Home / IB DP Biology- D3.2 Inheritance -FA 2025- IB Style Questions For HL Paper 2

IB DP Biology- D3.2 Inheritance -FA 2025- IB Style Questions For HL Paper 2

Question

All living organisms contain chromosomes. In addition to acting as stores of genetic
information, these chromosomes are involved in a range of active processes during the life
of a cell and of an organism.

(a) Outline the changes to chromosomes that occur during prophase in the first division of meiosis.
(b) Describe the processes that are carried out by enzymes that bind to DNA.
(c) Explain the effects that the environment can have on DNA in living organisms.

Answer/Explanation

Answer:

(a)
a. pairing/synapsis of homologous chromosomes / homologous chromosomes form bivalents;
b. crossing over / chromatid breaks then rejoins to non-sister chromatid;
c. exchange of DNA/alleles/genetic information between chromatids/chromosomes;
d. recombination / new combinations of alleles/genes generated;
e. condensation/shortening/thickening/supercoiling of chromatids/chromosomes;
f. formation of a chiasma where crossing over occurred;

(b)  replication
a. helicase unwinds the double helix/DNA;
b. helicase breaks hydrogen bonds between/separates/unzips DNA strands;
c. (DNA) gyrase/topoisomerase releases tensions in DNA as it unwinds;
d. (DNA) primase adds RNA primers (where DNA polymerase can bind);
e. DNA polymerase (III) replicates DNA/adds nucleotides (to make new strand);
f. DNA polymerase I replaces RNA (primers) with DNA;
g. DNA ligase seals nicks/joins sugar-phosphate backbones/joins (Okazaki) fragments; transcription
h. RNA polymerase used for transcription;
i. RNA polymerase unwinds / separates DNA strands / binds to the promoter;
j. RNA polymerase copies DNA base sequence of a gene/makes mRNA;
k. restriction enzymes/endonucleases cut DNA at specific base sequences;
l. telomerase adds nucleotides to the ends of chromosomes/makes telomeres;

(c)  Mutation
a. (environment can cause) mutation;
b. mutations are base sequence changes;
c. radiation/UV/gamma rays can cause mutations/changes to base sequences;
d. mutagenic/carcinogenic chemicals can cause mutations / mustard gas/another
example;
Epigenetics
e. (environment) can cause changes to gene expression;
f. methylation (patterns) in DNA changed (in response to environmental factors);
g. methylation inhibits (gene transcription) / acetylation promotes (gene
transcription);
h. body temperature/stress/diet (can affect gene expression);

Question

The spider Dolomedes plantarius usually has white bands down the left and right sides of its body, but some individuals lack these bands. The photograph shows the banded form of D. plantarius with a ball of spiderlings.

Crosses were performed to investigate the inheritance of this trait, by allowing specific males and females to mate. Numbers of banded and unbanded spiderlings that hatched out from all the eggs laid by the female were recorded. Results are shown in the table.

(a) Explain the conclusion that can be drawn from Cross 1.

(b) Deduce reasons for the difference between the results of Cross 2 and Cross 3.

(c) There were 79 progeny in Cross 4. Predict the expected results by completing the table.

(d) Identify, using one recognition feature visible in the photo, the phylum in which
D. plantarius is classified.

Answer/Explanation

Answer:

(a)
a. allele/trait/gene for banded is dominant / for unbanded is recessive;
b. because there is a ratio of 3 banded:1 unbanded
OR
because two banded spiders produced some unbanded offspring;
c. both parents are heterozygous;

(b)
a. (1:1 ratio) in cross 2 as banded parent is heterozygous/has one copy of each allele;
b. (no unbanded offspring) in cross 3 as banded parent is homozygous/has two alleles for banded;
c. (in crosses 2 and 3) banded parental phenotypes are the same but their genotypes are different;

(c)

(d)  arthropods (as spider has) segmentation/exoskeleton/jointed limbs/jointed
appendages/bilateral symmetry;

Question

Dachshunds have three basic coat types: wire-, smooth- or long-haired. These are affected by two genes, W and K. The presence of W always results in wire hair.

When long-haired dogs are crossed among themselves, they always produce long-haired puppies. When dogs heterozygous for both genes are crossed, they produce offspring in the ratio 12 wire-haired: 3 smooth-haired: 1 long-haired.

A male wire-haired dog was crossed with several long-haired females and the phenotypic ratio of the puppies was approximately 2 wire-haired:1 smooth-haired:1 long-haired.
a. Identify the genotype of the male wire-haired dog.
b. Using a Punnett square, determine how a smooth-haired puppy could be produced in the offspring.[2]

▶️Answer/Explanation

Ans:

a. WwKk

b. WK Wk wK and wk for male gamete genotypes and wk for female;

wwKk shown in Punnett square and identified as smooth;

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