Class 12 Biology Chapter 4 Principles of Inheritance and Variation

Key Concept: Polyploidy, Chemical Mutagens, Induction

b) Spindle fiber formation
[Solution Description]
Colchicine inhibits spindle fiber formation during cell division, preventing chromosomes from separating into daughter cells. This leads to chromosome doubling, as the cell cannot complete cytokinesis properly.

Key Concept: Polyploidy, Allopolyploidy, Speciation

a) $AABB$, 4n=30
[Solution Description]
The $F_1$ hybrid from $AA \times BB$ would have genome $AB$ and chromosome number n=7 + n=8 = 2n=15. Chromosome doubling in the $F_1$ hybrid results in an allotetraploid with genome $AABB$ and chromosome number 4n=30.

Key Concept: Polyploidy, Autopolyploidy, Allopolyploidy

a) Tetrasomic, 4n=40, genome $AAAA$
[Solution Description]
The original diploid species has genome $AA$ with 2n=20 chromosomes. If mitosis occurs but cytokinesis fails, the chromosome number doubles to 4n=40, resulting in an autotetraploid with genomic composition $AAAA$.

Key Concept: Mutagenic Agents, Nitrous Acid

d) Nitrous acid ($HNO_2$)
[Solution Description]
Nitrous acid ($HNO_2$) is known to cause deamination of bases, converting cytosine to uracil and adenine to hypoxanthine, which alters their base pairing properties and leads to mutations.

Key Concept: Autopolyploidy, Allopolyploidy

c) AABB
[Solution Description]
The $F_1$ hybrid from crossing $AA$ and $BB$ would have the genome $AB$. After chromosome doubling, the allopolyploid would have two sets of each genome, resulting in $AABB$.

Key Concept: Aneuploidy, Hyperploidy

c) 15
[Solution Description]
Trisomy involves the addition of one extra chromosome to the normal diploid set. Given the diploid number is $2n = 14$, a trisomic individual would have $2n + 1 = 15$ chromosomes.

Key Concept: Translocation, Aneuploidy

d) Translocation
[Solution Description]
Translocation involves the exchange of segments between non-homologous chromosomes. This can lead to partial sterility due to the loss of genes in some cells.

Key Concept: Inversion, Crossing-over

b) $A E D C B F G$
[Solution Description]
In a pericentric inversion, the centromere lies within the inverted segment. The segment $B C D E$ is rotated by 180°, changing the sequence to $A E D C B F G$.

Key Concept: Deletion, Duplication

b) $A B E F G$
[Solution Description]
An interstitial deletion involves the removal of a segment from within the chromosome. Here, the segment $C D$ is deleted, so the remaining sequence would be $A B E F G$.

Key Concept: Spontaneous Mutation, Induced Mutation

c) Induced mutations occur due to external mutagens, while spontaneous mutations arise naturally
[Solution Description]
Spontaneous mutations occur naturally without external factors, often due to replication errors or natural mutagens like cosmic rays. Induced mutations are caused by artificial mutagens like UV radiation, chemicals, or X-rays. UV-induced mutations are not spontaneous because they require an external mutagen. Spontaneous mutations arise randomly at low frequencies, while induced mutations can be controlled and have higher frequencies under exposure to mutagens.

Key Concept: Transition Mutation, Transversion Mutation

c) Transversion mutation
[Solution Description]
The given mutation involves a substitution of 'A' (adenine, a purine) by 'U' (uracil, a pyrimidine) in mRNA (or equivalently 'T' in DNA since A-T pairs exist in DNA). This is a transversion mutation because it involves replacing a purine with a pyrimidine. Transition mutations involve purine-purine or pyrimidine-pyrimidine substitutions, while transversions involve purine-pyrimidine swaps.

Key Concept: Substitution Mutation, Frame Shift Mutation

b) Deletion of the first 'A' in the first codon (ATG $\rightarrow$ TG)
[Solution Description]
Frame shift mutations are more severe than substitution mutations because they alter the entire reading frame downstream. Deletion or insertion of a nucleotide causes a frame shift, whereas substitution may only change one amino acid or have no effect (silent mutation). Here, the example shows a DNA segment where frame shift mutations would drastically alter the protein product compared to substitutions.
For instance, deleting the first 'A' in ATG changes the reading frame entirely:
Original codons: ATG (Met), CCA (Pro), GAT (Asp), TTA (Leu).
After deletion: TGC (Cys), CAG (Gln), ATT (Ile), etc. — completely different from the original.

Key Concept: Chromosomal mutations, Deletion

c) The normal chromosome forms a loop
[Solution Description]
When a chromosome has a deletion, its homologous chromosome cannot pair with the missing segment during meiosis. As a result, the normal chromosome forms a loop (deletion loop) to align properly with the homologous regions of the deleted chromosome. This loop corresponds to the region that is missing in the deleted chromosome.

Key Concept: Substitution mutation, Transition mutation

b) Glycine
[Solution Description]
The original codon is GAG (glutamic acid). A transition mutation replaces a purine with another purine or a pyrimidine with another pyrimidine. Changing the second base (A, a purine) to G (another purine) gives the codon GGG, which codes for glycine. Thus, the amino acid changes from glutamic acid to glycine.

Key Concept: Gene mutations, Frame shift mutation

b) AGC TAG AT
[Solution Description]
The original DNA sequence is ATG CTA GAT. Deleting the second nucleotide (T) gives the new sequence AGC TAG AT. This shifts the reading frame, changing all downstream codons. The original codons were ATG (start), CTA (Leu), GAT (Asp). After deletion: AGC (Ser), TAG (stop), AT (incomplete codon). This leads to premature termination of translation, resulting in a non-functional protein.

Key Concept: Induction of Polyploidy

c) By inhibiting spindle formation during mitosis or meiosis
[Solution Description]
Colchicine inhibits spindle fiber formation during cell division, preventing chromosomes from separating properly. This disruption leads to the doubling of chromosome numbers, resulting in polyploid cells.

Key Concept: Autopolyploidy vs. Allopolyploidy

c) Autotetraploids have multiple copies of the same genome, while allotetraploids have genomes from two different species.
[Solution Description]
An autotetraploid arises from the multiplication of the same genome (e.g., $AA$ becoming $AAAA$). In contrast, an allotetraploid results from hybridization between two distinct species followed by chromosome doubling (e.g., $A$ and $B$ genomes combining to form $AABB$).

Key Concept: Polyploidy and Speciation

d) Reproductive isolation due to differing chromosome numbers
[Solution Description]
Polyploidy leads to instant speciation because the new polyploid organism has a different chromosome number than its parent species, preventing successful reproduction with the original species. This reproductive isolation enables the polyploid population to evolve independently, forming a new species. Additionally, asexual reproduction allows the polyploid to propagate quickly.

Key Concept: Hyperploidy

d) 48 chromosomes
[Solution Description]
A tetrasomic individual has two extra copies of a specific chromosome (i.e., 2n + 2). For a human with a normal 2n = 46, the tetrasomic condition for chromosome 21 results in 46 + 2 = 48 chromosomes.

Key Concept: Euploidy

b) AAAA
[Solution Description]
An autopolyploid is formed when the same genome is multiplied. For a diploid species with genome AA, an autotetraploid will have four identical sets of the same genome (AAAA).

Key Concept: Aneuploidy

b) 6 chromosomes
[Solution Description]
A nullisomic condition refers to the loss of both homologous chromosomes in a pair. For a diploid organism with $2n = 8$, the nullisomic condition for chromosome 5 would result in $2n - 2 = 6$ chromosomes. The missing chromosomes are a homologous pair, so the total number of chromosomes is reduced by 2.

Key Concept: Translocation

c) Loss of genes in some gametes and partial sterility.
[Solution Description]
Reciprocal translocation involves the exchange of segments between non-homologous chromosomes. During gamete formation, independent segregation can lead to unbalanced gametes where some genes are lost or duplicated, reducing viability or causing partial sterility.

Key Concept: Inversion

b) $A-B-C-F-E-D-G$
[Solution Description]
In a pericentric inversion, the centromere lies within the inverted segment. If the original sequence is $A-B-C-D-E-F-G$, then inverting $D-E-F$ while including the centromere gives $A-B-C-F-E-D-G$.

Key Concept: Deficiency or Deletion

c) The normal chromosome will form a loop near the deficient region.
[Solution Description]
During meiotic pairing, the normal chromosome will form a loop where the deleted segment is absent in the other homologue because only identical regions pair. This loop appears near the deficient region.

Key Concept: Effects of gene mutations

d) Base substitution mutation
[Solution Description]
Sickle-cell anemia results from a base substitution mutation in the $\beta$-globin gene, where a single nucleotide change replaces glutamic acid with valine in the hemoglobin protein. This small change causes hemoglobin to crystallize at low oxygen levels, distorting red blood cells into a sickle shape.

Key Concept: Substitution mutations

c) Replacement of a purine with a pyrimidine or vice versa
[Solution Description]
A transversion mutation involves replacing a purine base (A or G) with a pyrimidine base (C or T) or vice versa. For example, an $AT$ pair could be replaced by $TA$, or $CG$ could be replaced by $GC$. This type of substitution alters the DNA sequence significantly compared to transitions (purine-purine or pyrimidine-pyrimidine changes).

Key Concept: Frame shift mutations

c) Deletion mutation
[Solution Description]
A deletion mutation is caused by the removal of a single nucleotide from a gene, altering the reading frame of codons from the point of deletion onwards. This causes all subsequent codons to be misread, leading to a completely different protein sequence. Such mutations are harmful because they disrupt the normal function of the protein.

Key Concept: Substitution Mutation

c) Transversion mutation
[Solution Description]
Substitution mutations involve replacing one nucleotide with another. A transversion mutation specifically refers to the replacement of a purine (A or G) with a pyrimidine (C or T) or vice versa. The correct answer identifies this type of substitution.

Key Concept: Chromosomal Aberration

c) It results in the loss of one or more genes, disrupting genic balance.
[Solution Description]
Chromosomal deletions involve the loss of a segment of a chromosome, which can result in the absence of one or more genes. This disturbs the genic balance and often leads to phenotypic abnormalities. The correct answer must describe the effect of gene loss due to deletion.

Key Concept: Frame Shift Mutation

c) It causes the reading frame of codons to shift due to the addition of a nucleotide.
[Solution Description]
A frame shift mutation occurs when one or more nucleotides are inserted into a DNA sequence, altering the reading frame of codons from the point of insertion onward. This changes the sequence of amino acids in the resulting protein, often leading to non-functional proteins. The correct answer is the option that correctly identifies this phenomenon.

Key Concept: Allopolyploidy

b) Doubling of chromosomes in a hybrid derived from two different species
[Solution Description]
Allopolyploidy results from the doubling of chromosomes in a hybrid derived from two different species. For example, if species X has genome A and species Y has genome B, their hybrid would have AB. Chromosome doubling in this hybrid results in an allopolyploid with AABB.

Key Concept: Autopolyploidy

b) Multiple copies of the same set of chromosomes
[Solution Description]
Autopolyploidy occurs when an organism has multiple copies of the same set of chromosomes. For instance, a diploid species with genome AA can become an autotetraploid with genome AAAA.

Key Concept: Polyploidy Definition

b) A condition where an organism has more than two complete sets of chromosomes
[Solution Description]
Polyploidy is a condition where an organism has more than two complete sets of chromosomes. For example, if a diploid cell (2n = 10) fails to divide during mitosis, it may result in a tetraploid cell with 4n = 20 chromosomes.

Key Concept: Hyperploidy - Trisomy

c) $2n + 1$
[Solution Description]
Trisomy is a hyperploid condition where an organism has an extra chromosome, resulting in $2n + 1$ chromosomes. For example, Down syndrome (Trisomy 21) occurs due to an extra copy of chromosome 21.
Nullisomy ($2n - 2$) and monosomy ($2n - 1$) are hypoploid conditions, whereas tetrasomy ($2n + 2$) involves an extra pair of homologous chromosomes. Thus, the correct answer is $2n + 1$.

Key Concept: Euploidy - Polyploidy

d) Allopolyploidy
[Solution Description]
Allopolyploidy results from the doubling of chromosome number in a hybrid derived from two different species. If species X and Y have genomes A and B respectively, their F₁ hybrid will have AB genome. Chromosome doubling produces an allopolyploid with AABB.
Autopolyploidy involves multiplication of genomes from the same species, while allopolyploidy involves genomes from different species. Therefore, the correct answer is "Allopolyploidy."

Key Concept: Aneuploidy - Monosomy

b) $2n - 1$
[Solution Description]
A monosomic individual has one less chromosome than the diploid set. If the diploid number is $2n$, then the monosomic condition is $2n - 1$.
For example, if a normal organism has $2n = 4$ chromosomes, a monosomic individual would have $3$ chromosomes ($4 - 1 = 3$). The correct option must represent this condition.

Key Concept: Inversion

b) The centromere is inside the inverted segment.
[Solution Description]
In a pericentric inversion, the centromere lies within the inverted segment. Crossing-over within the inversion loop results in chromatids with missing or duplicated genes, leading to cell inviability.

Key Concept: Translocation

c) Exchange of segments between non-homologous chromosomes.
[Solution Description]
Translocation involves the exchange of chromosome parts between two non-homologous chromosomes. This leads to partial sterility due to the loss of genes in some gametes and a reduction in viable gametes.

Key Concept: Deletion

c) Loss of a segment of a chromosome.
[Solution Description]
A chromosomal deletion involves the loss or absence of a segment of a chromosome, which can affect one or more genes. It disrupts the genic balance and alters the phenotype. Deletions can be identified by distortions in meiotic pairing, such as loop formation when homologous chromosomes pair.

Key Concept: Effect of Gene Mutations

c) Base substitution mutation
[Solution Description]
Sickle-cell anemia results from a base substitution mutation where valine replaces glutamic acid in the $\beta$ chains of hemoglobin due to a single nucleotide change.

Key Concept: Substitution Mutation

b) Transition mutation
[Solution Description]
In substitution mutations, a nucleotide is replaced by another. A transition mutation specifically replaces a purine (A or G) with another purine or a pyrimidine (C or T) with another pyrimidine.

Key Concept: Frame Shift Mutation

c) A deletion or insertion of nucleotides causes the reading frame to shift.
[Solution Description]
A frame shift mutation occurs when one or more nucleotides are inserted or deleted, causing the reading frame of codons to shift from the site of change onward. This alters the sequence of amino acids in the resulting protein.

Key Concept: Effects of Gene Mutations

c) Substitution mutation
[Solution Description]
Sickle-cell anaemia results from a single base substitution mutation where valine replaces glutamic acid in the $\beta$ chain of hemoglobin ($HbS$).

Key Concept: Types of Mutations

b) Substitution mutation
[Solution Description]
Substitution (replacement) mutations involve the replacement of one nucleotide by another, which can be either a transition or transversion.

Key Concept: Definition of Mutation

b) A heritable change in the structure of a gene or chromosome
[Solution Description]
A mutation is a heritable change in the structure of a gene or chromosome, altering DNA sequences and affecting the genotype and phenotype of an organism.

Key Concept: Synapsis, Crossing Over

a) Formation of the synaptonemal complex
[Solution Description]
Synapsis is the process where homologous chromosomes pair up during the zygotene stage of prophase I. This precise pairing ensures proper alignment of homologs, which is essential for subsequent crossing over during pachytene. Without synapsis, crossing over would not occur accurately.

Key Concept: Recombination Frequency, Genetic Mapping

b) The order is A-C-B
[Solution Description]
Given: RF(A-B) = 15%, RF(B-C) = 10%, RF(A-C) = 5%. The sum of RF(A-B) and RF(B-C) is 25%, but the actual RF(A-C) is 5%. This suggests that gene B is located outside genes A and C because the distance between A and C should be the difference of distances (15% - 10% = 5%). Hence, the correct order is A-C-B.

Key Concept: Mechanism of Recombination, Chiasmata Formation

b) 2
[Solution Description]
During prophase I, homologous chromosomes pair up and form chiasmata where crossing over occurs. If crossing over happens at one chiasma between two non-sister chromatids, it results in 2 recombinant chromatids and 2 non-recombinant chromatids. Since each meiosis produces 4 gametes, half of them (2) will be recombinant. Therefore, the number of recombinant gametes is 2.

Key Concept: Significance of Recombination

c) Ensuring identical genetic makeup in offspring
[Solution Description]
Genetic recombination has several significances: it increases genetic diversity, helps in chromosome mapping, and provides raw material for evolution. However, ensuring identical genetic makeup in offspring is not a result of recombination; rather, recombination ensures diversity.

Key Concept: Mechanism of Recombination

d) Prophase I
[Solution Description]
Crossing over occurs during prophase I of meiosis. In this stage, homologous chromosomes pair up and exchange genetic material through chiasmata formation, resulting in recombination.

Key Concept: Recombination Frequency Calculation

c) 25%
[Solution Description]
To calculate the recombination frequency, we use the formula:
$RF = \frac{\text{Number of recombinant offspring}}{\text{Total number of offspring}} \times 100$
Plugging in the given values:
Number of recombinant offspring = 120
Total number of offspring = 480
So,
$RF = \frac{120}{480} \times 100 = 25%$
The recombination frequency is 25%.

Key Concept: Significance of Recombination

c) Reduces mutation rates
[Solution Description]
Recombination increases genetic diversity, aids evolution by creating new gene combinations, and helps in chromosome mapping. It does not reduce mutation rates.

Key Concept: Mechanism of Recombination

c) Pachytene of Prophase I
[Solution Description]
Crossing over happens during prophase I of meiosis, specifically in the pachytene stage when homologous chromosomes exchange genetic material.

Key Concept: Recombinants Definition

c) Recombinants
[Solution Description]
Recombinants are organisms that show a mix of parental traits due to genetic recombination during meiosis.

Key Concept: Mutation types, Genic balance

d) Recessive and non-lethal
[Solution Description]
Since the mutation results in a non-functional protein but the organism survives, it indicates that the mutation is not lethal. Most mutations are recessive and affect genic balance, but they may or may not be lethal. Thus, this mutation is likely recessive and non-lethal.

Key Concept: Spontaneous mutation, Induced mutation

c) They originate without human intervention and have a lower frequency.
[Solution Description]
Spontaneous mutations occur naturally without human intervention, while induced mutations are artificially created. Spontaneous mutations have a lower frequency compared to induced mutations, which are purposefully caused at higher rates using mutagens.

Key Concept: Independent assortment, Crossing over, Random fertilization

c) 8
[Solution Description]
The number of possible combinations of maternal and paternal chromosomes in gametes due to independent assortment is given by $2^n$, where $n$ is the number of chromosome pairs. Here, $n = 3$. So, the number of combinations is $2^3 = 8$.

Key Concept: Types of Mutations

d) Mutations arise suddenly and randomly, affecting germplasm
[Solution Description]
According to Hugo de Vries' mutation theory, mutations arise suddenly and randomly, altering the germplasm. They can be dominant or recessive, lethal or non-lethal, and may be beneficial or harmful. Most mutations are recessive and lethal, disrupting genic balance. Mutations provide raw material for evolution but do not directly result in new species.

Key Concept: Mutations

c) Spontaneous mutations occur naturally, while induced mutations are man-made
[Solution Description]
Spontaneous mutations arise naturally without human intervention, often due to errors in DNA replication or environmental factors like radiation. Induced mutations are artificially created by exposing organisms to mutagens such as X-rays or chemicals. Spontaneous mutations occur continuously at a low frequency, while induced mutations are deliberately introduced and have a higher frequency of occurrence.

Key Concept: Independent Assortment of Chromosomes

a) Random alignment of homologous chromosomes leading to new allele combinations
[Solution Description]
The independent assortment of chromosomes during meiosis results in new combinations of alleles in gametes. This occurs because homologous chromosome pairs align randomly at the metaphase plate, leading to different segregation patterns. For a diploid organism with $n$ chromosome pairs, there are $2^n$ possible combinations of chromosomes in gametes.

Key Concept: Independent Assortment

b) 9:3:3:1
[Solution Description]
The $F_1$ generation produces all round yellow seeds ($RrYy$). When $F_1$ plants self-cross, the $F_2$ generation shows a phenotypic ratio of 9 round yellow : 3 round green : 3 wrinkled yellow : 1 wrinkled green, i.e., $9:3:3:1$.

Key Concept: Chromosomal Mutations

c) Inversion
[Solution Description]
Inversion is a chromosomal mutation where a segment of the chromosome rotates by $180^\circ$, rearranging the gene sequence but not altering the genetic material quantity. This can be paracentric (centromere outside inversion loop) or pericentric (centromere inside).

Key Concept: Mutations

b) Naturally occurring mutagenic agents like electric currents and temperature variations
[Solution Description]
Spontaneous mutations arise naturally due to factors like electric currents, atomic particles, and temperature variations as mentioned in the syllabus. They are not artificially induced.

Key Concept: Turner’s syndrome, Non-disjunction

b) Maternal meiosis I
[Solution Description]
Turner’s syndrome results from the absence of an X chromosome in the ovum (XO). This can happen if non-disjunction occurs during:
1. Meiosis I in the mother (homologous X chromosomes fail to separate) or
2. Meiosis II in the father (sister chromatids fail to separate).
However, maternal meiosis I errors are more common for X chromosome non-disjunction.
Step-by-step reasoning:
1. Non-disjunction in maternal meiosis I leads to ova missing an X chromosome.
2. Fertilization of such an ovum by a normal sperm (X or Y) results in XO zygote.
3. Paternal meiosis II errors are less likely to cause XO zygotes.

Key Concept: Turner’s syndrome, Genetic frequency

c) 4
[Solution Description]
Turner’s syndrome occurs in 1 out of every 2500 female births. To find the expected number of cases in 10,000 females:
Step-by-step calculation:
1. Divide the total number of females (10,000) by the frequency denominator (2500).
2. $10,000 \div 2500 = 4$.
3. Thus, 4 cases are expected.

Key Concept: Turner’s syndrome, Chromosomal non-disjunction

c) Webbed neck and short stature
[Solution Description]
The karyotype described indicates Turner’s syndrome (44 + XO). The clinical features include: immature female phenotype, webbed neck, short stature, underdeveloped breasts, and sterility due to poorly developed ovaries. Below-average intelligence and short 4th metacarpal/metatarsal bones are also common.
Step-by-step reasoning:
1. Karyotype shows 45 chromosomes (XO), confirming Turner’s syndrome.
2. Review key features: webbed neck, short stature, and sterility are hallmark signs.
3. Eliminate options not associated with Turner’s syndrome (e.g., male characteristics or extra X chromosomes).

Key Concept: Diagnostic Methods, Genetic Basis

c) Karyotyping to identify the chromosomal pattern.
[Solution Description]
Karyotyping is the definitive diagnostic method for Klinefelter's syndrome. It reveals the presence of an extra X chromosome ($47, XXY$). Hormonal assays can support the diagnosis by showing low testosterone and elevated FSH/LH levels, but karyotyping provides the genetic confirmation.

Key Concept: Clinical Implications, Management

b) Androgen replacement therapy to increase testosterone levels.
[Solution Description]
Individuals with Klinefelter's syndrome often have low testosterone due to underdeveloped testis. Androgen replacement therapy can help improve secondary sexual characteristics, muscle mass, and bone density. Fertility options are limited, but assisted reproductive technologies may be considered. Psychological support is also crucial due to associated mental health challenges.

Key Concept: Genetic Basis, Phenotypic Effects

b) The individual has 47 chromosomes with an extra X chromosome ($44 + XXY$).
[Solution Description]
Klinefelter's syndrome is caused by non-disjunction during meiosis, resulting in a karyotype of $47, XXY$. The presence of the Y chromosome leads to masculine development, but the extra X chromosome causes feminizing effects such as underdeveloped testis, sparse body hair, and some breast development. Individuals are sterile males and often have mild mental impairment.

Key Concept: Non-disjunction, Meiosis

b) Anaphase I
[Solution Description]
Non-disjunction occurs when homologous chromosomes fail to separate during meiosis I or sister chromatids fail to separate during meiosis II. In the case of sex chromosome abnormalities like Klinefelter’s (XXY) or Turner’s (XO) syndrome, non-disjunction can happen either in meiosis I (homologous X chromosomes fail to separate) or meiosis II (sister chromatids fail to separate). However, it most commonly occurs during meiosis I.

Key Concept: Turner’s syndrome, Monosomy

b) 45, XO
[Solution Description]
The symptoms—short stature, webbed neck, broad chest, and underdeveloped ovaries—are typical of Turner’s syndrome. This condition results from monosomy of the X chromosome, leading to a karyotype of 45, XO (44 autosomes + 1 X chromosome). Females normally have 46 chromosomes (44 + XX), but in Turner’s syndrome, one X chromosome is missing or altered.

Key Concept: Klinefelter’s syndrome, Non-disjunction

c) 47, XXY
[Solution Description]
The symptoms described—underdeveloped testis, sparse body hair, and some breast development—are characteristic of Klinefelter’s syndrome. This condition arises due to non-disjunction during meiosis, resulting in an extra X chromosome in males (47, XXY). Normal males have 46 chromosomes (44 + XY), but individuals with Klinefelter’s syndrome have 47 chromosomes (44 + XXY). Thus, the karyotype would show 47, XXY.

Key Concept: Genetic Mechanism, Risk Factors

c) Oocytes in older women are more likely to undergo non-disjunction
[Solution Description]
Advanced maternal age increases the risk of non-disjunction during meiosis I or II in oogenesis, leading to eggs with an incorrect number of chromosomes. Older oocytes are more prone to errors in chromosome segregation due to weakened spindle apparatus or accumulated DNA damage. This results in a higher likelihood of producing an egg with two copies of chromosome 21, which can lead to trisomy 21 upon fertilization.

Key Concept: Diagnosis, Amniocentesis

c) The fetus has Down's syndrome
[Solution Description]
The karyotype 47,XX,+21 indicates the presence of an extra chromosome 21 in a female fetus (XX), confirming a diagnosis of Down's syndrome. Amniocentesis is a prenatal test that detects chromosomal abnormalities by analyzing fetal cells from amniotic fluid. The "+21" denotes trisomy 21, which is the genetic cause of Down's syndrome.

Key Concept: Chromosomal Abnormality, Non-disjunction

c) 22 autosomes and 2 copies of chromosome 21
[Solution Description]
Down's syndrome is caused by trisomy of chromosome 21, which occurs when an egg or sperm carries an extra copy of chromosome 21 due to non-disjunction. If the mother contributed the abnormal gamete, her egg must have had two copies of chromosome 21 instead of one. When fertilized by a normal sperm with one chromosome 21, the zygote would have three copies (trisomy 21). The correct option should describe the egg having 24 chromosomes (23 + extra chromosome 21).

Key Concept: Genetic counseling, Risk assessment

d) Approximately 1%, accounting for potential undetected factors.
[Solution Description]
For parents with a child with standard trisomy-21 (non-translocation) and normal parental karyotypes, the recurrence risk is slightly higher (~1%) than the general population risk (~0.1% for maternal age <30). This slight increase is due to potential germline mosaicism or undetected predispositions.
Calculation:
1. General population risk for maternal age 28: ~1 in 1,000 (0.1%).
2. Recurrence risk after one affected child: ~1% (10-fold increase).
3. No familial translocation reduces the risk to baseline + minor factors.
4. Germline mosaicism (undetectable in blood karyotype) could elevate risk marginally.
5. Thus, the recurrence risk is approximately 1%.

Key Concept: Phenotypic traits, Chromosomal basis

b) Epicanthal eye folds and hypotonia (low muscle tone).
[Solution Description]
Down's syndrome (trisomy-21) has distinct phenotypic features. While other trisomies (e.g., Patau or Edwards syndrome) may share some traits, the combination of a transverse palmar crease, protruding tongue, and epicanthal eye folds is pathognomonic for Down's. Karyotyping revealing 47,XX/+21 or 47,XY/+21 is diagnostic.
Key steps:
1. Transverse palmar crease and protruding tongue are classic signs of Down's.
2. Congenital heart defects occur in ~50% of Down's cases (e.g., atrioventricular septal defects).
3. Epicanthal folds (eye folds) further differentiate Down's from other trisomies.
4. Other trisomies (e.g., trisomy-18) present with clenched fists or rocker-bottom feet, not seen here.
5. Thus, epicanthal folds would confirm Down's syndrome specifically.

Key Concept: Non-disjunction, Trisomy-21

d) Non-disjunction in the mother’s egg during oogenesis led to the extra chromosome 21.
[Solution Description]
Down's syndrome arises from non-disjunction during gametogenesis, leading to trisomy-21. Advanced maternal age (>35 years) is a significant risk factor due to increased meiotic errors in oogenesis. Here, the mother’s age (40 years) suggests higher likelihood of non-disjunction in her egg. The father’s age has minimal influence compared to maternal age. The karyotype confirms trisomy-21, consistent with Down's syndrome.
Step-by-step reasoning:
1. The karyotype shows 47 chromosomes (+1 chromosome 21), confirming trisomy-21.
2. Non-disjunction (failure of chromosome separation) during meiosis I/II in the mother’s egg is the primary cause.
3. Maternal age >35 elevates the risk of such errors.
4. Paternal age (35 years) is not a major contributing factor here.
5. Thus, the extra chromosome 21 most likely originated from the mother’s gamete.

Key Concept: Turner’s syndrome, Sex chromosome abnormalities

c) 45, XO
[Solution Description]
Turner’s syndrome is caused by monosomy of the X chromosome (45, XO), resulting in only one X chromosome instead of the usual two. This leads to features like short stature, webbed neck, and underdeveloped ovaries. The condition arises due to non-disjunction during meiosis, producing gametes lacking an X chromosome.

Key Concept: Klinefelter’s syndrome, Sex chromosome abnormalities

d) 47, XXY
[Solution Description]
Klinefelter’s syndrome is characterized by the presence of an extra X chromosome in males (47, XXY). This results from non-disjunction of sex chromosomes during meiosis, leading to abnormal gametes with either two X chromosomes or one Y chromosome. Symptoms include underdeveloped testes, sparse body hair, and sometimes gynecomastia (breast development).

Key Concept: Down's syndrome, Chromosomal abnormalities

d) Non-disjunction during meiosis in gamete formation
[Solution Description]
Down's syndrome occurs due to trisomy of chromosome 21, resulting in 47 chromosomes instead of the normal 46. This happens due to non-disjunction during meiosis, where chromosome 21 fails to separate properly during gamete formation. When such a gamete fuses with a normal gamete, the resulting zygote has an extra copy of chromosome 21.

Key Concept: Albinism, Enzyme Deficiency

b) Melanocyte tyrosinase
[Solution Description]
Albinism is caused by the absence or defect of the enzyme melanocyte tyrosinase. This enzyme is required for the conversion of tyrosine to melanin, the pigment responsible for skin and hair color. Without this enzyme, melanin cannot be produced, leading to albinism.

Key Concept: Thalassaemia, Autosomal Recessive Inheritance

b) 25%
[Solution Description]
Both parents are carriers, so their genotypes are Tt (where T is normal, t is thalassemia allele).
Possible gametes from each parent: T or t (each with 50% chance).
The possible offspring combinations are:
- TT (normal): 25% chance.
- Tt (carrier): 50% chance.
- tt (affected): 25% chance.
Thus, the probability of an affected child (tt) is 25%.

Key Concept: Haemophilia, Pedigree Analysis

b) 25%
[Solution Description]
The woman is a carrier, so her genotype is X$^H$X$^h$ (where X$^H$ is normal, X$^h$ is haemophilic). The man is normal, so his genotype is X$^H$Y.
Possible gametes from the mother: X$^H$ or X$^h$ (each with 50\% chance).
Possible gametes from the father: X$^H$ or Y (each with 50% chance).
The possible offspring combinations are:
- X$^H$X$^H$ (normal daughter): 25% chance.
- X$^H$X$^h$ (carrier daughter): 25% chance.
- X$^H$Y (normal son): 25% chance.
- X$^h$Y (haemophilic son): 25% chance.
Thus, the probability of a haemophilic son is 25%.

Key Concept: Turner's Syndrome, Chromosomal Abnormalities

c) Turner's syndrome
[Solution Description]
The symptoms described (short stature, webbed neck, underdeveloped ovaries, and karyotype 45, XO) are characteristic of Turner's syndrome, a sex chromosome abnormality due to monosomy of the X chromosome.

Key Concept: Thalassaemia, Autosomal Recessive Disorder

b) 25%
[Solution Description]
Both parents are carriers (HbA/HbS). The possible genotypes for their offspring are: HbA/HbA (normal), HbA/HbS (carrier), and HbS/HbS (affected). The probability of having a child with $\beta$-thalassemia major (HbS/HbS) is 25% (1 out of 4).

Key Concept: Haemophilia, Pedigree Analysis

b) 50%
[Solution Description]
The woman is a carrier (X$^H$X$^h$) and the man is normal (X$^H$Y). For a son to have haemophilia, he must inherit the X$^h$ chromosome from his mother. The possible combinations for the son are X$^H$Y (normal) or X$^h$Y (haemophilic). The probability is 50% since there are two equally likely outcomes.

Key Concept: Frequency and Management

c) 1 in 2500 females; estrogen therapy
[Solution Description]
Turner's syndrome affects approximately 1 in 2500 female births. While there is no cure, hormonal treatments like estrogen therapy are used to aid in developing secondary sexual characteristics such as breast development.

Key Concept: Phenotypic Features

b) Above-average intelligence
[Solution Description]
Turner's syndrome is marked by features such as webbed neck, short stature, poorly developed ovaries, and underdeveloped breasts. However, above-average intelligence is not associated with this condition; affected individuals typically have below-average intelligence.

Key Concept: Genetic Basis

b) Monosomy of X chromosome (XO)
[Solution Description]
Turner's syndrome arises due to monosomy of the X chromosome, resulting in a karyotype of 45 chromosomes (44 + XO). This occurs because of non-disjunction during gamete formation, leading to one less sex chromosome.

Key Concept: Epidemiology

b) 1 in 500
[Solution Description]
Klinefelter’s syndrome has a prevalence of approximately 1 in every 500 male births, as mentioned in the syllabus.

Key Concept: Clinical Features

c) Fully functional reproductive system
[Solution Description]
Klinefelter’s syndrome presents with underdeveloped testes, sparse body hair, possible breast development, and sterility. A fully functional reproductive system is not observed in affected individuals, making it the incorrect option.

Key Concept: Genetic Basis

c) 44 + XXY
[Solution Description]
Klinefelter’s syndrome arises due to non-disjunction during gametogenesis, leading to an extra X chromosome in males. The correct chromosomal composition is $44 + XXY$, resulting in a total of 47 chromosomes.

Key Concept: Sex chromosome abnormalities

c) Gametes with missing or extra sex chromosomes
[Solution Description]
Non-disjunction occurs when homologous chromosomes or sister chromatids fail to separate properly during meiosis. In sex chromosomes, this can result in gametes with either an extra or missing sex chromosome. Fertilization involving these gametes can produce conditions like Klinefelter’s syndrome (XXY) or Turner’s syndrome (XO).

Key Concept: Sex chromosome abnormalities

c) Webbed neck and short stature
[Solution Description]
Turner’s syndrome is characterized by monosomy (XO), where individuals have only one X chromosome. Common phenotypic features include a webbed neck, short stature, underdeveloped ovaries, and absence of secondary sexual characteristics like breast development. These individuals are sterile females.

Key Concept: Sex chromosome abnormalities

c) 44 + XXY
[Solution Description]
Klinefelter’s syndrome is characterized by trisomy of sex chromosomes, resulting in an extra X chromosome. The affected individual has 47 chromosomes (44 + XXY) instead of the normal 46 (44 + XY). This occurs due to non-disjunction during meiosis, leading to an abnormal number of chromosomes.

Key Concept: Clinical Features

c) Unusually tall height
[Solution Description]
Common physical features of Down's syndrome include a short stocky body, characteristic eye folds, lack of muscle tone, transverse crease on the palm, and flattened nasal bridge. However, unusually tall height is not associated with Down's syndrome, making it the correct answer.

Key Concept: Diagnosis and Risk Factors

c) Amniocentesis
[Solution Description]
Amniocentesis is a prenatal test that involves extracting a small amount of amniotic fluid for genetic analysis. It can detect chromosomal abnormalities like trisomy 21 (Down's syndrome) early in pregnancy. The other options listed are not standard methods for detecting Down's syndrome.

Key Concept: Chromosomal Abnormality

c) Trisomy of chromosome 21
[Solution Description]
Down's syndrome results from trisomy of chromosome 21, meaning there are three copies of chromosome 21 instead of the usual two. This occurs due to non-disjunction during gamete formation, leading to a total of 47 chromosomes (45 + XY or XX + extra chromosome 21). The correct answer corresponds to the description of this abnormality.

Key Concept: Symptoms of Down’s syndrome

b) High muscle tone
[Solution Description]
Down’s syndrome is characterized by features such as mental retardation, flattened nasal bridge, and lack of muscle tone. High muscle tone is not associated with Down’s syndrome; rather, affected individuals often have low muscle tone (hypotonia).

Key Concept: Diagnosis of Down’s syndrome

c) Amniocentesis
[Solution Description]
Amniocentesis is a prenatal test where a small amount of amniotic fluid is extracted and analyzed to detect chromosomal abnormalities, including Down’s syndrome. It is typically performed during the second trimester.

Key Concept: Down’s syndrome chromosomal abnormality

c) Trisomy of chromosome 21
[Solution Description]
Down’s syndrome is caused by trisomy of chromosome 21, resulting in a total of 47 chromosomes instead of the normal 46. This occurs due to non-disjunction during gamete formation, leading to an extra copy of chromosome 21 in the zygote.

Key Concept: Turner’s syndrome

b) Monosomy of the X chromosome (44 + XO)
[Solution Description]
Turner’s syndrome arises from monosomy of the X chromosome (44 + XO), leading to immature female development, webbed neck, short stature, and sterility.

Key Concept: Klinefelter’s syndrome

c) A condition where males have an extra X chromosome (44 + XXY)
[Solution Description]
Klinefelter’s syndrome is characterized by an extra X chromosome in males (44 + XXY), leading to underdeveloped testes, sparse body hair, and sometimes breast development. These individuals are sterile and may have mild mental impairment.

Key Concept: Down's syndrome

c) Trisomy of chromosome 21
[Solution Description]
Down's syndrome is caused by trisomy of chromosome 21, resulting in a total of 47 chromosomes (45 + XX or 45 + XY). This occurs due to non-disjunction during gamete formation.

Key Concept: Mendelian Disorder - Thalassaemia

b) Beta-globin chain
[Solution Description]
Thalassaemias are autosomal recessive disorders affecting haemoglobin synthesis. In $\beta$-thalassemia specifically, the production of the $\beta$-globin chain is impaired, leading to abnormal haemoglobin formation and subsequent anaemia.

Key Concept: Mendelian Disorder - PKU

b) Phenylalanine hydroxylase
[Solution Description]
Phenylketonuria (PKU) is caused by the absence or deficiency of the enzyme phenylalanine hydroxylase. This enzyme is responsible for converting phenylalanine to tyrosine. Without it, phenylalanine accumulates, leading to mental retardation and other developmental defects.

Key Concept: Mendelian Disorder - Haemophilia

c) 50%
[Solution Description]
Haemophilia is an X-linked recessive disorder. The mother is a carrier (X$^H$X$^h$), and the father is unaffected (X$^H$Y). The possible gametes from the mother are X$^H$ and X$^h$, while those from the father are X$^H$ and Y. The Punnett square shows that there is a 50% chance for a son to inherit the affected X chromosome (X$^h$Y), resulting in haemophilia.

Key Concept: Mendelian Disorders

b) 50%
[Solution Description] Haemophilia is an X-linked recessive disorder. A carrier female (X$^H$X$^h$) has a 50% chance to pass the defective X$^h$ allele to her son. Since males have only one X chromosome (inherited from the mother), the son will express the disorder if he receives X$^h$.

Key Concept: Chromosomal Disorders

c) Trisomy of chromosome 21
[Solution Description] Down's syndrome is caused by trisomy of chromosome 21, where there are three copies instead of the normal two. This results in a total of 47 chromosomes (45 + XX or XY).

Key Concept: Mendelian Disorders

c) Autosomal recessive
[Solution Description] Albinism is an autosomal recessive disorder. This means both parents must be carriers (heterozygous) for the trait to have an affected child. The child inherits one defective allele from each parent, making them homozygous recessive.

Key Concept: Prevalence

b) 1 in 2500
[Solution Description]
Turner’s syndrome occurs in approximately 1 out of every 2500 female births due to non-disjunction of sex chromosomes during gamete formation.

Key Concept: Clinical Feature

b) Webbed neck
[Solution Description]
Turner’s syndrome presents with several clinical features such as webbed neck, short stature, and underdeveloped breasts due to the absence of estrogens. The webbed neck is a distinctive diagnostic feature.

Key Concept: Chromosomal Abnormality

b) XO
[Solution Description]
Turner’s syndrome is caused by monosomy of the X chromosome, resulting in a karyotype of 45 chromosomes (44 + XO). This occurs due to non-disjunction during gamete formation.

Key Concept: Diagnosis

c) Karyotyping
[Solution Description]
Diagnosis of Klinefelter's syndrome is confirmed through karyotyping, which reveals the presence of an extra X-chromosome (47,XXY).

Key Concept: Clinical Features

c) Gynecomastia
[Solution Description]
Common clinical features of Klinefelter's syndrome include underdeveloped testes (sterility), sparse body hair, some degree of breast development (gynecomastia), longer than normal legs and arms, and possible mental retardation.

Key Concept: Genetic Basis

b) 47,XXY
[Solution Description]
Klinefelter's syndrome is caused by non-disjunction of sex chromosomes during gamete formation, leading to an extra X-chromosome. The resulting karyotype is 47,XXY instead of the normal 46,XY in males.

Key Concept: Klinefelter’s syndrome incidence

b) 1 in 500 male births
[Solution Description]
Klinefelter’s syndrome occurs in approximately 1 out of every 500 male births. It arises due to non-disjunction, giving rise to a 47,XXY karyotype.

Key Concept: Turner’s syndrome characteristics

c) 45, XO
[Solution Description]
Turner’s syndrome occurs due to monosomy, where one X chromosome is missing, resulting in a chromosomal complement of 45,XO. Affected individuals are sterile females with features like webbed neck, short stature, and underdeveloped ovaries.

Key Concept: Non-disjunction in meiosis

c) Klinefelter’s syndrome
[Solution Description]
Non-disjunction during meiosis can result in an individual having an extra X chromosome in males, leading to Klinefelter's syndrome (47, XXY). This condition is characterized by underdeveloped testes, sparse body hair, and often breast development and long limbs.

Key Concept: Physical Characteristics

d) Elongated limbs
[Solution Description]
While individuals with Down's syndrome exhibit flattened nasal bridge, transverse palm crease, and lack of muscle tone, elongated limbs are not a typical characteristic of this condition.

Key Concept: Diagnosis Method

c) Amniocentesis
[Solution Description]
Amniocentesis is a prenatal diagnostic test that can detect chromosomal abnormalities such as Down's syndrome early in pregnancy. The procedure involves taking a small sample of amniotic fluid for analysis.

Key Concept: Chromosomal Abnormality

c) Chromosome 21
[Solution Description]
Down's syndrome occurs due to the presence of an extra copy of chromosome 21, resulting in a total of 47 chromosomes instead of the usual 46. Therefore, the correct answer is Chromosome 21.

Key Concept: Physical Characteristics

d) Transverse crease on the palm
[Solution Description]
Individuals with Down's syndrome exhibit characteristic features such as a prominent forehead and flattened nasal bridge. Among the options, "Transverse crease on the palm" is a well-known feature. The correct answer is option d).

Key Concept: Non-disjunction

a) Non-disjunction of autosomes
[Solution Description]
Down's syndrome occurs due to non-disjunction, where homologous chromosomes fail to separate during meiosis. This leads to one gamete having an extra chromosome 21, resulting in trisomy. The correct answer is option a).

Key Concept: Down's Syndrome

c) Trisomy of chromosome 21
[Solution Description]
Down's syndrome is caused by trisomy of chromosome 21, which means there are three copies of chromosome 21 instead of the usual two. This results in a total of 47 chromosomes (45 + XX or XY). The correct answer is option c).

Key Concept: Turner’s syndrome

b) 45,XO
[Solution Description]
Turner’s syndrome is caused by monosomy of the X chromosome, where affected females have only one X chromosome (45,XO). This leads to underdeveloped ovaries and other physical characteristics like short stature and webbed neck.

Key Concept: Klinefelter’s syndrome

c) 47,XXY
[Solution Description]
Klinefelter’s syndrome is characterized by an extra X chromosome in males, leading to a karyotype of 47,XXY. This results from non-disjunction of sex chromosomes during meiosis.

Key Concept: Down's syndrome

c) Trisomy of chromosome 21
[Solution Description]
Down's syndrome occurs due to trisomy of chromosome 21, resulting in a total of 47 chromosomes instead of the normal 46. This happens due to non-disjunction during gamete formation.

Key Concept: Phenylketonuria

c) Lack of phenylalanine hydroxylase enzyme
[Solution Description]
PKU results from the absence of the enzyme phenylalanine hydroxylase, which converts phenylalanine to tyrosine, leading to accumulation of phenylalanine.

Key Concept: Colour Blindness

b) Mutation in genes on the X-chromosome
[Solution Description]
Red-green colour blindness is caused by a mutation in genes on the X-chromosome, leading to defective red or green cone cells in the eyes.

Key Concept: Haemophilia

b) A sex-linked recessive disorder impairing blood clotting
[Solution Description]
Haemophilia is a sex-linked recessive disorder where blood clotting is impaired due to a defective protein. It primarily affects males, and carrier females can pass it to their sons.

Key Concept: Mendelian Disorders

b) Tyrosinase
[Solution Description]
Albinism is caused by the absence or deficiency of the enzyme tyrosinase (melanocyte tyrosinase), which is required for the synthesis of melanin from tyrosine. Without this enzyme, melanin production is impaired, leading to the characteristic lack of pigmentation in albinism.

Key Concept: Chromosomal Disorders

b) XO
[Solution Description]
Turner's syndrome is characterized by monosomy of the X chromosome, resulting in a karyotype of 45,XO (44 autosomes + one X chromosome). This is different from Klinefelter's syndrome (47,XXY) and Down's syndrome (trisomy 21). The correct answer is XO.

Key Concept: Mendelian Disorders

b) Haemophilia
[Solution Description]
Sex-linked recessive disorders are those that are linked to the X chromosome and are recessive in nature. Examples include haemophilia and colour blindness. Albinism is an autosomal recessive disorder, not sex-linked. Down's syndrome is a chromosomal disorder. Therefore, the correct answer is haemophilia.

Key Concept: Mendelian Disorders, Pedigree Analysis

c) She is a heterozygous carrier.
[Solution Description]
Since the son is affected by an X-linked recessive disorder, he must have inherited the affected X chromosome from his mother (as fathers pass only Y to sons). The mother does not show symptoms, so she must be a carrier (heterozygous) for the disorder.
Genotypes:
- Affected son: $X^aY$
- Father: $X^AY$ (normal, as he is unaffected)
- Mother: $X^AX^a$ (carrier)
Conclusion: The mother is a heterozygous carrier of the disorder.

Key Concept: Haemophilia, Sex-linked Inheritance

c) 50%
[Solution Description]
The haemophilic man has genotype $X^hY$, where $X^h$ is the haemophilia allele. The normal woman must be a carrier ($X^HX^h$) because her father was haemophilic ($X^hY$).
Parental genotypes:
- Father: $X^hY$
- Mother: $X^HX^h$
Possible gametes:
- Father: $X^h$ or $Y$
- Mother: $X^H$ or $X^h$
Daughter's possible genotypes:
1. $X^HX^h$ (carrier)
2. $X^hX^h$ (haemophilic, rare but possible if both parents contribute $X^h$)
However, since the question asks for the probability of the daughter being a carrier ($X^HX^h$), we calculate:
- Probability mother passes $X^H$: 50%
- Probability father passes $X^h$: 50%
- Combined probability: $0.5 \times 0.5 = 0.5$ or 50%.

Key Concept: Sex-linked Inheritance, Colour Blindness

c) 50%
[Solution Description]
The woman is a carrier (heterozygous) for colour blindness, so her genotype is $X^CX^c$. The husband has normal vision, so his genotype is $X^CY$. The son inherits the Y chromosome from the father and one X chromosome from the mother. There is a 50% chance he will inherit the X chromosome with the colour blindness allele ($X^c$) from his mother, making him colour-blind.
Probability calculation:
- Mother's possible gametes: $X^C$ or $X^c$ (each with 50% chance).
- Father's gamete: Y (100%).
- Son's genotype: $X^CY$ (normal) or $X^cY$ (colour-blind), each with 50% probability.

Key Concept: Reciprocal cross, phenotypic ratio

c) 3 red-eyed : 1 white-eyed
[Solution Description]
The cross is between a homozygous red-eyed female ($X^RX^R$) and a white-eyed male ($X^wY$).
F1 generation: All females ($X^RX^w$) and males ($X^RY$) will be red-eyed.
Interbreeding F1: $X^RX^w$ × $X^RY$
Gametes:
Female: $X^R$, $X^w$
Male: $X^R$, $Y$
F2 combinations:
$X^RX^R$ (red-eyed female), $X^RX^w$ (red-eyed female), $X^RY$ (red-eyed male), $X^wY$ (white-eyed male).
Phenotypic ratio: 3 red-eyed (2 females + 1 male) : 1 white-eyed (male).

Key Concept: Sex-linked inheritance, wing length

d) 100%
[Solution Description]
Given that the male is heterozygous for normal wings ($X^LY$) and the female has miniature wings ($X^lX^l$). The possible gametes are:
Male: $X^L$, $Y$
Female: $X^l$
Possible offspring genotypes:
Females: $X^LX^l$ (normal wings)
Males: $X^lY$ (miniature wings)
Therefore, 50% of the total offspring are males, and all male offspring will have miniature wings. Since we are asked specifically about male offspring, 100% of them will have miniature wings.

Key Concept: Sex-linked inheritance, reciprocal crosses

b) 1 red-eyed female : 1 white-eyed female : 1 red-eyed male : 1 white-eyed male
[Solution Description]
The given cross is between a white-eyed female ($X^wX^w$) and a red-eyed male ($X^RY$). The F1 generation will have red-eyed females ($X^RX^w$) and white-eyed males ($X^wY$). When these F1 flies are intercrossed:
Females: $X^RX^w$ × Males: $X^wY$
Gametes from female: $X^R$, $X^w$
Gametes from male: $X^w$, $Y$
F2 generation combinations:
$X^RX^w$ (red-eyed female), $X^wX^w$ (white-eyed female), $X^RY$ (red-eyed male), $X^wY$ (white-eyed male).
Thus, the ratio of phenotypes in the F2 generation will be 1 red-eyed female : 1 white-eyed female : 1 red-eyed male : 1 white-eyed male.

Key Concept: Sex-linked inheritance, Drosophila wing length

c) 50%
[Solution Description]
Miniature wing ($X^m$) is recessive to wild-type ($X^M$).
1. Female parent: $X^mX^m$ (miniature-winged).
2. Male parent: $X^MY$ (wild-type).
3. F1 generation: Females ($X^MX^m$) are wild-type; Males ($X^mY$) are miniature-winged.
4. F1 cross: $X^MX^m$ × $X^mY$ produces gametes $X^M$, $X^m$ (female) and $X^m$, $Y$ (male).
5. F2 genotypes:
- Females: $X^MX^m$ (wild-type), $X^mX^m$ (miniature-winged).
- Males: $X^MY$ (wild-type), $X^mY$ (miniature-winged).
6. Proportion of miniature-winged flies = $\frac{2}{4}$ (one female and one male) = 50%.

Key Concept: Sex-linked inheritance, Human genetics

b) 50%
[Solution Description]
Color blindness is X-linked recessive.
1. The man's genotype is $X^cY$ (color-blind).
2. The woman’s father was color-blind ($X^cY$), so she must be a carrier ($X^CX^c$).
3. Possible gametes: Male gives $X^c$ or $Y$, Female gives $X^C$ or $X^c$.
4. For a son (XY), the mother contributes either $X^C$ or $X^c$.
5. Probability of son inheriting $X^c$ from the mother is $\frac{1}{2}$.
6. If he gets $X^c$, he will be color-blind (since the father always passes Y to the son).
The probability of the son being color-blind is 50%.

Key Concept: Sex-linked inheritance, Drosophila crosses

b) Red-eyed females and white-eyed males in equal proportion
[Solution Description]
The white-eyed trait is X-linked recessive in Drosophila.
1. The female parent (white-eyed) has genotype $X^wX^w$.
2. The male parent (red-eyed) has genotype $X^RY$.
3. The female offspring will receive $X^w$ from the mother and $X^R$ from the father, making them all $X^RX^w$ (red-eyed).
4. The male offspring will receive $X^w$ from the mother and Y from the father, making them all $X^wY$ (white-eyed).
Thus, the F1 generation consists of red-eyed females and white-eyed males in a 1:1 ratio.

Key Concept: Colour Blindness in Females

d) Father is colour-blind, mother is carrier or colour-blind
[Solution Description]
For a female to be homozygous for colour blindness (X$^c$X$^c$), she must inherit the defective X$^c$ chromosome from both parents. This requires:
1. Her father must be colour-blind (X$^c$Y).
2. Her mother must be either colour-blind (X$^c$X$^c$) or a carrier (X$^c$X).
Thus, the correct condition is when both parents carry at least one copy of the defective allele.

Key Concept: Haemophilia Inheritance

d) 100%
[Solution Description]
Haemophilia is an X-linked recessive disorder. The man has the genotype X$^h$Y (affected), and the woman is normal XX (not a carrier).
Their daughters will always receive one X chromosome from each parent. Since the father passes his affected X$^h$ chromosome to his daughters, while the mother contributes a normal X, all daughters will be carriers (X$^h$X). Hence, the probability is 100%.

Key Concept: Sex-linked Inheritance

b) 50\%
[Solution Description]
The gene for colour blindness is X-linked recessive. The mother is a carrier (X$^c$X), and the father has normal vision (XY). For a son to inherit colour blindness, he must receive the X chromosome carrying the defective allele from his mother.
Probability calculation: The mother can pass either her normal X (X) or the affected X$^c$ to her son. Therefore, the probability is 1/2 or 50\%.

Key Concept: Wing length inheritance

b) 1/2
[Solution Description]
Let X$^N$ represent the normal wing allele (dominant) and X$^n$ the miniature wing allele (recessive). The heterozygous female is X$^N$X$^n$, and the male is X$^n$Y.
Possible gametes: Female - X$^N$ or X$^n$; Male - X$^n$ or Y.
Male offspring can be either X$^N$Y (normal-winged) or X$^n$Y (miniature-winged). The probability of male offspring having normal wings is 1/2.

Key Concept: Sex-linked Traits

b) 1/2
[Solution Description]
The heterozygous red-eyed female has genotype X$^R$X$^w$, and the white-eyed male has genotype X$^w$Y. The possible gametes are:
Female: X$^R$ or X$^w$; Male: X$^w$ or Y.
Offspring possibilities:
X$^R$X$^w$ (red-eyed female), X$^w$X$^w$ (white-eyed female), X$^R$Y (red-eyed male), X$^w$Y (white-eyed male).
Out of the four possible outcomes, two result in white-eyed offspring (X$^w$X$^w$ and X$^w$Y). Thus, the probability is 2/4 = 1/2.

Key Concept: Sex-linked Inheritance

b) 1/4
[Solution Description]
The cross involves a white-eyed male (X$^w$Y) and a red-eyed female (X$^R$X$^R$). All F$_1$ offspring will be X$^R$X$^w$ (red-eyed females) and X$^R$Y (red-eyed males). Crossing these F$_1$ individuals gives:
Punnett Square for F$_2$:
Females (X$^R$X$^w$) x Males (X$^R$Y)
Possible gametes: X$^R$, X$^w$ from female; X$^R$, Y from male.
F$_2$ genotypic ratios:
1 X$^R$X$^R$ : 1 X$^R$X$^w$ : 1 X$^R$Y : 1 X$^w$Y.
Phenotypic ratios:
1 red-eyed female : 1 red-eyed female (carrier) : 1 red-eyed male : 1 white-eyed male.
Thus, 1/4 of the F$_2$ offspring will be white-eyed males.

Key Concept: Sex-linked inheritance in humans

d) 0%
[Solution Description]
The father has the genotype X$^{cb}$Y (color-blind), and the mother is X$^{+$X$^{+}$ (normal vision, not a carrier). Sons inherit their X-chromosome from their mother and Y from their father. Thus, all sons will receive X$^{+$ from the mother and Y from the father, making them X$^{+$Y (normal vision). So, the probability of a son being color-blind is zero.

Key Concept: Chromosomal theory of inheritance

c) The unequal distribution of eye color traits between sexes, matching X-chromosome inheritance
[Solution Description]
Morgan observed that the inheritance pattern of the white-eye trait in Drosophila followed the segregation of the X-chromosome. Since the trait was linked to the X-chromosome and not equally distributed among males and females, it confirmed that genes are located on chromosomes—a key tenet of the chromosomal theory of inheritance.

Key Concept: Sex-linked inheritance in Drosophila

c) 2 red-eyed females : 1 red-eyed male : 1 white-eyed male
[Solution Description]
The cross involves a white-eyed male (X$^{w}$Y) and a red-eyed female (X$^{+$X$^{+}$}). The F$_1$ generation will all be red-eyed (X$^{+$X$^{w}$ females and X$^{+$Y males). In the F$_2$ generation, the expected phenotypic ratio is 3 red-eyed : 1 white-eyed, but since the trait is sex-linked, it further divides as: 2 red-eyed females : 1 red-eyed male : 1 white-eyed male.

Key Concept: Haemophilia

d) Sons of carrier mothers have a 50% chance of being affected
[Solution Description]
Haemophilia is a sex-linked recessive disorder caused by a defective gene on the X-chromosome. Males are more likely to express the disorder because they have only one X-chromosome. Females need two copies of the defective gene to express haemophilia, which is rare. A carrier female (heterozygous) can pass the defective gene to her sons, who then have a 50% chance of being affected. Fathers cannot pass the disorder to their sons because they pass only the Y-chromosome to them.

Key Concept: Colour Blindness

c) 50%
[Solution Description]
Colour blindness is a sex-linked recessive trait located on the X-chromosome. A carrier mother has one normal X-chromosome (X$^N$) and one affected X-chromosome (X$^c$). She can pass either X$^N$ or X$^c$ to her son. Since sons receive their only X-chromosome from their mother, there is a 50% chance they will inherit X$^c$ and be colour blind.