B17: DNA and Genetics
DNA, genes, chromosomes and genetic inheritance
Dna And Genetics
📋 Key Concepts
DNA and Genetics: DNA carries the genetic code that determines an organism's characteristics. Understanding how genes and alleles are inherited allows us to predict the outcomes of genetic crosses using Punnett squares.
Key Terms
- DNA
- double helix
- genome
- chromosome
- gene
- allele
- dominant
- recessive
- homozygous
- heterozygous
- genotype
- phenotype
- Punnett square
- carrier
- polydactyly
- cystic fibrosis
📝 DNA, Genes and Chromosomes
DNA: A long molecule found in the nucleus of cells. It has a double helix structure (two strands twisted around each other). DNA carries the genetic code that controls what characteristics an organism has.
Genome: The entire genetic material of an organism. The Human Genome Project sequenced the complete human genome, which helps identify genes linked to diseases.
Chromosomes: DNA is packaged into chromosomes. Human body cells contain 46 chromosomes (23 pairs). Gametes (sex cells) contain 23 chromosomes (one of each pair).
Gene: A short section of DNA on a chromosome that codes for a specific protein (and therefore a specific characteristic).
Allele: Different versions of the same gene. For example, the gene for eye colour has alleles for blue eyes and brown eyes.
| Term | Definition |
|---|---|
| Dominant allele | Only one copy is needed for the characteristic to be expressed (shown). Represented by a capital letter (e.g. B). |
| Recessive allele | Two copies are needed for the characteristic to be expressed. Represented by a lower case letter (e.g. b). |
| Homozygous | Both alleles are the same (e.g. BB or bb). |
| Heterozygous | The two alleles are different (e.g. Bb). |
| Genotype | The combination of alleles an organism has for a gene (e.g. BB, Bb, or bb). |
| Phenotype | The physical characteristic that is expressed (what you see, e.g. brown eyes or blue eyes). |
Example 1
In pea plants, the allele for tall (T) is dominant over the allele for short (t). A plant has the genotype Tt. State the phenotype and explain why.
Solution:
Phenotype: tall.
The T allele is dominant, so only one copy is needed for the tall characteristic to be expressed. The t allele is recessive and is masked by the dominant T allele. Therefore the plant is tall.
📝 Punnett Squares
Punnett squares are used to predict the genotypes and phenotypes of offspring from a genetic cross.
Example 2
In pea plants, tall (T) is dominant over short (t). Cross two heterozygous tall plants (Tt × Tt).
Solution:
| T | t | |
| T | TT | Tt |
| t | Tt | tt |
Genotype ratio: 1 TT : 2 Tt : 1 tt
Phenotype ratio: 3 tall : 1 short
Probability of tall offspring = 3/4 = 75%
Probability of short offspring = 1/4 = 25%
Example 3
In pea plants, tall (T) is dominant over short (t). Cross a homozygous tall plant (TT) with a short plant (tt).
Solution:
| T | T | |
| t | Tt | Tt |
| t | Tt | Tt |
All offspring have genotype Tt (heterozygous).
All offspring have the tall phenotype (100% tall).
The short phenotype is not expressed because T is dominant over t.
Example 4
In mice, black coat (B) is dominant over brown coat (b). A black mouse is crossed with a brown mouse. Half the offspring are black and half are brown. What are the genotypes of the parents?
Solution:
The brown mouse must be bb (homozygous recessive, because brown is recessive).
The black mouse must be Bb (heterozygous). If it were BB, all offspring would be black.
Cross: Bb × bb
| B | b | |
| b | Bb | bb |
| b | Bb | bb |
Genotype ratio: 1 Bb : 1 bb
Phenotype ratio: 1 black : 1 brown (matches the observed results).
📝 Sex Determination
Sex chromosomes: In humans, sex is determined by the 23rd pair of chromosomes. Females have XX and males have XY.
Example 5
Show the Punnett square for sex determination and state the probability of having a boy or a girl.
Solution:
| X | Y | |
| X | XX | XY |
| X | XX | XY |
Probability of female (XX) = 2/4 = 50%
Probability of male (XY) = 2/4 = 50%
The sex of the baby is determined by whether the sperm carries an X or a Y chromosome.
📝 Genetic Disorders
| Feature | Polydactyly | Cystic Fibrosis |
|---|---|---|
| Type of allele | Dominant | Recessive |
| Effect | Extra fingers or toes | Thick sticky mucus in lungs, pancreas and digestive system; breathing and digestion problems |
| Inheritance | Only one copy of the allele needed (Dd or DD) | Two copies needed (ff); carriers have one copy (Ff) but are unaffected |
| Can carriers be identified? | No carriers — if you have the allele, you have the condition | Yes — carriers have genotype Ff and are unaffected but can pass on the allele |
| Treatment | Surgery to remove extra digits | Physiotherapy, antibiotics, enzyme supplements; no cure |
Example 6
Polydactyly is caused by a dominant allele (D). A person with polydactyly who is heterozygous (Dd) has a child with someone who does not have polydactyly (dd). What is the probability their child will have polydactyly?
Solution:
| D | d | |
| d | Dd | dd |
| d | Dd | dd |
Genotype ratio: 1 Dd : 1 dd
Phenotype ratio: 1 polydactyly : 1 no polydactyly
Probability of child having polydactyly = 2/4 = 50%
Example 7
Cystic fibrosis is caused by a recessive allele (f). Both parents are carriers (Ff). What is the probability their child will have cystic fibrosis? What is the probability the child will be a carrier?
Solution:
| F | f | |
| F | FF | Ff |
| f | Ff | ff |
Genotype ratio: 1 FF : 2 Ff : 1 ff
Probability of cystic fibrosis (ff) = 1/4 = 25%
Probability of being a carrier (Ff) = 2/4 = 50%
Probability of being unaffected and not a carrier (FF) = 1/4 = 25%
❓ Practice Questions
Q1: Define the terms: gene, allele, genotype, phenotype, dominant, recessive.
Q2: In rabbits, black coat (B) is dominant over white coat (b). A heterozygous black rabbit is crossed with a white rabbit. Use a Punnett square to show the expected offspring ratios.
Q3: Explain the difference between homozygous and heterozygous. Give an example of each using the allele for cystic fibrosis.
Q4: Polydactyly is caused by a dominant allele. Explain why two parents who both have polydactyly could have a child without polydactyly.
Q5: Both parents are carriers for cystic fibrosis (Ff). They already have one child with cystic fibrosis. What is the probability their next child will also have cystic fibrosis? Explain your answer.
Q6: Higher Explain how sex is determined in humans and why there is a 50% probability of each sex.
✅ Answers
- Gene: a short section of DNA that codes for a specific protein. Allele: different versions of the same gene. Genotype: the combination of alleles an organism has. Phenotype: the physical characteristic expressed. Dominant: allele that is expressed with only one copy. Recessive: allele that is only expressed with two copies.
- Bb × bb. Punnett square: Bb, Bb, bb, bb. Genotype ratio: 1 Bb : 1 bb. Phenotype ratio: 1 black : 1 white (50% each).
- Homozygous means both alleles are the same (e.g. FF = unaffected, ff = has cystic fibrosis). Heterozygous means the two alleles are different (e.g. Ff = carrier, unaffected but carries one recessive allele).
- Both parents could be heterozygous (Dd). If Dd × Dd, there is a 1/4 (25%) chance of a child being dd (no polydactyly). The recessive allele d from each parent combines to give dd.
- The probability remains 1/4 = 25%. Each pregnancy is an independent event — the genotype of previous children does not affect the probability for the next child. Ff × Ff always gives 1/4 chance of ff.
- Sex is determined by the 23rd pair of chromosomes: XX = female, XY = male. The mother always contributes an X chromosome. The father contributes either X or Y. Punnett square: XX, XX, XY, XY. Therefore 50% chance of female (XX) and 50% chance of male (XY). The sperm determines the sex of the offspring.
🎯 Exam Tips
- Always use the correct letter notation: capital for dominant, lower case for recessive (e.g. B and b)
- When drawing Punnett squares, put one parent's alleles across the top and the other down the side
- Always state BOTH the genotype ratio AND phenotype ratio in your answer
- Remember: polydactyly is DOMINANT (only one allele needed), cystic fibrosis is RECESSIVE (two alleles needed)
- Carriers have one recessive allele but do NOT show the characteristic (they are heterozygous)
- You cannot be a carrier for a dominant disorder — if you have the allele, you have the condition
- Each pregnancy is an INDEPENDENT event — previous children do not affect the probability for the next
- Sex is determined by the sperm (X or Y), not the egg (always X)
🔢 Maths Skills
Mathematical Skills
Punnett square probability: A Punnett square shows all possible offspring genotypes. To calculate probability: count the number of boxes showing a particular outcome and divide by 4 (total boxes). For Tt × Tt: 1 TT, 2 Tt, 1 tt → probability of tall = 3/4 = 75%, probability of short = 1/4 = 25%. Express probabilities as fractions, decimals or percentages. For carrier parents (Ff × Ff) of cystic fibrosis: probability of affected child (ff) = 1/4 = 25%, carrier (Ff) = 2/4 = 50%, unaffected non-carrier (FF) = 1/4 = 25%. Each pregnancy is an independent event.
⚠️ Common Misconceptions
Watch Out!
1. Dominant alleles are more common. Wrong: dominant alleles are always the most frequent in a population. Correct: dominant refers to how an allele is expressed (only one copy needed), not how common it is — many recessive alleles are more common than dominant ones (e.g. the allele for polydactyly is dominant but rare).
2. Being a carrier means having the disease. Wrong: a carrier of a genetic disorder has the condition. Correct: a carrier has one recessive allele for the disorder but does not show symptoms because the dominant allele masks the recessive one; they are heterozygous and unaffected but can pass the allele to offspring.
✍️ 6-Mark Question
Extended Answer
6 marks: Explain genetic inheritance using a Punnett square.
Cystic fibrosis is caused by a recessive allele (f), so two copies are needed for the condition. When both parents are carriers (Ff), a Punnett square can predict the offspring genotypes. The father's gametes carry either F or f, and the mother's gametes carry either F or f. The Punnett square shows: FF (25%), Ff (50%), ff (25%). The genotype ratio is 1 FF : 2 Ff : 1 ff. Phenotypically, 75% of children will not have cystic fibrosis (FF + Ff), while 25% will have cystic fibrosis (ff). There is a 50% chance of each child being a carrier (Ff). Each pregnancy is an independent event, so the probability remains the same regardless of previous children. Carriers are unaffected because the dominant F allele produces enough functional protein.
Mark scheme: 1 mark — correct parental genotypes identified; 1 mark — correct gametes shown; 1 mark — Punnett square drawn correctly; 1 mark — correct genotype ratio; 1 mark — correct phenotype ratio with percentages; 1 mark — explanation of carriers and independence of events.
📊 AO3: Analyse & Evaluate
Analysis and Evaluation
In a family, the parents are unaffected by cystic fibrosis but have a child with the condition. Draw a Punnett square to explain how this is possible. Their other two children are unaffected. What is the probability that each unaffected child is a carrier? If one of the unaffected children has a child with a partner who is also a carrier, what is the probability their child will have cystic fibrosis? Justify your answer using a second Punnett square.
📝 Exam Questions by Topic
🎬 Video Resources
Ready to ace your GCSE Combined Science exams?
Get the best revision books and guides to boost your grades.
🧠 Flashcards (Spaced Repetition)
📝 Exam Questions by Topic
🎯 Target Tests (Auto-Graded)
📝 Exam Questions by Topic
📄 Past Papers for Biology (AQA)
- AQA Past Papers & Mark Schemes
- Edexcel Past Papers & Mark Schemes
- OCR Past Papers & Mark Schemes
- Eduqas Past Papers & Mark Schemes
- CCEA Past Papers & Mark Schemes
For the most accurate and up-to-date past papers, always check the official exam board websites.