Heredity

Meiosis and Sexual Life Cycles (CED Unit 5)

Chromosome Terminology

• Somatic cells: Diploid ($2n$), containing two sets of chromosomes (one from each parent).
• Gametes: Haploid ($n$), containing one set of chromosomes.
• Homologous chromosomes: Pair of chromosomes (one maternal, one paternal) with the same genes at the same loci.
• Sister chromatids: Two identical copies of a chromosome joined at the centromere (produced by DNA replication).
• Karyotype: Organized display of an individual’s chromosomes.

Meiosis Overview

Meiosis reduces the chromosome number by half, producing four genetically distinct haploid cells From one diploid cell.

Feature	Mitosis	Meiosis
Divisions	1	2
Daughter cells	2	4
Ploidy	$2n$	$n$
Genetic variation	Identical	Different
Function	Growth, repair	Gamete production

Meiosis I (Reductional Division)

Prophase I:

• Chromosomes condense, homologous chromosomes pair up (synapsis) forming tetrads (bivalents).
• Crossing over occurs between non-sister chromatids at chiasmata, exchanging genetic material.
• Nuclear envelope breaks down, spindle forms.

Metaphase I:

• Tetrads align at the metaphase plate.
• Homologous pairs (not individual chromosomes) are oriented randomly (independent assortment).

Anaphase I:

• Homologous chromosomes separate (sister chromatids remain attached).
• Each daughter cell receives one chromosome from each homologous pair.

Telophase I:

• Two haploid cells form (each chromosome still has two sister chromatids).
• Brief interphase (no DNA replication).

Meiosis II (Equational Division)

Similar to mitosis but starting with haploid cells:

Prophase II: Chromosomes recondense, spindle forms. Metaphase II: Chromosomes align singly At the metaphase plate. Anaphase II: Sister chromatids separate. Telophase II: Four haploid Daughter cells form.

Sources of Genetic Variation

1. Crossing over (Prophase I): Recombination between non-sister chromatids creates new combinations of alleles on the same chromosome.
2. Independent assortment (Metaphase I): Random orientation of homologous pairs produces $2^n$ possible gamete combinations (where $n$ is the haploid number). For humans ($n = 23$): $2^{23} \approx 8.4 \times 10^6$ combinations.
3. Random fertilization: Any sperm can fertilize any egg, further multiplying genetic diversity.

Worked Example: Calculating the number of possible gamete combinations.

An organism has $n = 4$ (haploid number). The number of possible gamete combinations from Independent assortment alone is $2^4 = 16$.

If we also consider crossing over (which can occur at many points along each chromosome), the actual Number of genetically unique gametes is much larger. For humans ($n = 23$), independent assortment Alone produces $2^{23} \approx 8.4$ million combinations. With crossing over, the number of possible Gametes is effectively infinite.

Worked Example: The role of crossing over in producing new allele combinations.

Consider two genes on the same chromosome: gene A (alleles A and a) and gene B (alleles B and b). A Parent has the genotype AB/ab (A and B on one chromosome, a and b on the homologous chromosome).

Without crossing over, this parent can only produce two types of gametes: AB and ab (parental Types).

With crossing over between genes A and B, two additional gamete types are possible: Ab and aB (recombinant types). The frequency of recombinant gametes depends on the distance between the two Genes: the farther apart they are, the more likely crossing over is to occur between them.

If the recombination frequency between A and B is 20%, then 20% of the gametes will be recombinant (10% Ab and 10% aB) and 80% will be parental (40% AB and 40% ab).

Mendelian Genetics (CED Unit 5)

Mendel’s Laws

Law of Segregation: Each individual has two alleles for each gene, which separate during gamete Formation so each gamete receives one allele.

Law of Independent Assortment: Genes on different chromosomes assort independently during gamete Formation.

Monohybrid Cross

Cross between individuals differing in one trait.

Genotype	Phenotype
AA	Dominant
Aa	Dominant
aa	Recessive

Genotypic ratio (Aa $\times$ Aa): 1 AA : 2 Aa : 1 aa Phenotypic ratio: 3 dominant : 1 Recessive

Worked Example: A test cross.

A plant has purple flowers (dominant phenotype). To determine its genotype (PP or Pp), cross it with A white-flowered plant (pp).

If the plant is PP: all offspring will be Pp (purple). If the plant is Pp: approximately half the Offspring will be Pp (purple) and half pp (white).

If any white-flowered offspring are produced, the parent must be heterozygous (Pp).

Dihybrid Cross

Cross between individuals differing in two traits.

Genotypic ratio (AaBb $\times$ AaBb): 9:3:3:1 (phenotypic ratio for both dominant : first Dominant/second recessive : first recessive/second dominant : both recessive).

Worked Example: A complete dihybrid cross.

In peas, round seeds (R) are dominant over wrinkled (r), and yellow seeds (Y) are dominant over Green (y). Cross RrYy $\times$ RrYy.

Parental genotypes: RrYy $\times$ RrYy

Gametes from each parent: RY, Ry, rY, ry (four types, each equally likely)

Punnett square (4 $\times$ 4):

	RY	Ry	rY	ry
RY	RRYY	RRYy	RrYY	RrYy
Ry	RRYy	RRyy	RrYy	Rryy
rY	RrYY	RrYy	rrYY	rrYy
ry	RrYy	Rryy	rrYy	rryy

Phenotypic ratio: 9 round yellow : 3 round green : 3 wrinkled yellow : 1 wrinkled green.

Probability of a wrinkled green offspring: 1/16.

Probability of a round yellow offspring: 9/16.

Incomplete Dominance

Heterozygote has an intermediate phenotype. Example: red ($RR$) $\times$ white ($rr$) = pink ($Rr$).

Codominance

Both alleles are fully expressed in the heterozygote. Example: blood type AB expresses both A and B Antigens.

Multiple Alleles

More than two alleles exist for a gene in the population. Example: ABO blood group has three Alleles: $I^A$, $I^B$And $i$.

Genotype	Blood Type
$I^A I^A$ or $I^A i$	A
$I^B I^B$ or $I^B i$	B
$I^A I^B$	AB (codominance)
$ii$	O

Sex-Linked Inheritance

Genes on the X chromosome exhibit sex-linked inheritance. Males (XY) have only one X, so a single Recessive allele on the X chromosome will be expressed.

Example: Red-green color blindness and hemophilia are X-linked recessive.

Cross: Carrier female ($X^NX^n$) $\times$ normal male ($X^NY$):

	$X^N$	$Y$
$X^N$	$X^NX^N$ (normal female)	$X^NY$ (normal male)
$X^n$	$X^NX^n$ (carrier female)	$X^nY$ (affected male)

Phenotypic ratio: 1 normal female : 1 carrier female : 1 normal male : 1 affected male.

Linked Genes

Genes on the same chromosome tend to be inherited together (linked). The recombination frequency Measures the distance between linked genes:

\mathrm{Recombination frequency = \frac{\mathrm{Number of recombinant offspring}{\mathrm{Total offspring} \times 100

• $1\%$ recombination = 1 map unit (centimorgan, cM).
• Recombination frequency $\gt 50\%$: genes are unlinked (on different chromosomes or very far apart).

Worked Example: Gene mapping.

Genes A, B, and C are on the same chromosome. The recombination frequencies are:

A-B: 12% B-C: 8% A-C: 4%

The gene order is A—C—B (not A—B—C), because the recombination frequency between A and C (4%) is Less than between A and B (12%). If the order were A—B—C, the A-C distance would be 20%, not 4%.

The map is:

A ---4cM--- C ---8cM--- B

Non-Mendelian Genetics

Epistasis

One gene masks or modifies the expression of another gene.

Example: Coat color in Labrador retrievers. Gene B determines pigment (B = black, b = brown). Gene E determines deposition (E = pigment deposited, e = no pigment deposited).

	E_	ee
B_	Black	Yellow
bb	Brown	Yellow

Ratio: 9 black : 3 brown : 4 yellow.

Pleiotropy

One gene affects multiple, seemingly unrelated phenotypic traits. Example: Marfan syndrome affects The skeleton, eyes, and cardiovascular system.

Polygenic Inheritance

Multiple genes contribute to a single phenotypic trait, producing continuous variation. Examples: Height, skin color, eye color.

Environmental Influence on Gene Expression

Phenotype depends on both genotype and environment. Example: identical twins may have different Phenotypes due to different environments.

Chromosomal Abnormalities

Nondisjunction

Failure of homologous chromosomes or sister chromatids to separate during meiosis, producing gametes With abnormal chromosome numbers.

Condition	Cause	Description
Trisomy 21 (Down syndrome)	Extra chromosome 21	Intellectual disability, characteristic facial features
Trisomy 18 (Edwards)	Extra chromosome 18	Severe developmental delays
Trisomy 13 (Patau)	Extra chromosome 13	Severe birth defects
Klinefelter (XXY)	Extra X in male	Male with some female characteristics
Turner (XO)	Missing X in female	Female, sterile, short stature
XYY	Extra Y in male	Tall, normal

Structural Abnormalities

• Deletion: Loss of a chromosome segment.
• Duplication: Repeated segment.
• Inversion: Segment reversed in orientation.
• Translocation: Segment moved to a nonhomologous chromosome.

Common Pitfalls

1. Confusing mitosis and meiosis. Mitosis produces 2 identical diploid cells; meiosis produces 4 genetically distinct haploid cells.
2. Misidentifying when crossing over occurs. It occurs in Prophase I of meiosis, not mitosis.
3. Confusing sister chromatids with homologous chromosomes. Sister chromatids are identical copies; homologous chromosomes carry different alleles.
4. Forgetting that sex-linked traits are expressed differently in males and females. Males only need one copy of an X-linked recessive allele to express the trait.
5. Confusing incomplete dominance with codominance. In incomplete dominance, the heterozygote is intermediate (blended); in codominance, both phenotypes are fully expressed.
6. Incorrectly calculating recombination frequency. Only count recombinant (non-parental) offspring.
7. Confusing epistasis with dominance. Epistasis involves interactions between different genes; dominance involves alleles of the same gene.
8. Thinking that linked genes never recombine. Crossing over can separate linked genes; the probability depends on the distance between them.
9. Forgetting that nondisjunction can occur in meiosis I or meiosis II. Nondisjunction in meiosis I produces two abnormal gametes (both with extra or both missing a chromosome); nondisjunction in meiosis II produces one normal and one abnormal gamete.
10. Confusing trisomy and triploidy. Trisomy is one extra chromosome (2n+1); triploidy is a full extra set of chromosomes (3n).
11. Assuming the dominant phenotype always means homozygous dominant. A dominant phenotype can be either homozygous dominant or heterozygous — a test cross is needed to determine which.
12. Confusing genotype and allele frequencies. Genotype frequencies refer to proportions of specific genotypes (e.g., $p^2$, $2pq$, $q^2$); allele frequencies refer to proportions of specific alleles (e.g., $p$, $q$).
13. Thinking crossing over occurs at every chiasma. A chiasma is the visible result of crossing over; not all chiasmata result in recombination of every gene between them.

Practice Questions

1. In peas, tall (T) is dominant over short (t). A tall plant crossed with a short plant produces offspring that are all tall. What are the genotypes of the parents?

2. A woman who is a carrier for color blindness ($X^NX^n$) marries a man with normal vision ($X^NY$). What is the probability that their son will be colorblind?

3. Two genes, A and B, are 12 map units apart. If an individual with genotype $AB/ab$ is test-crossed, what percentage of the offspring will be $Ab/ab$?

4. Explain how nondisjunction during meiosis I differs from nondisjunction during meiosis II in terms of the resulting gametes.

5. In a dihybrid cross between $AaBb$ parents, what proportion of offspring will be homozygous recessive for both traits?

6. A man with blood type B (whose mother was type O) marries a woman with blood type AB. What are the possible blood types of their children?

7. Explain how epistasis differs from dominance, using coat color in Labrador retrievers as an example.

8. Why does independent assortment produce more genetic variation in organisms with higher haploid chromosome numbers? Calculate the number of possible gamete combinations for an organism with $n = 4$.

9. Explain the difference between incomplete dominance and codominance, giving an example of each.

10. A woman with Turner syndrome (XO) is colour blind. Explain which parent the colour blindness allele was inherited from and why.

11. Three genes (D, E, F) are linked. The recombination frequency between D and E is 20%, between E and F is 10%, and between D and F is 30%. Determine the order of the genes and draw a gene map.

12. Explain why linked genes do not always follow Mendel’s law of independent assortment.

13. A dihybrid cross between two pea plants (RrYy $\times$ RrYy) produces 800 offspring. How many offspring would you expect to have wrinkled green seeds?

14. Describe the process of crossing over during Prophase I and explain how it contributes to genetic variation.

15. Explain why the phenotypic ratio from a dihybrid cross with epistasis (9:3:4) differs from the standard Mendelian ratio (9:3:3:1).

16. A couple has a child with cystic fibrosis (autosomal recessive). Neither parent has the disease. What is the probability that their next child will also have cystic fibrosis?

17. Explain how nondisjunction in meiosis I leads to gametes with an extra chromosome, using a specific example.

18. Compare and contrast the inheritance patterns of autosomal dominant, autosomal recessive, and X-linked recessive traits.

19. In snapdragons, red flowers (RR) crossed with white flowers (WW) produce pink flowers (RW). What phenotypic ratio would you expect from crossing two pink-flowered plants?

20. Explain the concept of gene mapping and how recombination frequency is used to determine the distance between genes on a chromosome.

21. A man with blood type A (genotype unknown) marries a woman with blood type B (genotype unknown). Their first child has blood type O. What are the genotypes of the parents, and what blood types are possible for their future children?

22. Explain how X-inactivation (the Lyon hypothesis) affects the expression of X-linked genes in female mammals.

23. A species has $2n = 8$. How many chromosomes, chromatids, and DNA molecules are present in a cell during (a) G1, (b) G2, (c) metaphase of mitosis, (d) metaphase I of meiosis, and (e) metaphase II of meiosis?

24. Explain why linked genes on the same chromosome can still show independent assortment if they are very far apart.

25. Describe the inheritance pattern of mitochondrial DNA and explain why mitochondrial disorders are always maternally inherited.

Review: Chromosomal Theory of Inheritance and Linked Genes

The chromosomal theory of inheritance states that genes are located on chromosomes and that the Behaviour of chromosomes during meiosis accounts for inheritance patterns.

Linked genes: Genes located on the same chromosome tend to be inherited together because they Are physically connected. This violates Mendel’s law of independent assortment, which only applies To genes on different chromosomes.

Crossing over and recombination: During Prophase I of meiosis, homologous chromosomes pair up And exchange segments at chiasmata. Crossing over can separate linked genes if it occurs between Them. The probability of crossing over between two genes is proportional to the distance between Them: genes that are farther apart are more likely to be separated by crossing over.

Recombination frequency: The percentage of offspring that are recombinant (have a non-parental Combination of alleles).

\mathrm{Recombination frequency = \frac{\mathrm{Number of recombinant offspring}{\mathrm{Total offspring} \times 100

• 1% recombination = 1 centimorgan (cM) or 1 map unit.
• Maximum recombination frequency of 50% means the genes are effectively unlinked (on different chromosomes or very far apart on the same chromosome).

Worked Example: Three-point cross and gene mapping.

A three-point cross involves three linked genes and provides more accurate gene mapping than a Two-point cross because it can detect double crossovers.

Genes A, B, and C are linked in the order A—B—C. The parental gametes are ABC and abc. A test Cross produces the following offspring:

• ABC: 320 (parental)
• abc: 310 (parental)
• Abc: 30 (single crossover between A and B)
• aBC: 28 (single crossover between A and B)
• ABc: 15 (single crossover between B and C)
• abC: 17 (single crossover between B and C)
• AbC: 5 (double crossover)
• aBc: 3 (double crossover)

Total offspring: 728.

Recombination frequency between A and B:

$\frac{30 + 28 + 5 + 3}{728} \times 100 = \frac{66}{728} \times 100 \approx 9.1\%$

So A and B are approximately 9.1 cM apart.

Recombination frequency between B and C:

$\frac{15 + 17 + 5 + 3}{728} \times 100 = \frac{40}{728} \times 100 \approx 5.5\%$

So B and C are approximately 5.5 cM apart.

Gene map: A ---9.1 cM--- B ---5.5 cM--- C

The double crossovers (AbC and aBc) are the least frequent class, confirming that C is on the far Side from A.

Why double crossovers matter: Without the three-point cross, the observed recombination Frequency between A and C would be:

$\frac{30 + 28 + 15 + 17 + 5 + 3}{728} \times 100 = \frac{98}{728} \times 100 \approx 13.5\%$

But the actual distance is $9.1 + 5.5 = 14.6$ cM. The double crossovers make the A-C distance appear Shorter than it really is because they swap the alleles back to the parental arrangement. The Three-point cross corrects for this by counting the double crossovers twice.

Review: Sex Determination and Sex-Linked Traits in Detail

Mammalian sex determination: The SRY gene (Sex-determining Region Y) on the Y chromosome Triggers testis development. In the absence of SRY, ovaries develop. Therefore, the default pathway Is female development; male development requires the SRY gene product.

Dosage compensation (X-inactivation): Females have two X chromosomes while males have one. To Balance gene expression, one X chromosome in each female cell is randomly inactivated during early Embryonic development, forming a Barr body (a condensed, transcriptionally inactive X chromosome).

X-inactivation was discovered by Mary Lyon (1961) and is called the Lyon hypothesis. Because Inactivation is random, female mammals are mosaics: different cells express different X chromosomes. This is visible in calico cats (where coat colour is X-linked): only females can be calico because They have two X chromosomes with different colour alleles, and random X-inactivation produces Patches of different colours.

Worked Example: Inheritance of X-linked haemophilia through three generations.

Queen Victoria was a carrier of haemophilia ($X^HX^h$). Her husband was normal ($X^HY$).

Punnett square:

	$X^H$	$Y$
$X^H$	$X^HX^H$ (normal)	$X^HY$ (normal)
$X^h$	$X^HX^h$ (carrier)	$X^hY$ (haemophiliac)

Each daughter has a 50% chance of being a carrier. Each son has a 50% chance of having haemophilia. Historically, several of Queen Victoria’s male descendants were affected by haemophilia, which Spread through the royal families of Europe.

Y-linked inheritance: Genes on the Y chromosome are passed only from father to son. Examples Include the SRY gene (sex determination) and some genes involved in male fertility. Y-linked traits Are very rare because the Y chromosome is small and carries few genes.

Review: Polygenic Inheritance and Continuous Variation

Most traits are influenced by multiple genes (polygenic inheritance), producing continuous variation That follows a normal distribution (bell curve).

Examples of polygenic traits: Height, skin colour, eye colour, intelligence, risk of developing Certain diseases (e.g., type 2 diabetes, heart disease).

Worked Example: Skin colour as a polygenic trait.

Skin colour in humans is influenced by at least 3 genes, each with two alleles (contributing dark or Light pigmentation). If a person is homozygous for all dark alleles (AABBCC), they have very dark Skin. If homozygous for all light alleles (aabbcc), they have very light skin. Most people have Intermediate genotypes, producing a range of skin colours. When two people with intermediate skin Colour have children, the offspring show a continuous range of skin colours centred around the Parental average, following a normal distribution.

Environmental influence on polygenic traits: Even strongly genetic traits are influenced by the Environment. Height, for example, is approximately 80% heritable, but nutrition during childhood has A significant effect. A person with genes for tall stature who is malnourished during childhood will Not reach their full genetic potential height.

Review: Epigenetics and Gene Expression

Epigenetics refers to heritable changes in gene expression that do not involve changes to the DNA Sequence. Epigenetic modifications can be influenced by environmental factors and can sometimes be Passed to offspring.

DNA methylation: Methyl groups (-\mathrm{CH_3) are added to cytosine bases, in CpG Islands near gene promoters. Methylation generally silences gene expression by preventing Transcription factors from binding.

Histone modification: Histone proteins can be acetylated, methylated, or phosphorylated. Acetylation of histones generally loosens chromatin structure, promoting gene expression. Deacetylation (by histone deacetylases, HDACs) condenses chromatin and represses gene expression.

Worked Example: Epigenetic effects of diet on gene expression.

The Agouti mouse model demonstrates how maternal diet can affect offspring phenotype through Epigenetic changes. Pregnant mice fed a diet rich in methyl donors (folic acid, vitamin B12, Choline) produce offspring with brown coats and healthy weight. Mice fed a methyl-deficient diet Produce offspring with yellow coats, obesity, and increased cancer risk. The methyl-rich diet Increases DNA methylation at the Agouti gene locus, silencing the gene. This demonstrates that Environmental factors (diet) can alter epigenetic marks and affect phenotype.

Review: Mitochondrial Inheritance

Mitochondria are inherited almost exclusively from the mother. This is because:

1. The egg contributes the vast majority of cytoplasm (and therefore mitochondria) to the zygote.
2. The sperm contributes mainly its nucleus.
3. Sperm mitochondria are often destroyed after fertilisation.

Implications: Mitochondrial disorders (e.g., Leber hereditary optic neuropathy, mitochondrial Myopathy) are always maternally inherited. An affected mother passes the disorder to all of her Children, but an affected father does not pass it to any of his children. Mitochondrial DNA is also Used in tracing maternal ancestry (mitochondrial Eve).

Review: Summary Table of Inheritance Patterns

Pattern	Autosomal Dominant	Autosomal Recessive	X-Linked Recessive
Example	Huntington disease	Cystic fibrosis	Haemophilia, colour blindness
Affected individuals	Both sexes equally	Both sexes equally	Males much more frequently
Transmission	Affected parent to 50% offspring	Two carrier parents can have affected offspring	Carrier mother to 50% of sons
Skipping generations	No (vertical pattern)	Yes (horizontal pattern)	Yes (through carrier females)

Review: Genomic Imprinting

Genomic imprinting is an epigenetic phenomenon in which certain genes are expressed in a Parent-of-origin-specific manner. For some imprinted genes, only the paternal allele is expressed (the maternal allele is silenced); for others, only the maternal allele is expressed.

Mechanism: Imprinting occurs through DNA methylation during gamete formation. The methylation Marks are different in sperm and eggs, so the same gene can carry different epigenetic marks Depending on which parent it was inherited from.

Example: Prader-Willi syndrome and Angelman syndrome. Both conditions are caused by a deletion On chromosome 15, but the phenotype depends on which parent’s chromosome is deleted:

• If the paternal copy is deleted: Prader-Willi syndrome (characterised by constant hunger, obesity, intellectual disability, and short stature).
• If the maternal copy is deleted: Angelman syndrome (characterised by severe intellectual disability, seizures, ataxia, and a happy demeanour).

This demonstrates that the same chromosomal region can produce entirely different disorders Depending on which parent contributed the affected chromosome.

Worked Example: Understanding imprinting through inheritance patterns.

A woman with Angelman syndrome (deletion of maternal UBE3A region) has a child with a normal father. The child inherits a normal maternal chromosome 15 and a normal paternal chromosome 15. The child Will be unaffected because the paternal UBE3A allele is active in the relevant brain regions.

However, if the same woman’s brother (who has Prader-Willi syndrome due to the same deletion on his Paternal chromosome 15) has a child with a normal woman, the child will also be unaffected because It inherits a normal paternal chromosome 15 from the father.

Review: Variations in Inheritance — Lethal Alleles

Some alleles cause death when present in certain genotypes. These are called lethal alleles.

Recessive lethal alleles: Cause death only in the homozygous recessive condition. Heterozygotes Are unaffected carriers. Example: Tay-Sachs disease (autosomal recessive; homozygous recessive Individuals die in early childhood).

Dominant lethal alleles: Cause death even in the heterozygous condition. These are very rare Because they would kill the individual before they could reproduce. However, some dominant lethal Alleles act later in life, such as Huntington disease (onset after age 40), which allows The allele to be passed to offspring before symptoms appear.

Worked Example: How lethal alleles alter expected ratios.

In mice, the allele for yellow coat colour (Y) is dominant over grey (y). However, YY is lethal (embryos die), while Yy is yellow and yy is grey. Crossing two yellow mice (Yy $\times$ Yy) gives:

	Y	y
Y	YY	Yy
y	Yy	yy

Expected genotypes: 1 YY : 2 Yy : 1 yy.

Since YY is lethal, the observed ratio among live offspring is 2 Yy : 1 yy, which gives a phenotypic Ratio of 2 yellow : 1 grey, instead of the standard 3:1 ratio.

Review: Penetrance and Expressivity

Penetrance is the percentage of individuals with a particular genotype who show the expected Phenotype. If 80% of individuals with a dominant allele show the trait, the penetrance is 80%.

Complete penetrance: All individuals with the genotype show the phenotype (100%).

Incomplete (reduced) penetrance: Some individuals with the genotype do not show the phenotype. Example: BRCA1 gene mutations increase breast cancer risk, but not all carriers develop cancer.

Expressivity refers to the degree or severity with which a particular genotype is expressed in The phenotype. Variable expressivity means that individuals with the same genotype can show Different severities of the trait.

Worked Example: Neurofibromatosis type 1 (NF1).

NF1 is an autosomal dominant disorder. However, it shows both incomplete penetrance and variable Expressivity. Some individuals with the NF1 mutation show only a few cafe-au-lait spots (mild Expression), while others develop multiple neurofibromas, learning difficulties, and skeletal Abnormalities (severe expression). Rarely, an individual with the mutation shows no symptoms at all (incomplete penetrance).

Review: Uniparental Disomy

Uniparental disomy (UPD) occurs when an individual receives two copies of a chromosome (or part of a Chromosome) from one parent and no copy from the other parent. This can result in the expression of Recessive disorders or imprinting disorders.

Types of UPD:

• Heterodisomy: The individual receives two different homologous chromosomes from one parent (resulting from nondisjunction in meiosis I).
• Isodisomy: The individual receives two identical copies of the same chromosome from one parent (resulting from nondisjunction in meiosis II or from mitotic duplication of a monosomic chromosome).

Clinical significance: UPD can cause recessive disorders even if only one parent is a carrier. For example, if a child has maternal uniparental disomy for chromosome 7 and the mother is a carrier For cystic fibrosis (recessive), the child could be homozygous for the CF allele and have cystic Fibrosis, even if the father is not a carrier.

UPD for chromosomes with imprinted regions (e.g., chromosomes 7, 11, 15) can also cause imprinting Disorders. For instance, maternal UPD of chromosome 15 causes Prader-Willi syndrome, while paternal UPD of chromosome 15 causes Angelman syndrome.

Review: Telomeres and Ageing

Telomeres are repetitive nucleotide sequences (TTAGGG in humans) at the ends of chromosomes. They protect the chromosome ends from degradation and prevent chromosomes from fusing with each Other.

The end-replication problem: DNA polymerase cannot fully replicate the 3’ end of the lagging Strand, so telomeres shorten with each cell division. After approximately 50-70 cell divisions, Telomeres become critically short, triggering cellular senescence (the Hayflick limit).

Telomerase is an enzyme that adds telomeric repeats to chromosome ends. It is active in germ Cells, stem cells, and most cancer cells, but is inactive in most somatic cells. The reactivation of Telomerase in cancer cells allows them to divide indefinitely, making telomerase a target for cancer Therapy.

Worked Example: Telomere shortening and cellular ageing.

A skin cell divides 30 times. Each division shortens the telomere by approximately 50-200 base Pairs. After 30 divisions, the telomere has shortened by approximately 1500-6000 base pairs. When The telomere becomes too short, the cell enters senescence and can no longer divide. This limits the Regenerative capacity of tissues and contributes to ageing. In contrast, stem cells in the basal Layer of the epidermis express telomerase, maintaining their telomere length and allowing continuous Skin renewal throughout life.

Review: Chromosome Painting and Karyotyping Techniques

Karyotyping is the process of arranging chromosomes in a standard format to visualise Chromosomal abnormalities. Techniques include:

1. G-banding: Chromosomes are treated with the enzyme trypsin and stained with Giemsa, producing a characteristic banding pattern. Each chromosome has a unique banding pattern, allowing identification of individual chromosomes and detection of structural abnormalities (deletions, duplications, translocations).

2. Fluorescence in situ hybridisation (FISH): Fluorescently labelled DNA probes bind to specific chromosomal regions. This allows detection of specific chromosomal abnormalities, including microdeletions that are too small to see with standard karyotyping.

3. Comparative genomic hybridisation (CGH): Compares the DNA from a test sample with a reference sample to detect copy number variations (deletions and duplications) across the entire genome.

4. Spectral karyotyping (SKY): Uses multiple fluorescent probes, each painting a different chromosome in a different colour. This is particularly useful for identifying complex chromosomal rearrangements and translocations.

Worked Example: Detecting a translocation using FISH.

A patient has suspected chronic myelogenous leukaemia (CML), which is caused by a translocation Between chromosomes 9 and 22 (the Philadelphia chromosome). A FISH probe specific to the BCR gene on Chromosome 22 (labelled red) and a probe specific to the ABL gene on chromosome 9 (labelled green) Are applied. In a normal cell, the red and green signals appear on separate chromosomes. In a CML Cell, a fused yellow signal appears on the Philadelphia chromosome, confirming the t(9;22)(q34;q11) Translocation. This test is both sensitive and specific, and results can be obtained within 24-48 Hours.

Practice Problems

Question 1: Three-point cross and gene mapping

Genes D, E, and F are linked in the order D—E—F. A test cross of a trihybrid individual (\mathrm{DEF/def) produces the following offspring: \mathrm{DEF = 400, \mathrm{def = 380 \mathrm{DEf = 60$$\mathrm{deF = 70$$\mathrm{Def = 20$$\mathrm{dEF = 30 \mathrm{DeF = 15$$\mathrm{dEf = 25. Calculate the recombination frequencies between each pair Of genes and draw the genetic map.

Answer

Total offspring: $400 + 380 + 60 + 70 + 20 + 30 + 15 + 25 = 1000$.

Parental types (most frequent): \mathrm{DEF (400) and \mathrm{def (380).

Double crossovers (least frequent): \mathrm{DeF (15) and \mathrm{dEf (25).

Single crossover between D and E: \mathrm{DEf (60) and \mathrm{deF (70).

Single crossover between E and F: \mathrm{Def (20) and \mathrm{dEF (30).

Recombination frequency D-E: $\frac{60 + 70 + 15 + 25}{1000} \times 100 = \frac{170}{1000} \times 100 = 17.0\%$

Recombination frequency E-F: $\frac{20 + 30 + 15 + 25}{1000} \times 100 = \frac{90}{1000} \times 100 = 9.0\%$

Recombination frequency D-F: $\frac{60 + 70 + 20 + 30 + 15 + 25}{1000} \times 100 = \frac{220}{1000} \times 100 = 22.0\%$

Genetic map: D ---17.0 cM--- E ---9.0 cM--- F

Note that D-F distance (22.0 cM) equals D-E (17.0) + E-F (9.0) = 26.0 cM minus double crossovers ($2 \times 4.0 = 8.0\%$), because double crossovers restore the parental arrangement for the outer Markers. Using corrected distance: D-F = $22.0 + 2 \times 4.0 = 26.0$ cM.

Question 2: Hardy-Weinberg with selection

In a population, cystic fibrosis (autosomal recessive) has an incidence of 1 in 2500. If individuals With cystic fibrosis have a fitness of 0 (they do not reproduce), what will the frequency of the Cystic fibrosis allele be after one generation of selection?

Answer

Current frequency: $q^2 = 1/2500 = 0.0004$So $q = 0.02$ and $p = 0.98$.

Genotype frequencies before selection: p^2 = 0.9604$$2pq = 0.0392$$q^2 = 0.0004.

Fitness values: w_{AA} = 1$$w_{Aa} = 1$$w_{aa} = 0.

Mean fitness: $\bar{w} = p^2(1) + 2pq(1) + q^2(0) = 0.9604 + 0.0392 = 0.9996$.

After selection, the frequency of the recessive allele is: $q' = \frac{pq \cdot w_{Aa} + q^2 \cdot w_{aa}}{2\bar{w}} = \frac{0.98 \times 0.02 \times 1 + 0.0004 \times 0}{2 \times 0.9996} = \frac{0.0196}{1.9992} \approx 0.0098$

After one generation of selection, the allele frequency drops from $q = 0.02$ to $q' \approx 0.0098$. The frequency approximately halves because homozygous recessive individuals are Eliminated, removing two copies of the allele per affected individual.

Question 3: Epistasis dihybrid cross

In summer squash, white colour (W) is dominant over yellow (w), and disc-shaped fruit (D) is Dominant over sphere-shaped (d). When a white disc plant is crossed with a yellow sphere plant, all F1 offspring are white disc. The F2 generation shows a ratio of 12 white : 3 yellow disc : 1 yellow Sphere. Explain the genetic basis for this ratio and determine the genotypes of the F2 phenotypes.

Answer

This is an example of dominant epistasis, where the dominant allele at one gene (W) masks the Expression of alleles at the other gene (D/d).

The 12:3:1 ratio arises because:

• Any genotype with at least one W allele is white (regardless of D/d): 9 \mathrm{ W\_D\_ + 3 \mathrm{ W\_dd = 12 white.
• Genotypes that are \mathrm{wwD\_ are yellow disc: 3 yellow disc.
• Genotype \mathrm{wwdd is yellow sphere: 1 yellow sphere.

F2 genotypes:

• White: \mathrm{WWDD$$\mathrm{WWDd$$\mathrm{WWdd$$\mathrm{WwDD$$\mathrm{WwDd \mathrm{Wwdd
• Yellow disc: \mathrm{wwDD$$\mathrm{wwDd
• Yellow sphere: \mathrm{wwdd

The W gene is epistatic to the D gene. When the dominant W allele is present, the D gene’s Expression is completely masked, and all fruits appear white regardless of whether they carry D or d Alleles.

Question 4: Nondisjunction in meiosis I vs meiosis II

A woman with normal karyotype (46, XX) has a child with Klinefelter syndrome (47, XXY). Explain two Different mechanisms by which this could occur: (a) nondisjunction in meiosis I of oogenesis and (b) Nondisjunction in meiosis II. In each case, show the gametes produced and the resulting zygote.

Answer

(a) Nondisjunction in meiosis I: Homologous X chromosomes fail to separate. One secondary oocyte Receives both X chromosomes (XX) and the other receives none (O). After meiosis II, the XX cell Produces two XX gametes. When fertilised by a normal Y sperm, the zygote is XXY (Klinefelter).

Meiosis I: XX and O (instead of X and X). Meiosis II: XX $\to$ XX and XX; O $\to$ O and O. Gametes: XX, XX, O, O.

Zygote: XX (egg) + Y (sperm) = XXY.

(b) Nondisjunction in meiosis II: Meiosis I proceeds normally, producing secondary oocytes with one X each. In meiosis II, sister chromatids fail to separate in one cell, producing one XX gamete and One O gamete from that cell. When fertilised by a normal Y sperm, the XX gamete produces XXY.

Meiosis I: X and X (normal). Meiosis II: one X cell divides normally (X and X), the other has Nondisjunction (XX and O). Gametes: X, X, XX, O.

Zygote: XX (egg) + Y (sperm) = XXY.

Question 5: Sex-linked cross with probability

A woman who is a carrier for haemophilia ($X^HX^h$) and a man with normal clotting ($X^HY$) have two Children. What is the probability that both children are sons, both are colourblind, and both are Unaffected by haemophilia?

Answer

Each child’s sex is independent with P(son) = 1/2.

For each son, the probability of being affected (having haemophilia) is 1/2 (the son receives either $X^H$ or $X^h$ from the carrier mother and Y from the father).

For each son, the probability of being unaffected is 1/2.

Probability that both children are sons AND both are unaffected: $P = \left(\frac{1}{2}\right)^2 \times \left(\frac{1}{2}\right)^2 = \frac{1}{4} \times \frac{1}{4} = \frac{1}{16}$

Breaking this down: P(both sons) = $(1/2)(1/2) = 1/4$. Given both are sons, P(both unaffected) = $(1/2)(1/2) = 1/4$. Combined: $1/4 \times 1/4 = 1/16$.

Worked Examples

Example 1: Monohybrid inheritance

In pea plants, tall (T) is dominant over dwarf (t). A heterozygous tall plant is crossed with a dwarf plant. Show the cross and state the expected phenotypic ratio.

Solution:

Parental genotypes: $\text{Tt} \times \text{tt}$

Gametes: $\text{T}, \text{t}$ and $\text{t}, \text{t}$

Punnett square:

	T	t
t	Tt	tt
t	Tt	tt

Phenotypic ratio: 1 tall : 1 dwarf (or 50% tall, 50% dwarf)