MYP DP SAT AP Papers Pricing FAQ About
MYP 3 · Science

Reproduction and Genetics (intro)

40 questions across 2 sub-topics

Use the Sub-Topic filter above to focus on one.

Human Reproduction (basic) Variation and Inheritance (intro)

Human Reproduction (basic) 20 questions

QUESTION 1 2 marks Criterion A
Easy

State what is meant by the term gamete, and name the male and female gametes in humans.

Show complete worked solution

A gamete is a sex cell (reproductive cell) that joins with another gamete during fertilisation to form a new individual. Each gamete carries half the normal number of chromosomes.

The male gamete is the sperm (spermatozoon); the female gamete is the egg cell (ovum).

QUESTION 2 2 marks Criterion A
Easy
A B C D

The diagram shows a simplified front view of the human female reproductive system.

a. Identify the structure labelled A, and state its function.
[1]
b. Identify the structure labelled C, and state its function.
[1]
Show complete worked solution
(a)
A is the ovary. Its function is to produce egg cells (ova) and release one approximately every month (ovulation).
(b)
C is the uterus (womb). Its function is to hold and nourish a developing fetus during pregnancy; its lining thickens each month to prepare for a possible pregnancy.
QUESTION 3 2 marks Criterion A
Easy

State the approximate length of a full-term human pregnancy, and the approximate length of an average menstrual cycle.

Show complete worked solution

A full-term human pregnancy lasts about 40 weeks (around 9 months), counted from the first day of the last menstrual period.

An average menstrual cycle lasts about 28 days, although this varies between individuals (typically 21–35 days).

QUESTION 4 3 marks Criterion A
Medium
A B C

The diagram shows a simplified side view of the human male reproductive system.

a. Identify structure A and state its main function.
[1]
b. Identify structure B and state its function.
[1]
c. Identify structure C and state its function.
[1]
Show complete worked solution
(a)
A is the testis. It produces sperm cells and the hormone testosterone.
(b)
B is the sperm duct (vas deferens). It carries sperm from the testis towards the urethra.
(c)
C is the prostate gland. It produces a nutrient-rich fluid that is added to sperm to form semen, which helps to nourish and protect the sperm.
QUESTION 5 3 marks Criterion A
Medium

Fertilisation is the joining of a sperm cell and an egg cell to form a new individual.

a. Describe where fertilisation normally takes place in the human body, and what happens to the chromosomes of the sperm and egg when they join.
[2]
b. What is the name given to the single cell formed immediately after fertilisation?
[1]
Show complete worked solution
(a)
Fertilisation normally takes place in the oviduct (fallopian tube). When a sperm cell fuses with an egg cell, their nuclei combine so that the chromosomes from the sperm and the chromosomes from the egg join together, restoring the full number of chromosomes and forming a single new cell.
(b)
This cell is called a zygote.
QUESTION 6 4 marks Criterion A
Medium

Twins can form in two different ways.

a. Explain how identical twins form, and why they have the same genes.
[2]
b. Explain how non-identical (fraternal) twins form, and why they are no more genetically similar than any other siblings.
[2]
Show complete worked solution
(a)
Identical twins form when a single fertilised egg (zygote) splits into two separate embryos early in development. Because both embryos came from the same zygote, formed from the same sperm and the same egg, they have exactly the same genes (DNA).
(b)
Non-identical twins form when two separate eggs are released and fertilised by two separate sperm at around the same time, producing two separate zygotes. Since each twin develops from a different egg and a different sperm, they share genes in the same way as any other brother or sister, and can even be different sexes.
QUESTION 7 6 marks Criterion A
Hard

Doctors estimate a pregnant woman's due date by adding 40 weeks to the first day of her last menstrual period (LMP), since ovulation and fertilisation normally happen roughly 14 days after the start of a 28-day cycle.

a. A woman's last menstrual period began on 3 March. Assuming a regular 28-day cycle, on approximately what date would ovulation occur?
[2]
b. Using the 40-week rule, calculate the number of days from the LMP to the estimated due date.
[2]
c. If a different woman has a longer, 35-day cycle, explain why doctors might adjust her estimated due date to be later than the standard 280-day calculation.
[2]
Show complete worked solution
(a)
Ovulation occurs roughly $14$ days after the start of the cycle: $3\,\text{March} + 14\,\text{days} = 17\,\text{March}$. Ovulation would occur on approximately 17 March.
(b)
$$ 40\,\text{weeks} \times 7\,\text{days/week} = 280\,\text{days} $$ The estimated due date is $280$ days after the LMP.
(c)
In a longer cycle, ovulation happens later after the start of the period (around day $21$ instead of day $14$ for a $35$-day cycle), so fertilisation occurs later too. Because the standard $280$-day rule assumes ovulation on day $14$, using it unadjusted for a $35$-day cycle would underestimate the true age of the pregnancy, so doctors add the extra days (about a week) to give a more accurate, later due date.
QUESTION 8 6 marks Criterion B
Medium

A student wants to investigate whether the age at which the pubertal growth spurt begins differs between boys and girls, by surveying students at their school.

a. State the independent and dependent variables in this investigation.
[2]
b. State one variable that should be controlled, and explain why.
[2]
c. Describe a suitable method for collecting this data from a large sample of students.
[2]
Show complete worked solution
(a)
Independent variable: sex (boy or girl). Dependent variable: age (in years) at which the growth spurt began, as reported by each student.
(b)
The age range of students surveyed should be controlled (e.g. survey only students in the same year group). If the ages surveyed varied too widely, older students might simply be more likely to have already started their growth spurt regardless of sex, making the comparison unfair.
(c)
Distribute an anonymous questionnaire to a large sample of students (e.g. at least $30$ boys and $30$ girls) asking them to record their sex and, as accurately as they can recall or estimate with a parent's help, the age at which they noticed their growth spurt beginning. Collect the responses and calculate the mean age for boys and for girls separately, then compare the two means.
QUESTION 9 6 marks Criterion B
Medium

A student hypothesises that people who do regular intense exercise tend to have a longer average menstrual cycle length than those who do not.

a. Identify the independent and dependent variables for an investigation to test this hypothesis.
[2]
b. State two variables that should be controlled between the groups compared, and explain why for one of them.
[2]
c. Describe a method for collecting reliable data to test this hypothesis.
[2]
Show complete worked solution
(a)
Independent variable: level of regular exercise (e.g. hours of intense exercise per week). Dependent variable: average menstrual cycle length (in days).
(b)
Control age and use of hormonal contraception between the groups compared. Why control hormonal contraception: hormonal contraceptives can directly change cycle length regardless of exercise, so including people using them in either group would confuse the effect of exercise with the effect of the contraceptive.
(c)
Recruit a large sample of volunteers (e.g. at least $30$ per group) and ask each to record the first day of their period using a cycle-tracking diary or app for at least $3$ consecutive cycles, alongside a log of their weekly exercise. Calculate each person's mean cycle length over the $3$ cycles, then compare the mean cycle length of the high-exercise group with the low-exercise group.
QUESTION 10 8 marks Criterion B
Hard
GroupNumber of mothersMean birth weight (kg)
Smokers102.9
Non-smokers103.4

A researcher compared the average birth weight of babies born to 10 mothers who smoked during pregnancy with 10 mothers who did not, all recruited from a single small clinic in one town.

a. State one reason why using data from only 10 mothers in each group might make the conclusion unreliable.
[2]
b. State one variable, other than smoking, that should have been controlled between the two groups, and explain why.
[3]
c. Suggest one improvement to the method that would make the conclusion about smoking and birth weight more reliable.
[3]
Show complete worked solution
(a)
A sample of only $10$ per group is very small, so the results could easily be affected by chance (e.g. one or two unusually light or heavy babies), rather than truly reflecting the effect of smoking. A much larger sample is needed before a reliable conclusion can be drawn.
(b)
Maternal diet/nutrition should be controlled between the two groups. Poor nutrition, independently of smoking, is also known to reduce birth weight, so if the smokers in this study also happened to have poorer diets on average, the lower birth weight could be caused partly or wholly by diet rather than smoking, confusing the conclusion.
(c)
Recruit a much larger sample (e.g. hundreds of mothers) from multiple clinics in different towns, and match the smoking and non-smoking groups for other factors such as age, diet, and general health, so that smoking is the main difference between the groups. This would make it far more likely that any difference in birth weight is actually caused by smoking rather than by chance or another uncontrolled variable.
QUESTION 11 3 marks Criterion C
Easy
SexAverage age puberty begins (years)Average age of peak growth spurt (years)
Girls1012
Boys1114
a. According to the table, which sex generally begins puberty earlier?
[1]
b. Calculate the difference between the average age puberty begins and the average age of peak growth spurt, for boys.
[2]
Show complete worked solution
(a)
Girls generally begin puberty earlier (average age $10$) than boys (average age $11$).
(b)
$$14 - 11 = 3\,\text{years}$$ For boys, the peak growth spurt occurs on average $3$ years after puberty begins.
QUESTION 12 5 marks Criterion C
Medium
Day of cycle17142128
Oestrogen level (units)251042
a. On which day does oestrogen level peak, and what event in the cycle does this peak trigger?
[2]
b. Describe the overall pattern of oestrogen level shown by the table across the 28-day cycle.
[3]
Show complete worked solution
(a)
Oestrogen peaks on day 14. This peak triggers ovulation — the release of an egg from the ovary.
(b)
Oestrogen level rises from day $1$ to a peak around day $14$ (rising from $2$ to $10$ units), then falls from day $14$ to day $28$ (from $10$ back down to $2$ units), completing one full cycle before the pattern repeats in the next cycle.
QUESTION 13 4 marks Criterion C
Medium
AnimalHumanElephantDogMouse
Average gestation (days)2806406320
a. Calculate how many times longer an elephant's gestation period is than a mouse's.
[2]
b. Calculate the difference, in days, between the human and dog gestation periods.
[2]
Show complete worked solution
(a)
$$ \frac{640}{20} = 32 $$ An elephant's gestation period is $32$ times longer than a mouse's.
(b)
$$280 - 63 = 217\,\text{days}$$
QUESTION 14 5 marks Criterion C
Medium
Mother123456
Cigarettes per day051015200
Birth weight (kg)3.43.23.02.82.62.1
a. Identify the anomalous result in this table.
[1]
b. Explain how you identified it.
[2]
c. Suggest one other factor, besides smoking, that could explain mother 6's unusually low result.
[2]
Show complete worked solution
(a)
Mother $6$'s result is anomalous — birth weight of $2.1\,\text{kg}$ despite smoking $0$ cigarettes per day.
(b)
For mothers $1$ to $5$, birth weight decreases steadily as cigarettes per day increases ($3.4\to3.2\to3.0\to2.8\to2.6$). Mother $6$ smoked $0$ cigarettes, like mother $1$, so based on the pattern her birth weight should be close to $3.4\,\text{kg}$ — but it is much lower, at $2.1\,\text{kg}$, breaking the trend.
(c)
Other factors that affect birth weight include the baby being born premature (before full term), the mother's nutrition/diet during pregnancy, or a multiple pregnancy (twins) — any of these could explain a low birth weight unrelated to smoking.
QUESTION 15 5 marks Criterion C
Medium
Decade1980s1990s2000s2010s
Twin births per 1000 births19243033
a. Calculate the percentage increase in the twin birth rate from the 1980s to the 2010s.
[2]
b. Describe the trend shown by the data, and suggest one reason for it.
[3]
Show complete worked solution
(a)
$$ \frac{33-19}{19}\times100 = \frac{14}{19}\times100 = 73.7\%\ (\text{3 s.f.}) $$
(b)
The rate of twin births has steadily increased every decade shown, from $19$ per $1000$ births in the 1980s to $33$ per $1000$ births in the 2010s. One likely reason is the increased use of fertility treatments such as IVF, which sometimes involve implanting more than one embryo and can increase the chance of a multiple pregnancy.
QUESTION 16 8 marks Criterion C
Hard
HospitalABCD
Number of births recorded50040480510
Mean gestation length (days)279291280278
a. Identify which hospital's result is least reliable, and explain why.
[2]
b. Calculate the mean gestation length across hospitals A, C and D only, and compare it with the accepted average of 280 days.
[3]
c. Evaluate whether it would be reasonable to conclude that gestation length is genuinely longer at Hospital B than elsewhere.
[3]
Show complete worked solution
(a)
Hospital B's result is least reliable. It is based on only $40$ births, a much smaller sample than the other hospitals ($480$–$510$), so its mean is far more likely to be affected by chance and unusual cases.
(b)
$$ \text{mean} = \frac{279+280+278}{3} = \frac{837}{3} = 279\,\text{days} $$ This is very close to the accepted average of $280$ days, within $1$ day, showing good agreement between these three large-sample hospitals.
(c)
This conclusion would not be reasonable based on this data alone. Hospital B's much smaller sample size ($40$ compared to hundreds at the other hospitals) means its higher mean ($291$ days) could easily be due to chance variation rather than a real difference — for example, a few late deliveries recorded by coincidence. Before concluding a genuine difference exists, Hospital B would need to collect data from a much larger, comparable sample size.
QUESTION 17 7 marks Criterion C
Hard
Year1990200020102020
Average sperm count (million/mL)80685547
a. Calculate the percentage decrease in average sperm count from 1990 to 2020.
[2]
b. Describe the trend shown by the data.
[2]
c. A newspaper headline claims this data 'proves' that a single specific chemical is causing the decline. Evaluate this claim.
[3]
Show complete worked solution
(a)
$$ \frac{80-47}{80}\times100 = \frac{33}{80}\times100 = 41.25\% \approx 41\% $$
(b)
Average sperm count has steadily decreased over the three decades shown, falling from $80$ million/mL in 1990 to $47$ million/mL in 2020, a fall of roughly $10$–15 million/mL every decade.
(c)
This claim goes beyond what the data can show. The table only records a correlation between year and average sperm count — it gives no information about exposure to any specific chemical, or about other possible causes such as diet, obesity, smoking rates, or general lifestyle changes over the same 30 years. To support the claim that one specific chemical is responsible, a study would need to directly measure individuals' exposure to that chemical and control for these other factors, which this data does not do.
QUESTION 18 5 marks Criterion D
Medium
OutcomeBefore programmeAfter programme
Teenage pregnancy rate (per 1000)3018
Reported condom use (%)4568

A school district introduced a comprehensive sex education programme covering reproduction, contraception, and consent. The table shows data collected before and after the programme began.

Discuss one benefit and one drawback of the programme, using the data to support your discussion.

Show complete worked solution

Benefit: The teenage pregnancy rate fell substantially, from $30$ to $18$ per $1000$, and reported contraceptive (condom) use rose from $45\%$ to $68\%$. This means fewer teenagers are likely to face the health risks and disruption to education associated with an unplanned early pregnancy, and more are protecting themselves during sexual activity.

Drawback: Delivering a comprehensive programme requires significant teacher training, curriculum time, and funding, which must be taken from other parts of the timetable. In addition, reported contraceptive use, while much improved, is still not universal at $68\%$, meaning a substantial number of students may remain at risk of unplanned pregnancy or sexually transmitted infection without further improvement.

QUESTION 19 6 marks Criterion D
Medium
Maternal age (years)Under 3535–3738–3940+
Live birth rate per IVF cycle (%)3225155

IVF (in vitro fertilisation) is a fertility treatment in which eggs are fertilised with sperm outside the body before an embryo is placed in the uterus. The table shows how the chance of a live birth per IVF cycle changes with the mother's age.

Discuss one benefit and one drawback of IVF as a fertility treatment, referring to the data.

Show complete worked solution

Benefit: IVF gives people who cannot conceive naturally the chance to have a genetically related child. The data shows this chance is highest for younger mothers ($32\%$ per cycle under $35$), so many people can realistically expect success, especially if they start treatment while younger.

Drawback: Success rate falls sharply with age, down to just $5\%$ per cycle at $40$+, meaning many older patients need multiple costly cycles, each involving hormone injections and a minor surgical procedure, with no guarantee of success. The emotional and financial cost of repeated unsuccessful cycles can be considerable, and IVF is not affordable or accessible to everyone.

QUESTION 20 6 marks Criterion D
Hard

Non-invasive prenatal testing (NIPT) can screen a fetus's DNA from a sample of the mother's blood for certain chromosomal conditions, such as Down syndrome, early in pregnancy.

Evaluate the impact of this technology, discussing both a benefit and a concern it raises.

Show complete worked solution

Benefit: NIPT can detect certain chromosomal conditions from a simple maternal blood sample as early as $10$ weeks into pregnancy, without the small risk of miscarriage carried by more invasive tests such as amniocentesis. This gives parents earlier, safer information to prepare for their child's needs or to make informed decisions about their pregnancy and medical care.

Concern: Because the test provides very personal and sensitive genetic information early in a pregnancy, it raises ethical concerns about how the information might be used — including concern that it could increase pressure on parents to end pregnancies for conditions that would still allow the child a full and meaningful life. There are also concerns about the cost of the test and its availability differing between families and countries, which can make this information accessible to some parents but not others.

Variation and Inheritance (intro) 20 questions

QUESTION 1 2 marks Criterion A
Easy

State what is meant by variation, and give one example each of continuous and discontinuous variation in humans.

Show complete worked solution

Variation is the differences that exist between individuals of the same species.

Continuous variation can take any value within a range (e.g. height or body mass). Discontinuous variation falls into distinct, separate categories with no in-between values (e.g. blood group — A, B, AB or O).

QUESTION 2 3 marks Criterion A
Easy

Define the following terms used in genetics: chromosome, gene, and allele.

Show complete worked solution

A chromosome is a thread-like structure made of tightly coiled DNA, found in the nucleus of a cell, that carries genetic information.

A gene is a short section of DNA (found on a chromosome) that carries the instructions/code for a particular characteristic.

An allele is a different version of the same gene (e.g. the gene for seed colour might have a 'green' allele and a 'yellow' allele).

QUESTION 3 2 marks Criterion A
Easy

Explain the difference between genotype and phenotype, and between a dominant and a recessive allele.

Show complete worked solution

The genotype is the genetic make-up of an organism (the alleles it carries, e.g. $Rr$). The phenotype is the observable characteristic that results (e.g. round seeds).

A dominant allele is one whose characteristic is expressed in the phenotype whenever it is present, even with only one copy. A recessive allele is only expressed in the phenotype when two copies are present (i.e. no dominant allele is present).

QUESTION 4 4 marks Criterion A
Medium

In pea plants, the allele for round seeds ($R$) is dominant over the allele for wrinkled seeds ($r$). A pea plant with genotype $Rr$ is crossed with a pea plant with genotype $rr$.

a. Complete a Punnett square for this cross.
[2]
b. State the genotype ratio of the offspring.
[1]
c. State the phenotype ratio of the offspring, and name each phenotype.
[1]
Show complete worked solution
(a)
Rr
rRrrr
rRrrr

The parent $Rr$ contributes gametes $R$ and $r$; the parent $rr$ contributes only $r$ gametes. Combining these gives offspring genotypes $Rr$, $Rr$, $rr$, $rr$.

(b)
$1\,Rr : 1\,rr$ (i.e. $2\,Rr : 2\,rr$, simplified to $1:1$).
(c)
$1$ round : $1$ wrinkled — half the offspring are expected to have round seeds, half wrinkled.
QUESTION 5 3 marks Criterion A
Medium

A gene has two alleles, $B$ (dominant, brown fur) and $b$ (recessive, white fur). Three rabbits have genotypes $BB$, $Bb$, and $bb$.

a. Define the terms homozygous and heterozygous.
[1]
b. For each of the three genotypes given, state whether the rabbit is homozygous or heterozygous, and state its phenotype (fur colour).
[2]
Show complete worked solution
(a)
Homozygous means an organism has two identical alleles for a gene (e.g. $BB$ or $bb$). Heterozygous means an organism has two different alleles for a gene (e.g. $Bb$).
(b)
$BB$: homozygous, brown fur. $Bb$: heterozygous, brown fur (since $B$ is dominant). $bb$: homozygous, white fur.
QUESTION 6 4 marks Criterion A
Medium

Variation between individuals can be caused by genetic factors, environmental factors, or a combination of both.

a. Explain the difference between a genetic cause of variation and an environmental cause of variation.
[2]
b. Classify each of the following as mainly genetic, mainly environmental, or a combination of both: (i) natural eye colour, (ii) a scar from an injury, (iii) adult height.
[2]
Show complete worked solution
(a)
A genetic cause of variation comes from differences in the genes/DNA an individual inherits from its parents. An environmental cause of variation comes from an individual's surroundings or lifestyle (e.g. diet, climate, exercise) and is not inherited.
(b)
(i) Natural eye colour — genetic. (ii) A scar from an injury — environmental (caused entirely by an external event, not inherited). (iii) Adult height — a combination of both (genes set a potential range, but nutrition and health during childhood affect the final height reached).
QUESTION 7 6 marks Criterion A
Hard

In pea plants, purple flowers ($P$) are dominant over white flowers ($p$). Two heterozygous plants, both with genotype $Pp$, are crossed.

a. Draw a Punnett square for this cross, and give the genotype ratio of the offspring.
[2]
b. State the phenotype ratio of the offspring, and calculate the percentage of offspring expected to have white flowers.
[2]
c. If this cross produced 160 offspring plants in total, calculate the expected number with purple flowers.
[2]
Show complete worked solution
(a)
Pp
PPPPp
pPppp

Genotype ratio: $1\,PP : 2\,Pp : 1\,pp$.

(b)
Phenotype ratio $3$ purple : $1$ white. $$ \text{percentage white} = \frac{1}{4}\times100 = 25\% $$
(c)
$$ 160 \times \frac{3}{4} = 120 $$ Expected number with purple flowers $=120$.
QUESTION 8 6 marks Criterion B
Medium

A student wants to investigate whether hand span shows continuous variation within their class, by measuring every student.

a. State the variable being measured, and explain why a ruler or tape measure (rather than a simple 'yes/no' category) is an appropriate way to record it.
[2]
b. State two variables that should be kept the same for every student when taking the measurement, and explain why for one of them.
[2]
c. Describe how the collected data could be displayed to show that hand span is an example of continuous variation.
[2]
Show complete worked solution
(a)
The variable measured is hand span (distance from tip of thumb to tip of little finger, fingers spread, measured in cm). A ruler/tape measure is appropriate because hand span can take any value within a range (e.g. $17.3\,\text{cm}$), which is what makes it continuous variation — a category-based method could not record this level of detail.
(b)
Keep the hand used (e.g. always the right hand) and the way fingers are spread (fully stretched each time) the same for every student. Why keep the hand used the same: a person's dominant hand can be very slightly larger than the other, so measuring different hands for different students would not be a fair comparison.
(c)
Group the measurements into small, equal-width class intervals (e.g. $15$–$16\,\text{cm}$, $16$–$17\,\text{cm}$, etc.) and plot a histogram, with hand span on the x-axis and number of students on the y-axis, with bars touching (no gaps) since the values form a continuous range. A roughly bell-shaped distribution with no natural gaps between values supports that hand span is continuous variation.
QUESTION 9 6 marks Criterion B
Medium

A student wants to find out whether earlobe attachment (attached or free) and height both show the same type of variation.

a. State how you would record data for (i) earlobe attachment and (ii) height, for each student.
[2]
b. Explain why these two data-collection methods are different.
[2]
c. Suggest a suitable sample size for this investigation, and explain why a larger sample gives a more reliable picture of the variation in the class/population.
[2]
Show complete worked solution
(a)
(i) Earlobe attachment: record as one of two distinct categories — 'attached' or 'free' — by observation. (ii) Height: record as an exact numerical measurement in cm, using a height measure/stadiometer.
(b)
Earlobe attachment only has two possible distinct categories with no in-between values, so it can only be recorded as a category (discontinuous variation). Height can take any value across a continuous range, so it must be recorded as a precise numerical measurement rather than sorted into a small number of fixed categories.
(c)
A suitable sample would be the whole class or year group (at least 30+ students). A larger sample is more reliable because it is less likely to be skewed by a few unusual individuals, and gives a distribution (for height) or set of category proportions (for earlobes) that better represents the true pattern of variation in the wider population.
QUESTION 10 8 marks Criterion B
Hard
ConditionShade (cutting 1)Shade (cutting 2)Sun (cutting 1)Sun (cutting 2)Sun (cutting 3)
Height after 4 weeks (cm)1215222421

A student grew 5 genetically identical cuttings from the same parent plant. Two were grown in a shaded spot and three in direct sunlight, and their final heights after 4 weeks were recorded.

a. Explain why using genetically identical cuttings from the same parent plant was a good choice for this investigation.
[2]
b. Identify one weakness in this investigation's design, and explain why it reduces the reliability of the conclusion.
[3]
c. Suggest two improvements to this investigation that would make the conclusion more reliable.
[3]
Show complete worked solution
(a)
Because all the cuttings are genetically identical, any difference in height between the two groups cannot be due to genetic variation — it isolates environmental variation (light level) as the cause, making the comparison fairer.
(b)
The sample sizes are very small and unequal (only $2$ shaded cuttings vs $3$ sun cuttings). With so few repeats, the results are more easily affected by chance differences between individual cuttings (e.g. slightly different cutting size at the start), making it harder to be confident the height difference is really caused by light level rather than chance.
(c)
Use a larger, equal sample size in each group (e.g. at least $10$ cuttings per condition) to reduce the effect of chance variation between individual plants. Also control other variables that could affect growth, such as watering amount, pot size, and soil type, keeping these identical between the shade and sun groups so that light level is the only variable that differs.
QUESTION 11 3 marks Criterion C
Easy
Blood groupOABAB
Number of students12952
a. How many students in total were surveyed?
[1]
b. Calculate the percentage of students with blood group O.
[2]
Show complete worked solution
(a)
$12+9+5+2 = 28$ students.
(b)
$$ \frac{12}{28}\times100 = 42.9\%\ (\text{3 s.f.}) $$
QUESTION 12 5 marks Criterion C
Medium
Student12345678
Height (cm)152148161155158150163157
a. Calculate the mean height of these 8 students.
[2]
b. Calculate the range of the heights.
[1]
c. Explain why height is described as continuous, rather than discontinuous, variation, using this data as an example.
[2]
Show complete worked solution
(a)
$$ \text{mean} = \frac{152+148+161+155+158+150+163+157}{8} = \frac{1244}{8} = 155.5\,\text{cm} $$
(b)
$$163-148=15\,\text{cm}$$
(c)
Continuous variation can take any value within a range, with no fixed set of categories — as shown here, the heights take many different exact values ($148,150,152,\ldots,163\,\text{cm}$) across a range, rather than falling into a small number of distinct groups the way, e.g., blood group does. This is why height must be measured on a continuous numerical scale rather than sorted into categories.
QUESTION 13 5 marks Criterion C
Medium
PhenotypeGreenYellowTotal
Number observed582280

A student counted seed colour among 80 offspring from a cross between two heterozygous ($Gg$) pea plants, where green ($G$) is dominant over yellow ($g$).

a. For a cross between two heterozygous ($Gg \times Gg$) pea plants, state the expected phenotype ratio of green to yellow offspring.
[2]
b. Based on this expected ratio, calculate the expected number of yellow-seeded plants out of 80 offspring.
[1]
c. Compare the observed number of yellow plants (22) with your expected number, and comment on whether the observed data is a good match for the expected 3:1 ratio.
[2]
Show complete worked solution
(a)
Green ($G$) is dominant, so a $Gg\times Gg$ cross is expected to give a phenotype ratio of $3$ green : $1$ yellow.
(b)
$$ 80 \times \frac{1}{4} = 20 $$
(c)
The observed number of yellow plants ($22$) is very close to the expected number ($20$) — a difference of only $2$ plants out of $80$. This small difference is expected due to the natural randomness of which gametes combine at fertilisation, so the data is a good match for the expected $3:1$ ratio.
QUESTION 14 5 marks Criterion C
Medium
Family memberParent 1Parent 2Child 1Child 2Child 3
Cystic fibrosis?NoNoYesNoYes
a. Both parents are unaffected by cystic fibrosis, yet two of their children have the condition. Explain what this tells you about whether the cystic fibrosis allele is dominant or recessive.
[2]
b. State the genotype (in terms of a dominant allele $F$ and recessive allele $f$) of an affected child.
[1]
c. State the genotype of each unaffected parent, and explain your answer.
[2]
Show complete worked solution
(a)
Since both parents are unaffected but two children are affected, the allele for cystic fibrosis must be recessive. This is because both parents must be carrying one copy of the recessive allele without showing the condition themselves (they are heterozygous carriers); a child then only shows the condition if they inherit the recessive allele from both parents.
(b)
$ff$ (homozygous recessive).
(c)
Each unaffected parent must be $Ff$ (heterozygous carrier). They must carry the recessive $f$ allele, since two of their children are $ff$, but they are not affected themselves, which is only possible if they also carry the dominant $F$ allele masking it.
QUESTION 15 5 marks Criterion C
Medium
Person123456
Height (cm)145152149158151147
Number of fingers10109101010
a. Identify the anomalous result in this table.
[1]
b. Explain why this anomalous result is unlikely to represent normal genetic variation in finger number.
[2]
c. State which of the two variables in the table shows continuous variation, and which shows discontinuous variation, giving a reason for each.
[2]
Show complete worked solution
(a)
Person 3's finger count ($9$) is anomalous.
(b)
Humans normally and reliably develop $10$ fingers as a fixed characteristic of the species; genetic variation does not normally produce a range of different finger counts the way it does for height. Person 3's result of $9$ is far more likely explained by a non-genetic cause, such as an injury or accident resulting in the loss of a finger, rather than true inherited variation.
(c)
Height shows continuous variation — it takes a range of different exact values ($145$ to $158\,\text{cm}$) with no fixed categories. Number of fingers shows discontinuous variation — in a normally-developing population it takes only one fixed whole-number value ($10$), falling into a distinct category rather than a continuous range.
QUESTION 16 8 marks Criterion C
Hard
Height (cm)140–145145–150150–155155–160160–165
Number of students371262
a. How many students were included in this survey in total?
[2]
b. Estimate the mean height of the students, using the midpoint of each class interval.
[3]
c. Describe the shape of this distribution, and explain what it suggests about height as a form of variation.
[3]
Show complete worked solution
(a)
$3+7+12+6+2=30$ students.
(b)
$$ \text{estimated mean} = \frac{(142.5\times3)+(147.5\times7)+(152.5\times12)+(157.5\times6)+(162.5\times2)}{30} = \frac{4560}{30} = 152\,\text{cm} $$
(c)
The distribution is roughly bell-shaped (normal distribution): frequency is low at the extremes ($140$–$145\,\text{cm}$ and $160$–$165\,\text{cm}$) and highest in the middle class ($150$–$155\,\text{cm}$, with $12$ students), tailing off symmetrically either side. This bell-shaped pattern is typical of continuous variation controlled by many genes (and environmental factors) acting together, where most individuals cluster around an average value and progressively fewer individuals are found further from the average in either direction.
QUESTION 17 7 marks Criterion C
Hard
Experiment123Combined
Total offspring20202060
Purple flowers observed12181343

Three separate experiments each crossed two heterozygous purple-flowered pea plants, where the expected phenotype ratio is $3$ purple : $1$ white (i.e. $75\%$ purple).

a. Calculate the percentage of purple-flowered offspring observed in each of Experiment 1 and Experiment 2.
[2]
b. Calculate the percentage of purple-flowered offspring in the combined data (all 60 offspring), and compare it with the expected 75% for a 3:1 ratio.
[2]
c. Explain why the combined result from all three experiments gives a more reliable estimate of the true ratio than any single experiment on its own.
[3]
Show complete worked solution
(a)
$$ \text{Experiment 1: } \frac{12}{20}\times100=60\% \qquad \text{Experiment 2: } \frac{18}{20}\times100=90\% $$
(b)
$$ \frac{43}{60}\times100 = 71.7\%\ (\text{3 s.f.}) $$ This is close to the expected $75\%$, within about $3$ percentage points.
(c)
Each individual experiment has a small sample size ($20$ offspring), so chance alone can cause results to swing quite far from the expected ratio — Experiment $2$'s $90\%$ is well above the expected $75\%$, purely due to which gametes happened to combine. Combining all three experiments increases the total sample size to $60$, which averages out this random chance variation between individual experiments, giving a combined percentage ($71.7\%$) that sits much closer to the true expected value than any single small experiment.
QUESTION 18 5 marks Criterion D
Medium
Before selective breedingAfter several generations of selective breeding
Average yield (tonnes/hectare)4.27.8
Genetic variation in populationHighLow

Farmers have used selective breeding for many generations to increase the yield of a wheat crop, by always breeding from the highest-yielding plants. The table shows data before selective breeding began and after several generations.

Discuss one benefit and one drawback of this selective breeding programme, using the data given.

Show complete worked solution

Benefit: Yield increased substantially, from $4.2$ to $7.8$ tonnes/hectare — an increase of about $86\%$. This means significantly more food can be produced from the same area of farmland, which is important for feeding a growing population.

Drawback: The table shows genetic variation in the population has fallen from high to low. Because almost all the wheat now shares very similar genes, the whole crop is much more vulnerable to being wiped out by a single new disease, pest, or change in climate that this narrow gene pool cannot resist — there is little genetic variation left for natural selection to act on to produce resistant survivors.

QUESTION 19 6 marks Criterion D
Medium

Genetic testing can identify whether a person carries the recessive allele for certain inherited conditions, such as cystic fibrosis, even if they show no symptoms themselves.

Discuss one benefit and one concern of this technology for couples planning to have children.

Show complete worked solution

Benefit: If both partners are tested and found to be carriers of the same recessive condition, they can use this genetic information to make informed decisions before starting a family — for example, seeking genetic counselling, considering prenatal testing, or being prepared to arrange specialist medical care immediately after birth, all of which can improve outcomes for an affected child.

Concern: Genetic test results are highly personal and sensitive information; there are concerns about privacy, such as who else might gain access to the results, and about the emotional impact on a couple of learning they are carriers, which can create anxiety or difficult decisions about starting a family — even though being a healthy carrier of a single recessive allele does not affect the carrier's own health at all.

QUESTION 20 6 marks Criterion D
Hard

Genetic engineering can be used to transfer a gene from one species into a crop plant's genome. For example, a gene that produces a natural insect-repelling protein has been inserted into some maize (corn) varieties, so the plant no longer needs to be sprayed with as much insecticide to protect it from pests.

Evaluate the impact of this technology, discussing both a benefit and a concern it raises.

Show complete worked solution

Benefit: Because the modified maize produces its own natural pest resistance, farmers need to spray far less chemical insecticide on the crop. This reduces costs for farmers, reduces harm to helpful insects such as bees that are not the target pest, and reduces the amount of chemical residue that ends up in the environment and water supply.

Concern: Introducing new genes into crop species raises concerns about long-term ecological effects, such as the modified gene spreading to wild relative plants through cross-pollination, or insect pests eventually evolving resistance to the inserted protein — in the same way that overusing an antibiotic leads to resistant bacteria. There are also concerns among some consumers and countries about the safety and labelling of genetically modified food, which is why regulations around GM crops vary widely around the world.