IB Biology SL D3.2 Inheritance Questions

Practise IB Biology SL D3.2 by predicting inheritance from crosses and pedigrees while linking alleles, genotypes and phenotype.

Exam points

Identify haploid gametes and distinguish genotype, phenotype and allele combinations.
Use flowering-plant or autosomal Punnett grids to predict genotypes, phenotypes and ratios.
Infer dominance and unknown genotypes from test-cross results or offspring evidence.
Deduce ABO cross outcomes from IA, IB and i, including codominance and recessive alleles.
Distinguish codominance from incomplete dominance and predict or interpret offspring phenotype ratios.
Calculate XX/XY offspring probabilities; explain how Y-chromosome SRY/TDF initiates testis development.
Use X-linked haemophilia grids to calculate carrier and affected outcomes by sex.
Infer dominant, recessive or X-linked inheritance and genotype risks from pedigrees.
Explain recessive PKU from PAH deficiency to phenylalanine build-up and dietary treatment.
Explain how environment alters gene expression and phenotype without changing genotype.
Distinguish alleles by an SNP's single-base difference.
Read medians, quartiles, range and outliers from box plots to compare continuous data.

Covered syllabus points

D3.2.1—Haploid gametes + fusion = diploid zygote D3.2.2—Genetic crosses in flowering plants D3.2.3—Genotype D3.2.4—Phenotype D3.2.5—Dominant and recessive alleles D3.2.6—Phenotypic plasticity D3.2.7—Phenylketonuria (PKU)D3.2.8—SNPs and multiple alleles D3.2.9—ABO blood groups D3.2.10—Incomplete dominance and codominance D3.2.11—Sex determination D3.2.12—Haemophilia D3.2.13—Pedigree charts D3.2.14—Continuous variation D3.2.15—Box-and-whisker plots

[Maximum number: 1]

For what reason do gametes contain only one allele of each gene?

To prevent inbreeding in a population

Haploid cells contain only one set of chromosomes

The two alleles of a gene are separated during mitosis

Crossing over will always produce one allele of a gene

[Maximum number: 4]

The dolichos bean (Lablab purpureus) is cultivated as a food crop in tropical countries. Leaf cells in L. purpureus have 24 chromosomes.

(a)

L. purpureus can have purple or white flowers. Two pure-breeding varieties were crossed: HA 4 with white flowers and GL 424 with purple flowers. All of the $F_{1}$ plants had purple flowers. The $F_{1}$ plants were self-pollinated to produce an $F_{2}$ generation. There were 97 plants with purple flowers and 38 plants with white flowers in the $F_{2}$ generation.

Using a Punnett grid, explain the results of this cross.

[ 3 ]

Male and female gamete genotypes/alleles shown as P and p, or other letters following convention with a suitable key, in a Punnett grid.
$\mathrm{F}_{2}$ genotypes shown as $\mathrm{PP}$ , $\mathrm{Pp}$ , $\mathrm{pP}$ and $\mathrm{pp}$ .
$\mathrm{F}_{2}$ phenotypes indicated for each genotype on the Punnett grid / 3 purple to 1 white ratio indicated.

Punnett grid with P and p gametes along both edges, giving offspring genotypes PP, Pp, Pp and pp.

Official Punnett-grid answer image from the markscheme.

Accept other upper-case and lower-case letters for the alleles.
No punnett grid = 2 MAX.
Allow ECF if alleles incorrect.

(b)

Suggest a reason for the $F_{2}$ generation not corresponding exactly to the expected ratio of purple-flowered and white-flowered plants.

[ 1 ]

[Maximum number: 4]

(a)

Define the term genotype.

[ 1 ]

(b)

PKD (polycystic kidney disease) is an inherited disorder that causes the kidneys to become enlarged and lose function over time. PKD is caused by a dominant allele, D, which is not sex-linked.

The pedigree chart shows the inheritance of PKD in a family.

[ 3 ]

(i)

Calculate the probability of individuals 1 and 2 having a child with PKD.

[ 1 ]

0.5 / 50% / $\frac{1}{2}$ ;

(ii)

Explain how the pedigree chart shows that the dominant allele causing PKD is not on the X chromosome.

[ 2 ]

(If on the X chromosome)

ALTERNATIVE 1 Father evidenced route:

a. 1 has only one dominant allele on X (and not on the Y ) / would be $\mathrm{X}^{\mathrm{D}} \mathrm{Y}$ / OWTTE;

b. 1 passed his X chromosome/dominant allele to 3/7 I OWTTE OR

son/8 could not inherit the disease;

c. (so) all daughters would be affected / not possible for 3/7 to be healthy / OWTTE;

ALTERNATIVE 2 Mother evidenced route:

d. 2 does not have the dominant allele / is homozygous recessive / would be $\mathrm{X}^{\mathrm{d}} \mathrm{X}^{\mathrm{d}}$ IOWTTE;

e. 2 passed her X chromosome/ X ${ }^{d}$ /recessive allele(s) to 6/8;
OR

6/8/sons would receive the Y chromosome with no (dominant) allele / OWTTE;

f. (so) all sons/8 would be healthy / not possible for 8 to be affected / OWTTE;

Marking guidance:

Accept Punnett grids, providing they are
clearly annotated and identify specific
individuals.

[Maximum number: 1]

Define the term genotype.

[Maximum number: 1]

The graph shows a sigmoid population growth curve.

Sigmoid population-size curve against time. X points to its upper plateau; Y marks the early low section before the rapid rise.

The table summarizes the genome size of several organisms.

Genome-size table in base pairs: Helicobacter pylori 1,667,867; Drosophila melanogaster 130,000,000; Oryza sativa 420,000,000; Homo sapiens 3,200,000,000.

Distinguish between the terms genotype and phenotype.

[Maximum number: 3]

The most common form of Stargardt's disease is known to be autosomal recessive. Using a Punnett grid, deduce the probability of a child inheriting Stargardt's disease, if both of the parents are carriers of the disease but do not have the disease themselves.

	S	s
S	SS	Ss
s	Ss	ss

[Maximum number: 4]

Explain the causes and consequences of phenylketonuria (PKU).

[Maximum number: 1]

What is a difference between two alleles of a gene?

Their positions on homologous chromosomes

Their amino acid sequence

The characteristic they influence

Their base sequence

[Maximum number: 3]

Both the genome and the environment of an organism affect its growth and development. Variation between individuals of a species can be continuous or discrete (discontinuous). Mitosis produces identical cells for growth and meiosis allows for variation.

The micrographs show onion and whitefish cells in mitosis.

Micrographs compare onion and whitefish cells. X points to a large onion cell with visible condensed material; the whitefish cell shows two separated dark chromosome groups.

Blood group in humans is an example of discrete (discontinuous) variation. Explain the inheritance of ABO blood groups in humans.

[Maximum number: 1]

A Mirabilis jalapa plant with red flowers was crossed with one with white flowers. All plants in the F1 generation had pink flowers. What phenotype ratio would be expected in the F2 generation?

100 % pink

50 % red and 50 % white

25 % white, 50 % pink and 25 % red

75 % red and 25 % white

[Maximum number: 3]

In eukaryotes, the chromosomes are located in the nucleus, and the nucleus can divide by mitosis or meiosis.

Explain how the presence of a Y chromosome in the cells of a human embryo causes it to develop as a male.

[Maximum number: 3]

People with the inherited disease hemophilia have a deficiency in one of the proteins that act as clotting factors in blood.

A family has a history of hemophilia that is caused by a sex-linked recessive allele. A woman from this family is a carrier and marries a man who does not have the allele. Showing your working, determine the probability of their children having the disease.

Alleles clearly labelled for both parents, in a Punnett square or other format.
Correct genotypes for all four possibilities for children.

	Father: $X^{H}$	Father: $Y$
Mother: $X^{h}$	$X^{H}X^{h}$	$X^{h}Y$
Mother: $X^{H}$	$X^{H}X^{H}$	$X^{H}Y$

Phenotypes of the children:
The sons would have a $50\%$ chance of having hemophilia.
The daughters would have a $0\%$ chance of having hemophilia / $50\%$ chance of being a carrier.

[Maximum number: 4]

Thrombophilia is a human genetic condition where the blood has an increased tendency to clot. The condition is caused by a single base substitution mutation in DNA. If a person is homozygous for the gene, they are at greater risk for developing a blood clot than an individual who is heterozygous. The pedigree chart shows the inheritance of thrombophilia in a family.

(a)

Draw the symbol for individual X on the diagram.

[ 1 ]

(b)

Calculate the probability of male Y having an allele for the disorder.

[ 1 ]

(c)

Explain how the information in the box labelled B indicates that the gene is not sex-linked.

[ 2 ]

[Maximum number: 1]

The National Health and Nutrition Examination Survey in the USA measured heights of individuals aged 21-79. Height is an example of continuous variation.

Histogram of individual heights in centimetres, with a roughly bell-shaped distribution centred near 170 cm and few individuals at either extreme.

State the range of heights that is the mode.

[Maximum number: 2]

Dementia is a term used to describe a group of symptoms that are caused by a decline in brain function. People with dementia have problems with memory, language and behaviour.

Alzheimer's disease is the most common cause of dementia. Scientists investigated the effect of a drug called donanemab on people with Alzheimer's disease. They divided them into two groups:
- The treatment group received regular injections of a solution containing donanemab.
- The control group received regular injections of a solution without donanemab.

The scientists used the Alzheimer's Disease Rating Scale (ADRS) to measure brain function. The lower the score on this scale, the worse the brain function. The scientists calculated the mean ADRS score for each group at intervals over 76 weeks. They then calculated the mean change from their initial score at the start of the investigation (baseline).

Their results are shown in the graph.

(a)

SNAP-25 (Synaptosomal-Associated Protein) is a protein found in the brain.
In a different investigation, scientists measured the concentration of SNAP-25 in the cerebrospinal fluid (CSF) of three groups of people:
- 36 people without Alzheimer's disease (No AD)
- 18 people with Alzheimer's disease who had mild dementia (AD mild)
- 24 people with Alzheimer's disease who had severe dementia (AD severe)

Their results are shown in the box-and-whisker plot.

Estimate the median concentration of SNAP-25 in CSF for the AD mild group.
$\mathrm{ng} \mathrm{L}^{-1}$

[ 1 ]

280 ( $\mathrm{ng} \mathrm{L}^{-1}$ );

Accept answers in the range 270-280.

(b)

Using your knowledge of box-and-whisker plots, state what the data point shown as a triangle represents.

[ 1 ]

All question bank results loaded

a	if it was sex-linked it would be on the x chromosome;
b	there cannot be a heterozygous male if the trait is sex-linked
c	males would pass the allele to their daughter;
d	daughter is not shown as heterozygous so it is not sex linked;

a	outlier;
b	(both values) are more than 1.5 times the interquartile range/IQR above the third/upper quartile/maximum height;

Covered syllabus points

Question 1

Question 2

Question (a)

Question (b)

Question 3

Question (a)

Question (b)

Question (i)

Question (ii)

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

Question 11

Question 12

Question 13

Question (a)

Question (b)

Question (c)

Question 14

Question 15

Question (a)

Question (b)