypothesis: There is no difference in the proportion of red flowers and white flowers in the plant population. Alternative hypothesis: There is a difference in the proportion of red flowers and white flowers in the plant population.
(b) The total number of plants is 100. The proportion of red flowers is 45/100 = 0.45. The proportion of white flowers is 55/100 = 0.55. Therefore, the expected numbers are:
- Expected Red: 0.45 * 100 = 45
- Expected White: 0.55 * 100 = 55
(c) The chi-squared statistic is calculated as:
χ2 = Σ [(O - E)2 / E]χ2 = [(45 - 45)2 / 45] + [(55 - 55)2 / 55] = 0 + 0 = 0(d) The degrees of freedom (df) are calculated as (number of categories - 1). In this case, there are 2 categories (red and white flowers), so df = 2 - 1 = 1.
(e) The critical value for a chi-squared test with 1 degree of freedom at a 5% level of significance (alpha = 0.05) is 3.841.
(f) Since the calculated chi-squared statistic (0) is less than the critical value (3.841), we fail to reject the null hypothesis. Therefore, there is no statistically significant difference in the proportion of red flowers and white flowers in the plant population.
2.
A plant breeder is breeding pea plants. The allele for purple flowers (P) is dominant to the allele for white flowers (p). A plant with the genotype Pp is test crossed with a homozygous recessive plant (pp).
Question 1 Answer
Question 1 Answer:
a) Diagram of the test cross:
b) Expected genotypes and phenotypes of the offspring:
- Genotype: 50% Pp, 50% pp
- Phenotype: 50% purple flowers, 50% white flowers
c) Explanation: The test cross involves crossing a heterozygous individual (Pp) with a homozygous recessive individual (pp). This allows us to determine the genotype of the heterozygous parent. The Punnett square shows the possible allele combinations from each parent. Since the recessive allele (p) is only expressed when two copies are present, the offspring with the pp genotype will have white flowers. The offspring with the Pp genotype will have purple flowers because they carry at least one dominant P allele. Therefore, the phenotypic ratio is 1:1 (purple:white).
3.
A plant breeder is attempting to create a dwarf variety of a crop plant. The plant species has a single gene controlling plant height, with the allele for dwarfism (le) being recessive to the allele for normal height (Le). Describe the crosses the breeder could perform to produce dwarf plants, and predict the expected genotypes and phenotypes of the offspring for each cross.
To produce dwarf plants, the breeder needs to ensure that the offspring inherit two copies of the recessive le allele. Here are three possible crosses:
- Cross 1: LeLe x lele
Parental genotypes: LeLe (normal height) x lele (dwarf).
Punnett Square:
Expected offspring genotypes and phenotypes:
- Lele: 50% probability of being heterozygous (normal height).
- lele: 50% probability of being homozygous recessive (dwarf).
- Cross 2: Lele x Lele
Parental genotypes: Lele (normal height) x Lele (normal height).
Punnett Square:
Expected offspring genotypes and phenotypes:
- LL: 25% probability of being homozygous dominant (normal height).
- Le: 25% probability of being heterozygous (normal height).
- le: 25% probability of being homozygous recessive (dwarf).
- Cross 3: LeLe x Lele
Parental genotypes: LeLe (normal height) x Lele (normal height).
Punnett Square:
Expected offspring genotypes and phenotypes:
- LL: 25% probability of being homozygous dominant (normal height).
- Le: 25% probability of being heterozygous (normal height).
- le: 25% probability of being homozygous recessive (dwarf).
The breeder should choose Cross 1, as it guarantees the production of dwarf plants (lele genotype) in 50% of the offspring.