Genetics (Botany): genetic material and DNA replication, Mendelian genetics, linkage and crossing over, mutation and polyploidy

Crossing Over During Prophase-I of Meiosis-I

Crossing over is the exchange of corresponding segments between non-sister chromatids of homologous chromosomes. It occurs during the pachytene sub-stage of prophase-I and is the chief source of genetic recombination.

Sequence of Events

1. Leptotene (early prophase-I) Chromosomes condense and become visible as long, thin threads. Each chromosome already consists of two sister chromatids (formed in S-phase), but they are not yet distinct.

2. Zygotene Homologous chromosomes come close and pair lengthwise along their entire length. This pairing is called synapsis, and the paired structure is a bivalent (also called a tetrad, since it contains four chromatids). A proteinaceous synaptonemal complex forms between the homologues and holds them together.

3. Pachytene — crossing over occurs here The four chromatids of the bivalent are now clearly visible. The non-sister chromatids of the two homologues come into contact at certain points. At each such point, the chromatids break and the broken segments are exchanged and rejoined, so that each chromatid now carries a piece of its homologue. The points of exchange become visible in the next stage as chiasmata (singular: chiasma).

4. Diplotene The synaptonemal complex dissolves and the homologues begin to separate. They remain joined only at the chiasmata, which are the visible evidence that crossing over has taken place. A bivalent may show one or several chiasmata.

5. Diakinesis Chromosomes become maximally condensed, chiasmata move toward the ends (terminalisation), and the nuclear envelope breaks down, leading to metaphase-I.

Labelled Diagrams

   (a) Zygotene: synapsis          (b) Pachytene: crossing over
                                        (exchange of segments)

   ═══════════════                  ═══════╳═══════
   ═══════════════                  ═══════╳═══════
   Homologous pair (bivalent)       Non-sister chromatids break
   joined by synaptonemal           and exchange segments at the
   complex                          crossing point

   (c) Diplotene: chiasma visible   (d) Recombinant chromatids
                                        after separation

        ╲   ╱                            ═══════════
         ╳   ← chiasma                   ══════╳═════  ← recombinant
        ╱   ╲                            ══════╳═════  ← recombinant
                                         ═══════════

Labels to mark on the figure: homologous chromosomes, sister chromatids, non-sister chromatids, synaptonemal complex, point of crossing over, chiasma, recombinant chromatids.

Effect of Absence of Crossing Over on Genetic Variation

If crossing over does not occur:

1. No new gene combinations within a chromosome. Each homologous chromosome passes on intact to the gamete. Only the combinations already present in the parent's two chromosomes can be assorted.
2. Variation is limited to independent assortment. Genetic differences among offspring would arise only from the random separation of different homologous pairs at anaphase-I and from random fertilization. Genes lying on the same chromosome would always stay linked and would never be separated or recombined.
3. Linked genes remain linked forever. This reduces the number of possible gamete types drastically, so offspring would resemble one or the other parent rather than showing new trait combinations.
4. Reduced evolutionary adaptability. Lower genetic variation means the population has fewer variations on which natural selection can act, making it less able to adapt to a changing environment.

Thus, crossing over greatly increases genetic variation; its absence would make offspring far more uniform and genetically similar to the parents.

Biologically feasible model: Semiconservative replication

Of the three models proposed (conservative, semiconservative and dispersive), the semiconservative model of Watson and Crick (1953) is the one that actually occurs during the S-phase of interphase. It was proved experimentally by Meselson and Stahl (1958) using E. coli grown in $^{15}\text{N}$ medium and then transferred to $^{14}\text{N}$ medium; after one generation all DNA was of hybrid density, and after two generations half was hybrid and half light — results possible only under semiconservative replication. In this model, the two parental strands separate and each acts as a template, so each daughter DNA molecule has one old (parental) strand and one newly synthesised strand.

Mechanism of semiconservative replication

1. Initiation (origin and unwinding): Replication starts at a specific point called the origin of replication (ori). The enzyme helicase unwinds the double helix by breaking hydrogen bonds between base pairs, forming a replication fork. Single-strand binding (SSB) proteins keep the separated strands from re-pairing, and topoisomerase relieves the supercoiling/tension ahead of the fork.

2. Formation of the replication fork and RNA primer: The two separated strands act as templates. Since DNA polymerase cannot start a chain on its own, the enzyme primase synthesises a short RNA primer with a free $3'$-OH end on each template.

3. Elongation by DNA polymerase: DNA polymerase III adds deoxyribonucleoside triphosphates (dNTPs) complementary to the template, always in the $5' \to 3'$ direction, using the $3'$-OH of the primer/nascent strand. Because the two strands are antiparallel, the two templates are copied differently:

   • On the template running $3' \to 5'$, synthesis is continuous towards the fork — the leading strand.
   • On the template running $5' \to 3'$, synthesis is discontinuous, away from the fork, in short pieces called Okazaki fragments — the lagging strand. Each fragment needs its own primer.

4. Removal of primer and joining: DNA polymerase I removes the RNA primers and fills the gaps with DNA. DNA ligase then joins the Okazaki fragments and seals the nicks, forming a continuous strand.

5. Proofreading and termination: DNA polymerase has $3' \to 5'$ exonuclease (proofreading) activity, correcting wrong bases. Replication stops when the forks meet at the termination region; the two new daughter DNA molecules are then separated (in eukaryotes, along with the nucleus dividing in the M-phase).

Semi-discontinuous nature: One strand (leading) is synthesised continuously while the other (lagging) is synthesised discontinuously as Okazaki fragments; hence replication is called semi-discontinuous.

Diagrams

Replication fork (Y-shaped):

      parental double helix
        ┌───┐ ┌───┐
  3' ───┼───┼─┼───┼─── 5'   template (leading)
        └───┘ └───┘
          ↑     ↑
        helicase unwinds
        ┌───┐ ┌───┐
  5' ───┼───┼─┼───┼─── 3'   template (lagging)
        └───┘ └───┘
             │
        replication fork (Y-shape)

Leading and lagging strand synthesis:

 3' ────────────────────────────────► 5'  template
        leading strand  (continuous, 5'→3')
 5' ◄─── Okazaki ◄─── Okazaki ◄─── Okazaki   template
        fragments (discontinuous, 5'→3')
        joined by DNA ligase

Semiconservative result after one round of replication:

   Parent DNA        Daughter DNAs
   ═══════           ═══════   ═══════
   ───────           ───────   ───────
   (old–old)         (old–new) (old–new)

Key enzymes involved

Thus, during the S-phase each parental DNA molecule gives rise to two identical daughter molecules, each conserving one parental strand — the semiconservative mechanism.

(a) X-linked inheritance of eye colour (2 marks)

In Drosophila melanogaster the gene for eye colour lies on the X chromosome, and the Y chromosome carries no corresponding allele. Red eye (W) is the dominant wild-type allele; white eye (w) is recessive.

A male has only one X (he is hemizygous), so a single recessive allele is enough to make him white-eyed; a female must carry it on both X chromosomes. A male always receives his X from his mother and passes it only to his daughters.

(b) Genetic crosses and phenotypic outcomes (4 marks)

Cross 1: white-eyed female × red-eyed male

P: XʷXʷ × XᵂY. Gametes: Xʷ | Xᵂ, Y

F₁: all daughters red-eyed (carriers), all sons white-eyed, i.e. 1 red female : 1 white male. The mother's trait appears in the sons and the father's trait in the daughters.

Cross 2: red-eyed (pure) female × white-eyed male (Morgan's original cross)

P: XᵂXᵂ × XʷY. Gametes: Xᵂ | Xʷ, Y

F₁: XᵂXʷ (red female) and XᵂY (red male), so all F₁ are red-eyed.

F₁ × F₁: XᵂXʷ × XᵂY

F₂: 2 red females : 1 red male : 1 white male, i.e. 3 red : 1 white, and the white-eyed flies are all males. This sex-linked pattern differs from the autosomal 3 : 1, which appears equally in both sexes.

(c) Why this is criss-cross inheritance (2 marks)

In criss-cross inheritance a character is passed from a parent to the offspring of the opposite sex, and then on to the next generation in the same crossing pattern. In Cross 2 the white-eye allele goes from the white-eyed father to his daughters (carriers, which look red), and from these daughters to half of their sons, who are white-eyed: father → daughter → grandson. In Cross 1 the white-eyed mother passes the trait to her sons, while her daughters get the dominant allele from the father and show the father's trait. Because a son's X always comes from his mother and a daughter's from both parents, the trait crosses between the sexes in alternate generations. This is criss-cross inheritance.

Incomplete dominance is a pattern of inheritance in which neither allele of a pair is completely dominant over the other. The heterozygote (F1) shows an intermediate phenotype between the two parental (P) phenotypes, instead of resembling either parent.

Mechanism up to F2 generation — example: flower colour in Snapdragon (Antirrhinum majus)

1. Parental (P) generation: A pure red-flowered plant (RR) is crossed with a pure white-flowered plant (rr).
2. F1 generation: All offspring are heterozygous (Rr) and bear pink flowers — intermediate between red and white. Thus neither R nor r is completely dominant; each expresses partially.
3. F2 generation: F1 plants (Rr) are self-pollinated. The gametes R and r separate during meiosis and recombine at random:

Result: F2 phenotypic ratio = 1 red : 2 pink : 1 white (1 : 2 : 1), which is the same as the genotypic ratio. This is because each genotype has its own distinct phenotype — there is no masking of one allele by another.

Difference from complete dominance:

Thus in complete dominance the dominant allele fully masks the recessive allele, whereas in incomplete dominance the heterozygote shows a blended, intermediate trait.

Chromosomal Mutations

Introduction. Chromosomal mutations (chromosomal aberrations) are changes that affect the number or the structure of whole chromosomes, unlike gene mutations which affect a single gene. They arise mainly from errors in cell division (non-disjunction, breakage and faulty reunion) and from mutagens such as radiation and chemicals. They are of two broad types — numerical and structural.

1. How chromosomal mutations alter chromosomes (1 mark)

• Number: During meiosis, failure of homologous chromosomes or sister chromatids to separate (non-disjunction) produces gametes with an extra or a missing chromosome. Fertilisation of such gametes changes the chromosome number of the zygote — either by whole sets (euploidy) or by one or a few individual chromosomes (aneuploidy).
• Structure: Agents like X-rays, UV light and chemicals break the chromosome. If the broken ends rejoin abnormally — or fail to rejoin — the chromosome loses, gains, inverts or exchanges segments, altering its structure.

2. Types of chromosomal mutations (4 marks)

A. Numerical mutations (change in chromosome number)

(i) Euploidy — change in the number of complete chromosome sets (n).

Polyploidy may be autopolyploidy (extra sets from the same species) or allopolyploidy (extra sets from different species, e.g. Triticale, bread wheat). Polyploids are common in plants but usually lethal or sterile in animals.

(ii) Aneuploidy — gain or loss of one or a few chromosomes.

B. Structural mutations (change in chromosome structure)

1. Deletion (deficiency): A segment of a chromosome is lost. If lost at the end it is terminal deletion; if lost internally it is interstitial deletion. Example: Cri-du-chat syndrome (deletion in chromosome 5).
2. Duplication: A segment is repeated twice or more in the same chromosome, so genes are present in extra copies. Example: Bar-eye in Drosophila.
3. Inversion: A segment breaks, rotates 180° and rejoins. Paracentric inversion lies within one arm; pericentric inversion includes the centromere. It does not change gene number but changes gene order.
4. Translocation: A segment is transferred to a non-homologous chromosome. Reciprocal translocation involves exchange between two non-homologous chromosomes; Robertsonian translocation involves fusion of two acrocentric chromosomes at the centromere. Example: Philadelphia chromosome in chronic myeloid leukaemia.

3. Significance of chromosomal mutations (3 marks)

Harmful / negative significance

• Genetic disorders: Aneuploidy causes serious human disorders — Down's syndrome (trisomy 21), Turner's syndrome (XO), Klinefelter's syndrome (XXY), Edward's syndrome (trisomy 18).
• Reduced fertility and lethality: Most aneuploids and many structural mutants are sterile or die early; deletions and duplications upset gene balance (dosage effect) and are often lethal.
• Cancer: Translocations can activate oncogenes, e.g. Philadelphia chromosome causes leukaemia.

Useful / positive significance

• Evolution and speciation: Polyploidy and structural rearrangements create reproductive isolation and new species; allopolyploidy has given rise to crops like wheat and Triticale.
• Agriculture and horticulture: Polyploidy increases vigour, size and yield — triploid seedless banana and watermelon, tetraploid potato, hexaploid wheat.
• Genetic research: Deletions, duplications and inversions are used to map genes and study chromosome behaviour.
• Variation: They are an important source of genetic variation on which natural selection acts.

Conclusion. Chromosomal mutations thus change chromosome number (euploidy, aneuploidy) or structure (deletion, duplication, inversion, translocation). Though mostly harmful, causing disorders and sterility, they are of great evolutionary and agricultural value as a source of variation and new species.

Dihybrid cross between round-yellow and wrinkled-green pea plants

Parental generation (P₁):

• Round-yellow (pure-breeding): RRYY
• Wrinkled-green (pure-breeding): rryy

Here R = round (dominant), r = wrinkled (recessive); Y = yellow (dominant), y = green (recessive).

Gametes of parents:

• RRYY produces only RY
• rryy produces only ry

F₁ generation:

$RRYY \times rryy \rightarrow \text{all } RrYy$

All F₁ plants are round-yellow (heterozygous for both traits), showing that round and yellow are dominant.

F₂ generation — gametes of F₁:

The F₁ hybrid RrYy forms four types of gametes in equal proportion:

$RY,\ Ry,\ rY,\ ry \quad (1:1:1:1)$

Punnett square (4 × 4):

Classification of F₂ offspring:

• Round-yellow: RRYY, RRYy, RrYY, RrYy → 9
• Round-green: RRyy, Rryy → 3
• Wrinkled-yellow: rrYY, rrYy → 3
• Wrinkled-green: rryy → 1

Phenotypic ratio of F₂:

$\boxed{9 : 3 : 3 : 1}$

(round-yellow : round-green : wrinkled-yellow : wrinkled-green)

Demonstration of the law of independent assortment:

Mendel's law of independent assortment states that when two or more pairs of contrasting characters are considered together, the alleles of each pair segregate independently of the alleles of the other pair during gamete formation, and the factors assort independently.

In this cross:

1. The F₁ hybrid RrYy produces four types of gametes — RY, Ry, rY and ry — in equal proportion (1:1:1:1). This is possible only if the R/r pair and the Y/y pair separate independently of each other during meiosis.
2. If the two gene pairs were linked or dependent, only the parental combinations RY and ry would appear. Instead, the new combinations Ry and rY also appear in equal numbers, showing independent assortment.
3. The F₂ phenotypic ratio 9:3:3:1 is obtained by multiplying the two independent monohybrid ratios: $(3:1) \times (3:1) = 9:3:3:1$ This confirms that seed shape and seed colour are inherited independently of each other.
4. The appearance of the two new (recombinant) phenotypes — round-green and wrinkled-yellow — in the F₂ generation is direct proof of independent assortment.

Thus, the dihybrid cross demonstrates that the inheritance of one pair of characters is independent of the inheritance of the other pair, verifying Mendel's law of independent assortment.

Double helix structure of DNA (Watson and Crick, 1953)

• DNA consists of two polynucleotide strands coiled around the same axis to form a right-handed double helix.
• Each nucleotide is made of deoxyribose sugar, a phosphate group and a nitrogenous base (adenine, guanine, cytosine or thymine). Nucleotides join by phosphodiester bonds into a sugar-phosphate backbone on the outside.
• The two strands are antiparallel ($5' \to 3'$ and $3' \to 5'$).
• The bases lie inside and are joined by hydrogen bonds between complementary pairs: A=T (two bonds) and G$\equiv$C (three bonds). Hence the amount of A equals T, and G equals C (Chargaff's rule).
• The helix has a diameter of 2 nm, a pitch of 3.4 nm per turn, 10 base pairs per turn, and 0.34 nm between adjacent base pairs.
• It has a major groove and a minor groove.

Diagram: draw two intertwined strands with rungs showing base pairs, and label the backbone, sugar, phosphate, bases, hydrogen bonds, $5'$ and $3'$ ends, and the pitch.

If assigned to make a model of DNA, essential components needed

1. Deoxyribose sugars and phosphate groups to make two antiparallel backbones.
2. The four bases A, T, G and C in the correct pairing (A with T, G with C), joined with the right number of hydrogen bonds (2 and 3).
3. A right-handed twist with 10 base pairs per turn, and the correct dimensions (2 nm width, 3.4 nm pitch).

(d) Criss-cross inheritance

An X-linked trait passes from a father to his daughter (a normal carrier) and then to her son (the grandson), so it crosses from one sex to the other in alternate generations.

The chromosome numbers are missing from the question. Reading it as a plant with $2n$ chromosomes and a mutant with $2n+1$ (one extra chromosome):

Type of mutation: The mutant has gained a single chromosome, so it is aneuploidy (trisomy, $2n+1$). It arises by non-disjunction of a homologous pair in meiosis.

Difference between aneuploidy and euploidy

DNA as the universal genetic material (2 marks)

1. Occurrence: DNA is the genetic material in almost all organisms — viruses, bacteria, plants and animals — except some plant viruses (e.g. TMV) and retroviruses (e.g. HIV), where RNA acts as the genetic material. Hence DNA is called universal.
2. Stability and suitability: DNA is chemically more stable and less reactive than RNA because of the presence of deoxyribose sugar (which lacks the 2′–OH group) and the nitrogenous base thymine instead of uracil. It can also replicate with high fidelity and undergo mutation, so it can store and transmit genetic information reliably from generation to generation. RNA, being reactive and unstable, cannot serve as the universal genetic material.

Chemical composition of DNA (1 mark)

DNA is a polynucleotide. Its hydrolytic products are:

• Pentose sugar: 2′-deoxy-D-ribose
• Nitrogenous bases: Purines — Adenine (A) and Guanine (G); Pyrimidines — Cytosine (C) and Thymine (T)
• Phosphate groups

A nitrogenous base joined to deoxyribose forms a nucleoside; nucleoside + phosphate forms a nucleotide. Nucleotides are linked by 3′–5′ phosphodiester bonds to form a polynucleotide chain.

Watson and Crick's model of DNA (4 marks)

Watson and Crick (1953) proposed the double-helix model on the basis of X-ray diffraction data of Wilkins and Franklin and Chargaff's rules.

Salient features:

1. DNA consists of two polynucleotide chains coiled around a common axis in a right-handed helix (B-DNA).
2. The two chains are antiparallel — one runs 5′→3′ and the other 3′→5′.
3. The sugar–phosphate backbone lies on the outside and the bases project inward.
4. The two strands are held together by hydrogen bonds between complementary bases: A=T (two H-bonds) and G≡C (three H-bonds). This is the basis of complementarity.
5. One complete turn of the helix is 3.4 nm (34 Å) long and contains 10 base pairs, so the distance between two successive base pairs is 0.34 nm (3.4 Å).
6. The diameter of the helix is 2 nm (20 Å).
7. The two strands are not identical but complementary, so each strand can act as a template — the basis of semi-conservative replication.

Labelled diagram:

       3'                     5'
        |                      |
  5' —A═T—A═T—G≡C—C≡G—T═A— 3'
        |                      |
  3' —T═A—T═A—C≡G—G≡C—A═T— 5'
        |                      |
       5'                     3'

  A = Adenine   T = Thymine   G = Guanine   C = Cytosine
  ═ = 2 hydrogen bonds      ≡ = 3 hydrogen bonds
  Backbone = deoxyribose + phosphate (outside)
  One turn = 3.4 nm, 10 bp; rise per bp = 0.34 nm; diameter = 2 nm

Functions of DNA (1 mark)

1. It stores genetic information in the sequence of its nucleotides.
2. It replicates (semi-conservatively) and transmits genetic information to daughter cells and offspring.
3. It controls all cellular activities and protein synthesis through transcription (DNA → RNA → protein).
4. It undergoes mutation, which produces variations needed for evolution.

(c) Sickle Cell Anaemia

It results from a point mutation — the substitution of a single base (GAG → GTG) in the β-globin gene, replacing glutamic acid with valine at the 6th position. The other three are chromosomal (numerical) disorders.

The process is linkage.

Linkage is the tendency of genes lying on the same chromosome to be inherited together from one generation to the next. Such linked genes do not assort independently, so they mostly appear in the parental combinations. Non-parental (recombinant) combinations appear in a small number because crossing over in prophase-I of meiosis exchanges segments between non-sister chromatids and separates some linked genes.

• The closer two genes lie on a chromosome, the stronger the linkage and the fewer the recombinants.
• The farther apart they are, the weaker the linkage and the more recombinants.

Example: Bateson and Punnett's sweet pea cross, with blue flower (B) and long pollen (L) dominant to red flower (b) and round pollen (l).

Parents: $BBLL \times bbll$ gives $F_1$: $BbLl$ (blue, long).

Test cross of $F_1$ ($BbLl \times bbll$) gave mainly the parental types (blue long and red round) and only a few non-parental types (blue round and red long), instead of the 1:1:1:1 ratio expected from independent assortment. This shows that $B$ with $L$ and $b$ with $l$ stay linked.

Colour blindness as a sex-linked trait

The gene for colour vision (red–green colour blindness) is located on the X chromosome and is recessive to the normal allele. The Y chromosome carries no corresponding allele. Let the normal allele be $X^C$ and the recessive (colour-blind) allele be $X^c$.

• Females (XX): need both X chromosomes to carry $X^c$ to be colour blind ($X^cX^c$). A heterozygous female ($X^CX^c$) is a carrier — she has normal vision but transmits the allele.
• Males (XY): have only one X chromosome, so a single $X^c$ makes them colour blind ($X^cY$). Thus males are affected far more often than females.

---

Cross 1: Carrier mother × Normal father

Result: 50% of sons are colour blind; all daughters have normal vision (half of them carriers). The trait appears in sons even though neither parent is affected.

---

Cross 2: Normal mother × Colour-blind father

Result: All sons are normal; all daughters are carriers. The father's defect is not expressed in his children but is hidden in his daughters.

---

Cross 3: Carrier mother × Colour-blind father

Result: 50% of daughters and 50% of sons are colour blind — the only cross producing affected females.

---

Analysis and criss-cross inheritance

1. The trait is never transmitted from father to son, because a son receives his only X from his mother and his Y from his father.
2. A colour-blind father passes his $X^c$ only to his daughters, who become carriers; a carrier mother passes $X^c$ to half her sons, who become colour blind.
3. Hence the allele travels from father to grandson through the daughter, i.e. it crosses from one sex to the other in successive generations — father → daughter → grandson. This alternating passage of a sex-linked gene from father to daughter and from mother to son is called criss-cross inheritance.
4. Because males carry only one X, they cannot mask the recessive allele, so the defect is expressed in them much more frequently than in females.

(a) Translation would terminate prematurely.

A sense codon changed into a stop codon is a nonsense mutation, so the ribosome stops at that point and releases a shortened (truncated) polypeptide.

A mutation is a sudden, heritable change in the DNA sequence. It arises spontaneously or through mutagens (UV, X-rays, chemicals), and it has both useful and harmful effects.

Beneficial effects (2 marks)

1. Variation and evolution: Mutations produce new alleles for natural selection to act on. Example: the dark form of the peppered moth (Biston betularia) survived on soot-darkened trees in industrial areas.
2. Crop improvement: Induced mutations (gamma rays, chemicals) have given high-yielding, disease-resistant or early-maturing varieties. Example: the wheat variety Sharbati Sonora, produced by irradiating Sonora 64. Colchicine-induced polyploidy gives seedless watermelon.
3. Natural advantage in humans: The sickle-cell allele protects heterozygous carriers from malaria, and adult lactase persistence lets adults digest milk.
4. Industry and medicine: Mutant strains of Penicillium give much higher penicillin yields.

Harmful effects (2 marks)

1. Genetic diseases: Sickle-cell anaemia (point mutation, glutamic acid to valine), phenylketonuria, thalassaemia and colour blindness.
2. Chromosomal disorders: Down's syndrome (trisomy 21), Turner's syndrome (45, XO), Klinefelter's syndrome (47, XXY) and cri-du-chat syndrome (deletion in chromosome 5).
3. Cancer: Mutations in proto-oncogenes (e.g. ras) or tumour-suppressor genes (e.g. p53) cause uncontrolled cell division. UV rays can cause skin cancer and tobacco smoke lung cancer.
4. Antibiotic resistance: Mutations make bacteria resistant, so infections become hard to treat.

Most mutations are harmful or neutral, but the rare beneficial ones drive evolution and breeding.

Law of independent assortment: When two pairs of contrasting characters are considered together in a cross, the alleles of one pair segregate (assort) independently of the alleles of the other pair during gamete formation. Each gamete receives one allele from each pair, and the two pairs combine at random, so the inheritance of one character is independent of the inheritance of the other.

Cross: RrYy × RrYy (R = round, r = wrinkled; Y = yellow, y = green)

Gametes from each parent: RY, Ry, rY, ry (four types, in equal proportion)

Punnett square (F₂):

Phenotypic ratio (F₂):

Phenotypic ratio = 9 : 3 : 3 : 1

Genotypic ratio (F₂):

Genotypic ratio = 1 : 2 : 2 : 4 : 1 : 2 : 1 : 2 : 1

How the ratios support the law:

If the two pairs of characters were inherited together (linked), the F₂ offspring would show only the two parental combinations (round-yellow and wrinkled-green) in a 3 : 1 ratio. Instead, four phenotypic classes appear in a 9 : 3 : 3 : 1 ratio. The two new combinations — round-green and wrinkled-yellow — arise because the alleles R/r and Y/y separate independently of each other during gamete formation. Each parent produces four types of gametes (RY, Ry, rY, ry) in equal numbers, and these combine at random. The appearance of the two non-parental (recombinant) types in predictable proportions therefore proves that the two pairs of characters assort independently, confirming the law of independent assortment.

DNA as the genetic material

DNA qualifies as the genetic material of most organisms because:

1. It carries coded information. The sequence of bases (A, T, G, C) along the polynucleotide chain stores the information for the structure, function and development of the organism.
2. It replicates faithfully. Complementary base pairing and semi-conservative replication give exact copies, so information passes unchanged from cell to cell and parent to offspring.
3. It can mutate. Rare changes in base sequence create variation, the raw material of evolution.
4. It is chemically stable. Deoxyribose, hydrogen-bonded base pairs and base stacking make DNA far more stable than RNA, whose 2′-OH group makes it easily hydrolysed.
5. Experimental proof. Griffith's transformation experiment, Avery–MacLeod–McCarty's identification of DNA as the transforming principle, and the Hershey–Chase bacteriophage experiment showed that DNA is hereditary material. (In some viruses such as TMV and HIV, RNA is the genetic material.)

Double helical structure of DNA

Watson and Crick (1953) proposed the double-helix model of B-DNA, using X-ray data of Wilkins and Franklin and Chargaff's base-composition rules.

1. DNA has two polynucleotide chains coiled around a common axis in a right-handed helix, like a twisted ladder.
2. The chains are antiparallel: one runs $5' \to 3'$, the other $3' \to 5'$.
3. The sugar–phosphate backbone lies outside; the nitrogenous bases point inwards.
4. Bases pair by hydrogen bonds: A = T (two bonds) and G ≡ C (three bonds), so A = T and G = C (Chargaff's rule). A purine always pairs with a pyrimidine.
5. One complete turn is 3.4 nm long and has 10 base pairs, so adjacent base pairs are 0.34 nm apart.
6. The helix is 2 nm in diameter and has a wider major groove and a narrower minor groove.
7. The two strands are complementary, so each can serve as a template for the other.

Labelled diagram (simplified)

 5' P-S-P-S-P-S-P-S-P-S 3'      sugar-phosphate backbone (strand 1)
      |   |   |   |   |
      A = T   G ≡ C   A = T ...  base pairs (H-bonds)
      |   |   |   |   |
 3' P-S-P-S-P-S-P-S-P-S 5'      sugar-phosphate backbone (strand 2)

 P = phosphate, S = deoxyribose sugar
 A=T : two H-bonds;  G≡C : three H-bonds
 1 turn = 3.4 nm = 10 bp; diameter = 2 nm; major and minor grooves

Similarities between DNA and RNA

• Both are nucleic acids, polymers of nucleotides (pentose sugar + phosphate + nitrogenous base).
• Both contain the bases adenine, guanine and cytosine.
• Both have a sugar–phosphate backbone joined by $3' \to 5'$ phosphodiester bonds.
• Both take part in the flow of genetic information and protein synthesis.

Differences between DNA and RNA

Significance in genetic processes

DNA is the store of hereditary information and, by replication, passes it to daughter cells and offspring. By transcription it makes mRNA, and by translation (using mRNA, tRNA and rRNA) the message is turned into proteins, which determine the phenotype. DNA and RNA together are the molecular basis of inheritance, variation and gene expression.

(b) Pentose sugar and phosphate groups

A phosphodiester bond is formed when the phosphate group at the 5′ carbon of one nucleotide joins the hydroxyl group at the 3′ carbon of the pentose sugar of the adjacent nucleotide. Thus each phosphate links two sugars, giving the sugar–phosphate backbone of DNA and RNA.

Allopolyploidy is the condition in which an organism possesses two or more complete sets of chromosomes that come from two or more different species (i.e., the chromosome sets are of hybrid origin). It arises when two different species hybridise, and the resulting sterile hybrid later undergoes chromosome doubling (by failure of meiosis or by colchicine treatment), producing a fertile individual with a doubled chromosome number.

Steps involved:

1. Two different species (say A and B) cross to give an interspecific hybrid (AB). This hybrid is usually sterile, because the chromosomes of species A and species B are not homologous and cannot pair properly during meiosis.
2. The chromosome number of this hybrid is doubled (e.g., by colchicine, which blocks spindle formation). Now every chromosome has a homologous partner.
3. The resulting plant is an allopolyploid (AABB) and is fertile, because normal pairing and segregation can occur at meiosis.

Suitable example — Triticale: Triticale is a man-made allopolyploid obtained by crossing wheat (Triticum, AABB, $2n = 28$) with rye (Secale, RR, $2n = 14$). The sterile hybrid (ABR, $2n = 21$) is treated with colchicine to double the chromosome number, giving a fertile allopolyploid with $2n = 42$ (AABBRR). It combines the high yield and grain quality of wheat with the hardiness and disease resistance of rye.

Other examples: hexaploid bread wheat (Triticum aestivum, $2n = 42$), cultivated cotton (Gossypium hirsutum), and Raphanobrassica (radish × cabbage).

Significance: Allopolyploidy is an important mechanism of speciation in plants and is widely used in plant breeding to create new crop varieties with combined desirable traits.

Complete vs incomplete linkage

Complete linkage: two genes lie so close on the same chromosome that crossing over never separates them. Only the parental combinations enter the gametes, so a dihybrid test cross gives just two parental types in a 1 : 1 ratio and no recombinants. It is rare, e.g. in male Drosophila (no crossing over) and in a few maize genes.

Incomplete linkage: the genes are far enough apart for crossing over to occur between them occasionally. Besides the parental types, a few recombinant types appear, so the test cross gives four classes, with parental types far more numerous than recombinants. The recombination percentage is used to map the distance between genes.

Experiment showing incomplete linkage: Hutchinson's maize cross (C and Sh genes)

In maize, C (coloured aleurone) is dominant over c (colourless), and Sh (full, plump grain) over sh (shrunken). Both genes lie on the same chromosome.

1. P: coloured-plump (CCShSh) × colourless-shrunken (ccshsh).
2. F₁: all coloured-plump (CcShsh).
3. Test cross: F₁ (CcShsh) × double recessive (ccshsh).

If the genes assorted independently, the ratio would be 1 : 1 : 1 : 1. The observed result was:

Interpretation: the two parental classes are very much commoner than expected, so the genes are linked and tend to stay together. The few recombinants (301 plants) show that crossing over occasionally separates them, so linkage is incomplete.

$\text{Recombination frequency} = \frac{149 + 152}{8368} \times 100 \approx 3.6\%$

The genes are therefore about 3.6 map units apart.

Cross in which they are the same: the cross between two heterozygous plants (F1 x F1, a monohybrid cross) in a character showing incomplete dominance, as in Mirabilis jalapa (four o'clock plant).

Explanation

In Mirabilis, the red-flowered ($RR$) and white-flowered ($rr$) parents give an F1 generation in which all plants are pink ($Rr$), because neither allele completely dominates the other.

When the pink F1 plants are self-pollinated or crossed with each other:

• Genotypic ratio: $RR : Rr : rr = 1 : 2 : 1$
• Phenotypic ratio: red : pink : white $= 1 : 2 : 1$

Since the heterozygote has a distinct phenotype of its own, each genotype produces a different phenotype, so the two ratios are identical (1 : 2 : 1). In contrast, in complete dominance the same cross gives a genotypic ratio of 1 : 2 : 1 but a phenotypic ratio of 3 : 1.

Watson and Crick's model of DNA (1953)

Watson and Crick proposed a double helix model using X-ray diffraction data of Wilkins and Franklin and Chargaff's base-pairing rules.

• DNA is made of two polynucleotide strands wound around a common axis to form a right-handed helix.
• The strands are antiparallel: one runs $5' \to 3'$ and the other $3' \to 5'$.
• Each nucleotide consists of a deoxyribose sugar, a phosphate and a nitrogenous base. The sugar-phosphate backbone is on the outside and the bases are inside.
• The bases are paired by hydrogen bonds: adenine with thymine (A=T, two bonds) and guanine with cytosine (G$\equiv$C, three bonds). The two strands are therefore complementary.
• One complete turn is 3.4 nm long and contains 10 base pairs, with 0.34 nm between consecutive base pairs. The diameter of the helix is 2 nm.
• Major and minor grooves are present on the surface.

Diagram: draw two intertwined ribbons (backbones) joined by horizontal rungs, with A-T joined by two lines and G-C by three, labelled with the sugar, phosphate, bases, hydrogen bonds, $5'$ and $3'$ ends, 3.4 nm pitch and 2 nm width.

Three differences between DNA and RNA

(a) Two genes are far apart on the same chromosome.

The frequency of crossing over depends on the distance between the genes: the farther apart they are, the more likely a chiasma forms between them. Genes near the centromere or close together are linked tightly, and genes on different chromosomes assort independently.

(d) Addition or deletion of one set or more than one set of chromosomes in diploid chromosome

Euploidy is a change in the number of complete chromosome sets (e.g. haploid $n$, triploid $3n$, tetraploid $4n$). Changes of single chromosomes (options a-c) are aneuploidy.

Genetic material (1 mark)

Genetic material is the molecule that carries hereditary information and passes it from parents to offspring; in most organisms it is DNA, and in some viruses it is RNA.

Structure of RNA (2 marks)

• RNA is generally a single-stranded polynucleotide chain.
• Each nucleotide has a ribose sugar, a phosphate group and a nitrogen base: adenine, guanine, cytosine and uracil (uracil replaces thymine).
• Nucleotides are joined by phosphodiester bonds. The strand may fold back to form short double-helical regions.
• Three types: mRNA (linear, carries the code), tRNA (cloverleaf shape, anticodon loop) and rRNA (forms ribosomes with proteins).

Functions (1 mark)

• mRNA carries the genetic message from DNA for protein synthesis.
• tRNA carries amino acids to the ribosome and matches codons with its anticodon.
• rRNA forms the structure of ribosomes, where proteins are made.

(c) Frame shift

Adding or deleting a single nitrogenous base shifts the reading frame of the codons after that point, changing every following amino acid. Transition and transversion are types of base substitution, where one base is replaced by another without changing the number of bases.