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GCSE · Biology

B6a DNA, Genetics and Inheritance

AQA 8464 · Calculator allowed · about 100 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

Mendel's Peas: The Birth of Genetics

Original text written for Revision Library.

In the 1860s, an Austrian monk named Gregor Mendel spent several years breeding pea plants in the garden of his monastery, carefully recording how characteristics such as flower colour and stem height were passed from one generation to the next. By counting thousands of offspring, Mendel showed that characteristics are controlled by discrete 'factors' (now called alleles) that are inherited from each parent, rather than blending together as many scientists of his time believed. He also showed that some factors are dominant over others, and that the ratios of characteristics in offspring follow predictable patterns of probability. Mendel's work was largely ignored during his lifetime, but was rediscovered decades later and became the foundation of modern genetics. Today, the same principles Mendel discovered - alleles, dominance and predictable ratios - are used to understand how genes are inherited, how sex is determined, and how genetic disorders such as cystic fibrosis and sickle-cell anaemia are passed from parents to their children.

Deoxyribonucleic acid (DNA) is the molecule that carries genetic information in almost every living cell.

(a)State the shape formed by a molecule of DNA.(1)

(b)Define the term 'genome'.(1)

(c)Define the term 'gene'.(1)

(d)State the name of the structures, found in the nucleus of a cell, on which genes are located.(1)

(Total for Question 1 is 4 marks)

Different forms of the same gene are called alleles.

(a)Define the term 'allele'.(1)

(b)Define the term 'genotype'.(1)

(c)Define the term 'phenotype'.(1)

(Total for Question 2 is 3 marks)

In a genetic cross, the probability that an offspring has a particular genotype is 3 in 8.

(a)Write this probability as a fraction, in its simplest form.(1)

(b)Write this probability as a decimal.(1)

(c)Write this probability as a percentage.(1)

(Total for Question 3 is 3 marks)

In guinea pigs, black fur is caused by a dominant allele, B. White fur is caused by the recessive allele, b. Priya owns two guinea pigs that are both heterozygous (Bb) for fur colour, and breeds them together.

(a)Complete a genetic cross diagram (Punnett square) to show the possible genotypes of the offspring.(2)

Figure (to be drawn): Blank Punnett square grid provided: parents Bb x Bb, for the student to complete.

(b)Using your diagram, state the genotype ratio of the offspring.(1)

(c)Calculate the percentage of offspring expected to have white fur.(1)

(Total for Question 4 is 4 marks)

Gametes (sex cells) are produced by a type of cell division called meiosis. A normal human body (somatic) cell contains 46 chromosomes, arranged in 23 pairs.

(a)Name the type of cell division that produces gametes.(1)

(b)State the number of chromosomes found in a normal human sperm cell.(1)

(c)State the number of chromosomes that will be present in a fertilised egg cell (zygote) formed from this sperm cell and a normal human egg cell.(1)

(d)Explain why gametes must contain half the number of chromosomes of a normal body cell.(1)

(Total for Question 5 is 4 marks)

Twins can be identical (monozygotic) or non-identical (dizygotic).

(a)Describe how identical (monozygotic) twins form.(1)

(b)Describe how non-identical (dizygotic) twins form.(1)

(c)Explain why identical twins always have the same genotype, but non-identical twins do not.(2)

(Total for Question 6 is 4 marks)

In humans, sex is determined by one pair of chromosomes. Females have the genotype XX; males have the genotype XY. A couple from Leeds are planning to have three children.

(a)State the sex chromosome(s) found in a normal human egg cell.(1)

(b)State the sex chromosome(s) found in human sperm cells.(1)

(c)State the probability that any one of their children will be a girl.(1)

(d)Calculate the probability that all three of the couple's children will be girls.(2)

(Total for Question 7 is 5 marks)

Some people can taste a bitter chemical called PTC (phenylthiocarbamide), found in vegetables such as broccoli and Brussels sprouts; other people cannot taste it at all. The ability to taste PTC is caused by a dominant allele, T. The inability to taste PTC is caused by the recessive allele, t. Priya can taste PTC. Her partner, Daniel, cannot taste PTC. Priya and Daniel already have one child, Freya, who cannot taste PTC.

(a)State the genotype of Daniel, who cannot taste PTC.(1)

(b)Priya's genotype could be TT or Tt. Using the information about Freya, explain which genotype Priya must have.(2)

(c)Priya and Daniel plan to have another child. Calculate the probability, as a percentage, that this child will be able to taste PTC.(2)

(Total for Question 8 is 5 marks)

In Labrador dogs, black coat colour is caused by a dominant allele, B. Chocolate (brown) coat colour is caused by the recessive allele, b. A dog breeder has a black Labrador, but does not know whether its genotype is BB or Bb.

(a)State the possible genotype(s) of the black Labrador.(1)

(b)Describe how the breeder could carry out a test cross to find out the black Labrador's genotype.(2)

(c)If the black Labrador's genotype is Bb, state the expected ratio of black to chocolate puppies produced by this test cross.(1)

(d)State what result from the test cross would show that the black Labrador's genotype is BB.(1)

(Total for Question 9 is 5 marks)

Required practical: A student extracts DNA from kiwi fruit. The method is: (1) mash the kiwi fruit with a pestle and mortar; (2) mix the pulp with a detergent and salt solution; (3) filter the mixture through muslin into a clean test tube; (4) gently pour ice-cold ethanol down the side of the tube, so that it forms a layer on top of the filtered extract.

(a)State the purpose of adding detergent to the fruit pulp.(1)

(b)State the purpose of adding salt to the mixture.(1)

(c)State the purpose of filtering the mixture through muslin.(1)

(d)Explain why the ethanol used in this practical should be ice-cold.(2)

(e)Name the general group of enzymes that could break down DNA if the extract is not kept cold.(1)

(f)Describe what the student should observe at the boundary between the ethanol layer and the filtered extract.(1)

(g)Explain why the extracted DNA appears as a stringy, clumped precipitate, rather than as separate, individual DNA molecules.(2)

(h)Suggest one variable that should be kept the same to fairly compare the amount of DNA extracted from different fruits.(1)

(i)Suggest one reason why kiwi fruit is often chosen for this practical.(1)

(j)State one safety precaution that should be taken when carrying out this practical.(1)

(k)The student repeats the practical using strawberry instead of kiwi fruit, and finds that a smaller mass of DNA precipitate is produced from the same mass of fruit. Suggest one reason for this difference.(1)

(l)The student weighs 5 g of mashed kiwi fruit and extracts a DNA precipitate with a mass of 0.004 g. Calculate the percentage of the fruit's mass, by mass, that was extracted as DNA precipitate.(2)

(Total for Question 10 is 15 marks)

Sickle-cell anaemia is an inherited blood disorder caused by a recessive allele, a. The dominant allele, A, produces normal red blood cells. A person with genotype Aa is described as having 'sickle-cell trait'; they do not have sickle-cell anaemia, but carry one copy of the sickle-cell allele. Kwame and Naomi are both known carriers (Aa) of the sickle-cell allele, and are expecting their first child.

(a)Complete a genetic cross diagram to find the possible genotypes of their child, and use it to state the probability, as a fraction, that the child will have sickle-cell anaemia.(2)

Figure (to be drawn): Blank Punnett square grid provided: parents Aa x Aa, for the student to complete.

(b)State the probability that the child will have sickle-cell trait (genotype Aa) rather than sickle-cell anaemia.(1)

(c)State the probability that the child will have genotype AA (unaffected, and not a carrier).(1)

(d)Higher tier only. Kwame and Naomi plan to have a second child. Calculate the probability, as a percentage, that both of their children will have sickle-cell anaemia.(2)

(Total for Question 11 is 6 marks)

Albinism is a condition in which a person's skin, hair and eyes lack the pigment melanin. In one family, both parents, Grace and Michael, are unaffected by albinism. They have three children: Aisha (unaffected), Tomasz (who has albinism), and Ellie (unaffected).

(a)Use this information to explain whether the allele that causes albinism is dominant or recessive.(2)

(b)Using A for the dominant (normal) allele and a for the recessive (albinism) allele, state the genotypes of Grace and Michael.(2)

(c)State the genotype of Tomasz.(1)

(d)Aisha does not have albinism. State the possible genotype(s) of Aisha.(1)

(Total for Question 12 is 6 marks)

In fruit flies, grey body colour is caused by a dominant allele, G. Black body colour is caused by the recessive allele, g. A genetics student crosses two heterozygous fruit flies (Gg x Gg) and rears 320 offspring from the cross.

(a)State the expected phenotype ratio of the offspring.(1)

(b)Calculate the expected number of black-bodied flies among the 320 offspring.(2)

(c)The student actually counted 68 black-bodied flies. Calculate the percentage difference between the expected number of black-bodied flies (from part (b)) and the actual number counted, as a percentage of the expected number.(2)

(Total for Question 13 is 5 marks)

Higher tier only. Red-green colour blindness is caused by a recessive allele carried on the X chromosome. The dominant allele, X^B, gives normal colour vision; the recessive allele, X^b, causes colour blindness. Sofia has normal colour vision but is a carrier (genotype X^B X^b). Her partner, Tom, has normal colour vision (genotype X^B Y).

(a)State the genotype of a man with red-green colour blindness.(1)

(b)Complete a genetic cross diagram to show the possible genotypes of Sofia and Tom's children.(2)

Figure (to be drawn): Blank Punnett square grid provided: mother X^B X^b x father X^B Y, for the student to complete.

(c)Calculate the probability that a son of Sofia and Tom will be colour-blind.(1)

(d)Calculate the probability that a daughter of Sofia and Tom will be colour-blind.(1)

(e)Explain why red-green colour blindness is much more common in men than in women.(2)

(Total for Question 14 is 7 marks)

Huntington's disease is an inherited disorder that damages the nervous system. It is caused by a dominant allele, H; the recessive allele, h, does not cause the disease. Symptoms do not usually appear until a person is 30 to 50 years old. Elin is heterozygous (Hh) for the Huntington's allele. Her partner, Rhys, does not carry the allele (hh).

(a)Complete a genetic cross diagram and use it to calculate the probability, as a percentage, that a child of Elin and Rhys will inherit the Huntington's allele.(2)

Figure (to be drawn): Blank Punnett square grid provided: parents Hh x hh, for the student to complete.

(b)Predictive genetic testing can tell a person whether they have inherited the Huntington's allele, years before any symptoms would appear. State one possible advantage, and one possible disadvantage, of a person choosing to have this test.(2)

(Total for Question 15 is 4 marks)

Phenylketonuria (PKU) is an inherited metabolic disorder caused by a recessive allele, p. The dominant allele, P, does not cause PKU. Josh and Chloe are both known carriers (Pp) of the PKU allele.

(a)Complete a genetic cross diagram to find the possible genotypes of their children, and state the probability, as a fraction, that a child will have PKU.(2)

Figure (to be drawn): Blank Punnett square grid provided: parents Pp x Pp, for the student to complete.

(b)Josh and Chloe's first child does not have PKU. State the probability that their second child will have PKU, and explain your answer.(1)

(c)Higher tier only. Calculate the probability, as a percentage, that if Josh and Chloe have three children in total, none of the three will have PKU.(2)

(Total for Question 16 is 5 marks)

Higher tier only. Cystic fibrosis is an inherited disorder caused by a recessive allele. When both parents are carriers, each of their children has a probability of 1/4 of having cystic fibrosis. A couple, who are both carriers, are expecting a baby.

(a)State the rule used to combine the probabilities of two independent genetic events, such as a baby's sex and whether it inherits a genetic disorder.(1)

(b)Using the probability that the baby has cystic fibrosis (1/4) and the probability that the baby is born male (1/2), calculate the probability that the baby will be a boy with cystic fibrosis.(2)

(c)Explain why this combined probability is lower than the probability of the baby having cystic fibrosis alone.(1)

(Total for Question 17 is 4 marks)

Explain how the processes of meiosis and fertilisation can produce brothers and sisters (siblings) who are genetically different from each other, even though they share the same two parents.

(Total for Question 18 is 6 marks)

Mark scheme · B6a DNA, Genetics and Inheritance

Question 1

(a) B1 double helix
(a) Answer: A double helix.
(b) B1 the entire genetic material (all of the DNA) of an organism, oe
(b) Answer: The entire genetic material of an organism.
(c) B1 a short section of DNA (found on a chromosome) that codes for a particular protein, and so controls a specific characteristic, oe
(c) Answer: A short section of DNA that codes for a particular protein and controls a characteristic.
(d) B1 chromosomes
(d) Answer: Chromosomes.

Question 2

(a) B1 one of the alternative/different forms of a gene, oe
(a) Answer: One of the different forms of a gene.
(b) B1 the combination of alleles an organism has for a particular characteristic, oe
(b) Answer: The combination of alleles an organism has for a characteristic.
(c) B1 the observable physical characteristics of an organism, resulting from its genotype (and, in some cases, the environment), oe
(c) Answer: The observable physical characteristics of an organism.

Question 3

(a) B1 3/8 cao (already in simplest form)
(a) Answer: 3/8
(b) B1 0.375 cao
(b) Answer: 0.375
(c) B1 37.5% cao, ft from (b)
(c) Answer: 37.5%

Question 4

(a) M1 correct gametes shown for each parent: B and b
(a) A1 correct offspring genotypes shown: BB, Bb, Bb, bb
(a) Answer: Offspring genotypes: BB, Bb, Bb, bb.
(b) B1 1 BB : 2 Bb : 1 bb (accept 1:2:1) ft from (a)
(b) Answer: 1 BB : 2 Bb : 1 bb
(c) B1 25% cao, ft from (a)
(c) Answer: 25%

Question 5

(a) B1 meiosis
(a) Answer: Meiosis.
(b) B1 23
(b) Answer: 23 chromosomes.
(c) B1 46, ft from (b) doubled
(c) Answer: 46 chromosomes.
(d) B1 so that when two gametes fuse at fertilisation, the normal (diploid) chromosome number is restored in the offspring, rather than the chromosome number doubling every generation, oe
(d) Answer: So that fertilisation restores the normal chromosome number, instead of it doubling every generation.

Question 6

(a) B1 one egg cell is fertilised by one sperm cell, forming a single zygote, which then splits to form two separate embryos, oe
(a) Answer: One egg is fertilised by one sperm, forming a single zygote that later splits into two embryos.
(b) B1 two separate egg cells are released and each is fertilised by a different sperm cell, forming two separate zygotes, oe
(b) Answer: Two separate eggs are each fertilised by a different sperm cell, forming two separate zygotes.
(c) B1 identical twins develop from a single zygote (one fertilisation event) that splits, so both embryos contain exactly the same DNA/genotype
(c) B1 non-identical twins develop from two separate zygotes, formed from two genetically different eggs and two genetically different sperm cells, so, like ordinary siblings, they share only some of their alleles, oe
(c) Answer: Identical twins come from one zygote splitting, so they share identical DNA; non-identical twins come from two separate zygotes formed from different egg and sperm cells, so they share alleles only as ordinary siblings do.

Question 7

(a) B1 X (only)
(a) Answer: X.
(b) B1 X or Y
(b) Answer: X or Y.
(c) B1 1/2 (50%), ft from (a) and (b)
(c) Answer: 1/2 (50%).
(d) M1 1/2 x 1/2 x 1/2, ft from (c)
(d) A1 1/8 (12.5%) cao
(d) Answer: 1/8 (12.5%)

Question 8

(a) B1 tt
(a) Answer: tt
(b) B1 Freya cannot taste PTC, so Freya's genotype must be tt, meaning Freya inherited a t allele from each parent
(b) B1 Daniel (tt) can only pass on a t allele; for Freya to be tt, Priya must also have passed on a t allele, so Priya's genotype must be Tt (not TT, as a TT parent could only pass on a T allele)
(b) Answer: Priya must be Tt, because Freya (tt) must have inherited a t allele from Priya as well as from Daniel; if Priya were TT she could not pass on a t allele.
(c) M1 Tt x tt gives offspring genotypes Tt, Tt, tt, tt, ft from (b)
(c) A1 50% cao
(c) Answer: 50%

Question 9

(a) B1 BB or Bb
(a) Answer: BB or Bb.
(b) B1 breed the black Labrador with a chocolate (bb) Labrador
(b) B1 look at the coat colours (phenotypes) of the puppies produced, oe
(b) Answer: Breed the black Labrador with a chocolate (bb) Labrador, then look at the coat colours of the puppies produced.
(c) B1 1 black : 1 chocolate (1:1)
(c) Answer: 1 black : 1 chocolate.
(d) B1 all/every puppy produced is black, with no chocolate puppies, oe
(d) Answer: If every puppy produced is black, with no chocolate puppies.

Question 10

(a) B1 breaks down the cell membranes (and nuclear membrane) of the cells, releasing the DNA into solution, oe
(a) Answer: It breaks down the cell membranes, releasing the DNA into solution.
(b) B1 causes the DNA molecules to clump/stick together, helping the DNA to precipitate out of solution later, oe
(b) Answer: It causes the DNA molecules to clump together, helping the DNA to precipitate later.
(c) B1 removes/separates solid pieces of fruit debris (cell wall/pulp fragments) from the liquid extract containing the dissolved DNA, oe
(c) Answer: It removes solid fruit debris, leaving the liquid extract containing the DNA.
(d) B1 DNA is not soluble in cold ethanol (unlike in water), so keeping the ethanol ice-cold ensures the DNA precipitates out of solution rather than dissolving, oe
(d) B1 the cold temperature also slows/reduces the activity of any enzymes present that could break down the DNA, oe
(d) Answer: DNA is insoluble in cold ethanol, so it precipitates rather than dissolving; the cold temperature also slows enzymes that could break the DNA down.
(e) B1 DNases (DNA-degrading enzymes)
(e) Answer: DNases.
(f) B1 a white, cloudy/stringy precipitate of DNA forms at the boundary between the two layers, oe
(f) Answer: A white, stringy precipitate of DNA forms at the boundary between the two layers.
(g) B1 each cell contains a very large number of DNA molecules, and the salt and ethanol cause many of these molecules to stick/clump together
(g) B1 a single DNA molecule is far too thin to be seen without a microscope, so only the large clump of many strands together is visible to the naked eye, oe
(g) Answer: Many thousands of DNA molecules clump together due to the salt and ethanol; a single DNA molecule is far too thin to see, but the clump of many strands is visible.
(h) B1 any one of: mass of fruit used, volume/concentration of the detergent-salt solution, temperature, or time allowed for each step, oe
(h) Answer: For example, the mass of fruit used should be kept the same for each sample.
(i) B1 it is soft and easy to mash, so its cells are broken open quickly and easily, releasing DNA efficiently, oe
(i) Answer: It is soft and easy to mash, so its cells release DNA easily.
(j) B1 e.g. wear eye protection, and keep the ethanol away from any naked flame, as it is flammable, oe
(j) Answer: Wear eye protection and keep the flammable ethanol away from any naked flame.
(k) B1 e.g. different fruits contain different amounts of DNA per gram, or their cells are broken open with different efficiency during mashing, oe
(k) Answer: Different fruits can contain different amounts of DNA per gram, or their cells may break open less easily when mashed.
(l) M1 (0.004 / 5) x 100, oe
(l) A1 0.08% cao
(l) Answer: 0.08%

Question 11

(a) M1 correct offspring genotypes shown: AA, Aa, Aa, aa
(a) A1 1/4 cao
(a) Answer: 1/4
(b) B1 1/2 (50%), ft from (a)
(b) Answer: 1/2 (50%)
(c) B1 1/4 (25%), ft from (a)
(c) Answer: 1/4 (25%)
(d) M1 1/4 x 1/4, ft from (a)
(d) A1 1/16 (6.25%) cao
(d) Answer: 1/16 (6.25%)

Question 12

(a) B1 the allele must be recessive
(a) B1 because two unaffected parents (who do not show albinism) have had a child, Tomasz, who does have albinism; this can only happen if the allele is recessive and both parents are unaffected carriers (heterozygous), masking the allele, oe
(a) Answer: The allele must be recessive, because two unaffected parents produced a child with albinism, which can only happen if both parents are unaffected carriers masking a hidden recessive allele.
(b) B1 Grace: Aa
(b) B1 Michael: Aa
(b) Answer: Both Grace and Michael are Aa.
(c) B1 aa
(c) Answer: aa
(d) B1 AA or Aa, ft from (b)
(d) Answer: AA or Aa

Question 13

(a) B1 3 grey : 1 black
(a) Answer: 3 grey : 1 black
(b) M1 (1/4) x 320, ft from (a)
(b) A1 80 cao
(b) Answer: 80
(c) M1 (80 - 68) / 80 x 100, ft from (b)
(c) A1 15% cao
(c) Answer: 15%

Question 14

(a) B1 X^b Y
(a) Answer: X^b Y
(b) M1 correct gametes shown: X^B and X^b from Sofia, X^B and Y from Tom
(b) A1 correct offspring genotypes shown: X^B X^B, X^B X^b, X^B Y, X^b Y
(b) Answer: Offspring genotypes: X^B X^B, X^B X^b, X^B Y, X^b Y.
(c) B1 1/2 (50%), ft from (b)
(c) Answer: 1/2 (50%)
(d) B1 0%, ft from (b)
(d) Answer: 0%
(e) B1 colour blindness is caused by a recessive allele on the X chromosome; a man has only one X chromosome, so if he inherits the recessive allele he will be colour-blind, as there is no second X chromosome carrying a dominant allele to mask it
(e) B1 a woman has two X chromosomes, so she would need to inherit the recessive allele on both X chromosomes to be colour-blind; if she inherits just one recessive allele she is an unaffected carrier, because the dominant allele on her other X chromosome is expressed instead, oe
(e) Answer: Men have only one X chromosome, so a single recessive allele causes colour blindness; women have two X chromosomes, so they need the recessive allele on both to be affected, otherwise they are unaffected carriers.

Question 15

(a) M1 correct offspring genotypes shown: Hh, Hh, hh, hh
(a) A1 50% cao
(a) Answer: 50%
(b) B1 advantage, e.g. allows the person to plan for the future (family, finances, career) with accurate information, or allows relatives to make informed decisions about having children, oe
(b) B1 disadvantage, e.g. may cause significant psychological distress/anxiety, especially as there is currently no cure or treatment to prevent the disease developing, or could affect the person's ability to get life insurance, oe
(b) Answer: Advantage: allows informed planning for the future. Disadvantage: may cause significant anxiety, especially as there is no cure available.

Question 16

(a) M1 correct offspring genotypes shown: PP, Pp, Pp, pp
(a) A1 1/4 cao
(a) Answer: 1/4
(b) B1 still 1/4, ft from (a); because each pregnancy is an independent event, so the genotype of an earlier child does not affect the probability for a later child, oe
(b) Answer: Still 1/4, because each pregnancy is an independent event and is not affected by the outcome of an earlier pregnancy.
(c) M1 3/4 x 3/4 x 3/4, ft from (a)
(c) A1 awrt 42.2% (27/64) cao
(c) Answer: 27/64 (approximately 42.2%)

Question 17

(a) B1 multiply the individual probabilities together, oe
(a) Answer: Multiply the individual probabilities together.
(b) M1 1/4 x 1/2, ft from (a)
(b) A1 1/8 (12.5%) cao
(b) Answer: 1/8 (12.5%)
(c) B1 the combined event requires two separate, independent conditions to both occur; multiplying two probabilities that are each less than 1 always produces a smaller value than either probability alone, oe
(c) Answer: Because it requires two independent conditions to both happen, and multiplying two probabilities that are each less than 1 always gives a smaller result than either probability on its own.

Question 18

Level 3 (5-6): A detailed, logically sequenced answer that correctly explains how independent segregation during meiosis produces genetically different gametes AND how random fertilisation combines gametes from each parent, clearly linking both processes to why siblings (other than identical twins) are genetically different from one another.
Level 2 (3-4): The answer describes meiosis producing genetically different gametes, or random fertilisation, with only partial or one-sided linkage to why siblings differ genetically; some scientific detail may be missing or muddled.
Level 1 (1-2): One or two isolated, relevant points are made about meiosis or fertilisation, with little or no linkage to genetic differences between siblings.
Level 0 (0): No relevant content.
Indicative content:
- Meiosis is the type of cell division that produces gametes (sperm cells and egg cells).
- During meiosis, each pair of chromosomes separates (segregates) independently and randomly into different gametes.
- This means each gamete produced contains a different, random mixture of the parent's chromosomes (and therefore alleles), so gametes produced by the same parent are genetically different from one another.
- Meiosis also halves the chromosome number, producing haploid gametes (23 chromosomes in a human gamete) from a diploid parent cell (46 chromosomes).
- Fertilisation is the fusion of a haploid sperm cell with a haploid egg cell, which restores the full (diploid) chromosome number in the resulting zygote.
- Fertilisation is random: any one of millions of genetically different sperm cells could fuse with the egg cell, adding a further random/unpredictable element.
- Because each parent produces many genetically different gametes, and fertilisation randomly combines one gamete from each parent, each zygote receives a unique combination of alleles.
- This is why siblings, although they inherit their alleles from the same two parents, are not genetically identical to one another (except identical twins, who develop from a single zygote).

DNA, Genetics and Inheritance - Worksheets, Questions and Revision