Year 10 Science · Unit 1 · Lesson 4
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Learning Goals
Punnett square practice
Complete each Punnett square and fill in the phenotype ratio. Use B = brown eyes (dominant) and b = blue eyes (recessive); C = normal allele, c = cystic fibrosis allele.
| Cross 1: BB × bb (brown × blue) | ||
|---|---|---|
| b | b | |
| B | ||
| B | ||
| Phenotype ratio: | ||
| Cross 2: Bb × Bb (both carriers for blue) | ||
|---|---|---|
| B | b | |
| B | ||
| b | ||
| Phenotype ratio: | ||
| Cross 3: Cc × cc (carrier × affected, CF) | ||
|---|---|---|
| C | c | |
| c | ||
| c | ||
| Phenotype ratio: | ||
| Cross 4: Cc × Cc (two CF carriers) | ||
|---|---|---|
| C | c | |
| C | ||
| c | ||
| Phenotype ratio: | ||
Real-world context
Cystic fibrosis (CF) is the most common life-threatening genetic disorder in Australia, affecting approximately 1 in 2,500 newborns. About 1 in 25 Australians carries a single recessive CF allele (c) without showing symptoms. Two carrier parents (Cc × Cc) have a chance of having an affected child (cc) with each pregnancy.
(a) Using the Punnett square from Cross 4 above, state the probability (as a fraction and a percentage) that a child of two CF carriers will: (i) have CF; (ii) be a carrier; (iii) be unaffected and not a carrier.
(b) A CF-affected child is born to parents who both appear healthy. Explain why neither parent showed symptoms of CF, even though they each passed on a CF allele. Use the terms dominant, recessive, genotype and phenotype in your answer.
1. Huntington's disease is caused by a dominant allele (H). A person with genotype Hh has the disease. Construct a Punnett square for Hh × hh and state the probability that a child will inherit Huntington's disease.
2. Blood group AB results from codominance, neither the A nor the B allele is dominant or recessive. A person with blood group AB has one A allele and one B allele. Explain how codominance is different from dominant–recessive inheritance, using blood groups as your example.
Wrap Up
In one sentence, what was the main idea of this lesson?