S5 Biology Paper: Inheritance, Sex Linkage and Meiosis (100 marks)

2026-11-10

S5 Biology – Inheritance and meiosis

S5 Biology paper: 6 questions, 22 parts, 100 marks, 60 min. Monohybrid and test crosses, X-linked inheritance, linkage, DNA replication and meiosis; mark schemes included.

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Section A – Basic genetics and Mendel law

Answer in full sentences and give genotypes whenever a ratio is required.

#	Question	Answer and marking notes
1	Distinguish between a gene and an allele, giving one example of each.	A gene is a length of DNA on a chromosome that controls a characteristic; alleles are different versions of the same gene at the same locus on a pair of homologous chromosomes. Example: the gene for pea stem height has a tall allele T and a dwarf allele t. (4 marks: gene definition 1; allele definition 1; locus 1; example 1)
2	Explain the terms homologous chromosomes, homozygote and heterozygote.	Homologous chromosomes are a pair of chromosomes of similar size and shape, one from each parent, carrying the same loci. A homozygote has two identical alleles (TT or tt); a heterozygote has two different alleles (Tt). (4 marks: homologous definition 2; homozygote 1; heterozygote 1)
3	How does Mendel law of segregation explain an all-tall F1 and a 3 tall : 1 dwarf F2?	Parents are TT x tt, so F1 is all Tt; T is completely dominant to t, hence all tall. At meiosis in F1 the homologous chromosomes separate, so half the gametes carry T and half carry t. Random fertilisation gives 1 TT : 2 Tt : 1 tt, that is 3 tall : 1 dwarf. (5 marks: F1 genotype 1; dominance 1; gamete ratio 1; genotype ratio 1; phenotype ratio 1)
4	Explain how a test cross finds the genotype of a tall pea plant and state the two possible outcomes.	Cross the unknown tall plant with a dwarf homozygote tt. If all offspring are tall the plant is TT; if offspring are about 1 tall : 1 dwarf it is Tt. Since tt gives only t gametes, the offspring phenotypes reveal the gametes of the tested plant. (5 marks: method 1; TT outcome 1; Tt outcome 1; reason 2)

Section B – Monohybrid crosses and test crosses

T = tall, t = dwarf; T is completely dominant to t.

#	Question	Answer and marking notes
5	For the cross Tt x Tt, give the genotype ratio and the phenotype ratio of the offspring.	Each parent gives half T and half t gametes; the genotype ratio is 1 TT : 2 Tt : 1 tt and the phenotype ratio is 3 tall : 1 dwarf. (4 marks: gametes 1; genotype ratio 2; phenotype ratio 1)
6	The cross above produces 400 offspring. What is the expected number of dwarf plants?	A dwarf plant must be tt, probability 1/4; expected number = 400 x 1/4 = 100 plants (expected tall = 300). (4 marks: probability 1; working 1; value 1; unit 1)
7	For the test cross Tt x tt, give the offspring ratio and the expected number of dwarf plants among 400 offspring.	tt gives only t gametes, so offspring are 1 Tt : 1 tt, that is 1 tall : 1 dwarf; probability 1/2 and expected number = 400 x 1/2 = 200. (4 marks: ratio 1; probability 1; working and value 1; unit 1)
8	Two tall pea plants are crossed and 100 of 400 offspring are dwarf. Deduce the parental genotypes and explain.	A dwarf offspring is tt, so each parent contributed one t; both parents are tall, so both are Tt. Expected dwarf = 400 x 1/4 = 100, which matches the observed number. (4 marks: Tt x Tt 1; reason 1; expected value 1; conclusion 1)

Section C – X-linked inheritance

Red-green colour blindness is caused by recessive allele b on the X chromosome; the normal allele is B and the Y chromosome carries no such gene.

#	Question	Answer and marking notes
9	A carrier mother (X^B X^b) marries a normal father (X^B Y). What is the probability that a son is colour blind?	A son takes X from the mother and Y from the father; one of the mother two X chromosomes is X^b, so the probability is 1/2, that is 50%. (4 marks: chromosome source 1; maternal gametes 1; probability 1; percentage 1)
10	For the same marriage, what are the probabilities that a daughter is colour blind and that a daughter is a carrier?	A daughter must receive X^B from her father, so she has at least one normal allele and the probability of colour blindness is 0.0%. The chance she takes X^b from her mother is 1/2, so 50% are carriers X^B X^b. (5 marks: paternal X^B 1; probability and reason 2; carrier 1; percentage 1)
11	For the same marriage with two sons, find the probability that neither is colour blind and that at least one is.	Each son has probability 1/2 of being normal; both normal = (1/2)^2 = 1/4, that is 25%. At least one affected = 1 – 1/4 = 3/4, that is 75%. (4 marks: single probability 1; product 1; at least one 1; percentages 1)
12	If the father is colour blind (X^b Y) and the mother is a carrier (X^B X^b), what is the probability that a daughter is colour blind? Explain the difference from the previous case.	A daughter must receive X^b from this father; the chance she also takes X^b from her mother is 1/2, so the probability is 1/2, that is 50%. In the previous case the father was X^B Y, so every daughter received X^B and could not be affected. (5 marks: paternal gamete 1; maternal gamete 1; probability 2; comparison 1)

Section D – Linkage and recombination frequency

The data below come from a test cross between a tall red double heterozygote and a dwarf white double recessive.

#	Question	Answer and marking notes
13	Among 1000 offspring: tall red 420, dwarf white 430, tall white 70, dwarf red 80. Calculate the recombination frequency.	Parental types = 420 + 430 = 850; recombinant types = 70 + 80 = 150; recombination frequency = 150 / 1000 x 100% = 15.0%. (5 marks: parental 1; recombinant 1; working 2; value and unit 1)
14	Use the recombination frequency to find the map distance and decide whether the two genes are linked.	Map distance = recombination frequency = 15.0 cM. Since 15.0% is well below 50%, the two genes are on the same chromosome and are linked. (5 marks: value 1; unit 1; comparison with 50% 1; conclusion 1; reason 1)
15	Explain why there are fewer recombinant offspring, and give the expected number of recombinants if the genes assorted independently.	The genes lie on the same chromosome and tend to be inherited together, keeping the parental combinations; only crossing over between homologous chromosomes produces recombinants, so they are fewer. With independent assortment the expected number of recombinants = 1000 x 1/2 = 500. (5 marks: linkage 2; crossing over 1; expectation idea 1; value 1)

Section E – DNA structure, semi-conservative replication and mutation

A double-stranded DNA fragment is 200 base pairs long and adenine (A) makes up 30% of all bases.

#	Question	Answer and marking notes
16	Find the percentage of guanine (G) and the number of A, T, G and C bases.	Total bases = 200 x 2 = 400; A = 120 and by complementary pairing T = 120; the remaining 160 are shared equally, so G = 80 and C = 80, that is 20.0% G. (4 marks: total 1; A and T 1; G and C 1; percentage 1)
17	Calculate the total number of hydrogen bonds in the fragment (2 between A and T, 3 between G and C).	There are 120 A-T pairs giving 240 bonds and 80 G-C pairs giving 240 bonds; total = 480. (4 marks: A-T term 1; G-C term 1; sum 1; value 1)
18	One strand has 200 nucleotides. How many phosphodiester bonds are in this strand?	Within one strand each nucleotide is joined to the next, so 200 nucleotides have 200 – 1 = 199 phosphodiester bonds. (4 marks: linear chain 2; minus one 1; value 1)
19	If the fragment replicates semi-conservatively for 3 rounds, how many DNA molecules are produced and how many contain an original strand?	Number of molecules = 2^3 = 8. In semi-conservative replication each original strand is retained in one molecule, so 2 molecules contain an original strand. (4 marks: formula 1; number 1; original-containing 1; explanation 1)

Section F – Meiosis and gamete variation

A human body cell has 46 chromosomes (23 pairs).

#	Question	Answer and marking notes
20	After meiosis I and II, how do chromosome number and DNA content change in human spermatocytes?	A primary spermatocyte has 46 chromosomes. After meiosis I each secondary spermatocyte has 23 chromosomes, each with two chromatids; after meiosis II each spermatid has 23 chromosomes as single chromatids, with one quarter of the original DNA content. (5 marks: number after I 2; two chromatids 1; number after II 1; DNA content 1)
21	How many kinds of gamete can independent assortment of 23 human chromosome pairs produce? Explain how crossing over adds further variation.	Number of gamete kinds = 2^23 = 8,388,608. In crossing over, non-sister chromatids of homologous chromosomes exchange segments, recombining linked alleles and producing even more gamete types. (6 marks: formula 2; value 2; mechanism 1; variation 1)
22	Compare the products of meiosis in one primary oocyte and one primary spermatocyte.	One primary oocyte gives one ovum and three polar bodies, which later degenerate; one primary spermatocyte gives four functional sperm. All products have 23 chromosomes. (6 marks: one ovum 1; three polar bodies 2; four sperm 2; chromosome number 1)

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