Q.Conditions of a karyotype 2n + 1, 2n – 1 and 2n + 2, 2n – 2 are called:

Prophase I is the longest and most complex phase of meiosis, crucial for genetic recombination and variation. It is subdivided into five distinct subphases, each characterized by specific chromosomal behaviors. Understanding the sequence and significance of these events is key to grasping how genetic diversity is generated during sexual reproduction.

Here's a breakdown of each subphase and its defining characteristics:

1. Leptotene (Leptonema):

   • This is the first stage of Prophase I.
   • During leptotene, the chromatin material begins to condense, making the individual chromosomes visible under a light microscope.
   • Each chromosome consists of two sister chromatids, but they are so tightly associated that they appear as single, long, thin threads.
   • The chromosomes are attached to the nuclear envelope at their telomeres.
   • Matching with List-2: The description "Chromosome looks like thin threads" (III) perfectly fits Leptotene.
   • Therefore, D-III.

2. Zygotene (Zygonema):

   • Following leptotene, homologous chromosomes begin to pair up, a process called synapsis.
   • This pairing is highly specific and precise, aligning homologous genes.
   • A protein structure called the synaptonemal complex forms between the paired homologous chromosomes, holding them together. The paired homologous chromosomes are now called a bivalent or a tetrad (because it consists of four chromatids).
   • Matching with List-2: "Synaptonemal complex formation" (I) is the hallmark of Zygotene.
   • Therefore, C-I.

3. Pachytene (Pachynema):

   • This is typically the longest subphase of Prophase I.
   • The homologous chromosomes remain tightly paired, and the bivalents become more condensed and distinctly visible.
   • The most significant event of pachytene is crossing over, the exchange of genetic material between non-sister chromatids of homologous chromosomes.
   • This exchange occurs at specific sites called recombination nodules, which are visible structures along the synaptonemal complex.
   • Matching with List-2: "Appearance of recombination nodule" (IV) directly corresponds to Pachytene.
   • Therefore, B-IV.
   Important

   Crossing over is a critical event that shuffles genetic information, leading to new combinations of alleles on chromosomes and increasing genetic diversity.

4. Diplotene (Diplonema):

   • Although not present in List-1, it's important to understand this stage before Diakinesis.
   • In diplotene, the synaptonemal complex dissolves, and the homologous chromosomes begin to separate from each other.
   • However, they remain attached at specific points where crossing over occurred. These points of attachment are called chiasmata (singular: chiasma).
   • The chiasmata represent the physical manifestation of previous crossing over events.

5. Diakinesis:

   • This is the final stage of Prophase I.
   • During diakinesis, the chromosomes become fully condensed and appear as distinct, compact structures.
   • The chiasmata move towards the ends of the chromosomes, a process called terminalization of chiasmata. This effectively "unzips" the homologous chromosomes from the centromere outwards.
   • The nuclear envelope breaks down, and the nucleolus disappears. The meiotic spindle begins to form.
   • Matching with List-2: "Completion of terminalization of chiasmata" (II) is the defining event of Diakinesis.
   • Therefore, A-II.

Combining these matches:

• A (Diakinesis) matches II (Completion of terminalization of chiasmata)
• B (Pachytene) matches IV (Appearance of recombination nodule)
• C (Zygotene) matches I (Synaptonemal complex formation)
• D (Leptotene) matches III (Chromosome looks like thin threads)

This corresponds to option (D).

The table in the question lists four processes in meiosis I, each paired with a subphase and a character (the event that occurs). Your job is to spot which rows are factually correct. This is a classic memory-and-concept check from the chapter on cell division — you need to know the precise sequence of prophase I and what happens in each substage.

Let’s go row by row.

1. Row I: Synapsis → Zygotene → Bivalent formation

   Synapsis is the pairing of homologous chromosomes. It begins in zygotene and results in structures called bivalents (or tetrads). This is textbook correct. So I is correct.

2. Row II: Crossing over → Pachytene → Recombination of genes

   Crossing over — the exchange of genetic material between non-sister chromatids — occurs in pachytene. This leads to recombination of genes. This is also correct. So II is correct.

3. Row III: Replication → Diplotene → Segregation of genes

   Here’s the trap. DNA replication happens in the S phase of interphase, not in prophase I. Diplotene is the stage where chiasmata become visible and homologous chromosomes begin to separate — but segregation of genes (i.e., separation of alleles) occurs later, during anaphase I. So both the subphase and the character are wrong. III is incorrect.

4. Row IV: Terminalization → Diakinesis → Division of chromosome

   Terminalization refers to the movement of chiasmata toward the ends of chromosomes, which does happen in diakinesis. However, the character “Division of chromosome” is misleading — chromosome division (separation of sister chromatids) occurs in anaphase II, not in diakinesis. Diakinesis is the final stage of prophase I, where the nuclear envelope breaks down and the spindle forms. So the character is mismatched. IV is incorrect.

Only rows I and II are correct.

The Concept: X-linked Recessive Inheritance

Colour blindness (red-green) is caused by a recessive allele on the X chromosome. This is crucial: males have only one X (XY), so a single recessive allele makes them colour blind. Females have two X's, so they need two recessive alleles to be affected; with one recessive they are carriers (unaffected).

The father in this problem is colour blind — his genotype is XᶜY (where Xᶜ = recessive allele for colour blindness). The mother is a carrier — her genotype is XᶜX (one normal X, one recessive X). Neither parent is normal; the father is affected, the mother is unaffected but carries the allele.

Step-by-Step Cross

1. Write the parental genotypes

   Father: XᶜY

   Mother: XᶜX

2. Determine the possible gametes

   Father produces two types of sperm: Xᶜ and Y (each with equal probability).

   Mother produces two types of eggs: Xᶜ and X (each with equal probability).

3. Make the Punnett square

4. Interpret each outcome

   • XᶜXᶜ: daughter, colour blind (both X's carry the recessive allele)
   • XᶜX: daughter, carrier (unaffected, but can pass the allele)
   • XᶜY: son, colour blind (his only X is recessive)
   • XY: son, normal (his X is normal)

5. Count the affected children

   Out of 4 equally likely outcomes, 2 are colour blind: the daughter XᶜXᶜ and the son XᶜY. That's 2 out of 4, or 50%.

The core concept here is understanding the historical development of genetics and the specific contributions of key scientists who shaped our understanding of heredity. Each scientist listed made a foundational discovery or proposed a significant theory that advanced the field. By recalling their primary work, we can accurately match them to their respective contributions.

Here's how we match each scientist to their contribution:

1. Matching A. Sutton and Boveri:

   • Walter Sutton and Theodor Boveri independently proposed the Chromosomal Theory of Inheritance in the early 1900s. They observed that the behaviour of chromosomes during meiosis (reduction division) and fertilization perfectly paralleled the behaviour of Mendel's "factors" (genes). This led them to conclude that chromosomes are the carriers of genetic material.
   • Therefore, A matches iii. Chromosomal theory of inheritance.

2. Matching B. Alfred Sturtevant:

   • Alfred Sturtevant was a student of T.H. Morgan. Based on Morgan's work on linkage and recombination in Drosophila, Sturtevant realized that the frequency of recombination between two linked genes is directly proportional to the distance between them on the chromosome. He used this principle to construct the first chromosome maps (also known as genetic maps), showing the relative positions of genes on a chromosome.
   • Therefore, B matches i. Chromosome map.

3. Matching C. Hugo de Vries:

   • Hugo de Vries was one of the three scientists (along with Carl Correns and Erich von Tschermak) who independently rediscovered Mendel's laws of inheritance in 1900. He also proposed the concept of mutation, observing sudden, heritable changes in the evening primrose (Oenothera lamarckiana). He suggested that these mutations were the source of new variations and the raw material for evolution.
   • Therefore, C matches ii. Mutation.

4. Matching D. T.H. Morgan:

   • Thomas Hunt Morgan conducted extensive experiments on the fruit fly (Drosophila melanogaster). His work provided strong experimental evidence for the chromosomal theory of inheritance. He discovered the phenomenon of linkage, demonstrating that genes located on the same chromosome tend to be inherited together. He also explained recombination as a result of crossing over between homologous chromosomes.
   • Therefore, D matches iv. Linkage.

Combining these matches, we get:

A - iii

B - i

C - ii

D - iv

Comparing this with the given options:

(A) A-iii, B-i, C-ii, D-iv

(B) A-ii, B-i, C-iii, D-iv

(C) A-iii, B-ii, C-i, D-iv

(D) A-ii, B-iv, C-ii, D-iii

The correct combination is found in option (A).

The question tests your recall of which chromosomes carry the defective genes for three well-known autosomal recessive disorders. This is a standard fact in human genetics, often asked in exams to check whether you’ve memorized the chromosomal locations of common disease genes. The key is to associate each disorder with its specific gene and chromosome number — not just the disease name.

Let’s break it down one by one.

1. Phenylketonuria (PKU)

   PKU is caused by a mutation in the gene for the enzyme phenylalanine hydroxylase, which converts phenylalanine to tyrosine. This gene is located on chromosome 12. Without the enzyme, phenylalanine builds up and causes intellectual disability if untreated.

2. Sickle cell anemia

   This is a hemoglobin disorder — a point mutation in the beta-globin gene leads to abnormal hemoglobin S. The beta-globin gene is on chromosome 11. The mutation changes a single amino acid (glutamic acid to valine), causing red blood cells to sickle under low oxygen.

3. Cystic fibrosis

   Cystic fibrosis results from mutations in the CFTR gene (cystic fibrosis transmembrane conductance regulator), which codes for a chloride ion channel. This gene is found on chromosome 7. Defective CFTR leads to thick mucus in the lungs, pancreas, and other organs.

Now match the order given in the question: phenylketonuria, sickle cell anemia, cystic fibrosis. That’s chromosome 12, then 11, then 7.

The key is to recall the specific events of meiosis and the parts of a chromosome. Each term in List-I describes a distinct process or structure, and you need to pair it with the correct phase or component from List-II.

1. A) Separation of homologous chromosomes

   Homologous chromosomes (one from each parent) separate during Anaphase I of meiosis. This is the reductional division where the chromosome number is halved. So A matches III) Anaphase - I.

2. B) Initiation of chromosomal pairing

   The pairing of homologous chromosomes begins in Prophase I and is called synapsis. This forms bivalents. So B matches IV) Synapsis.

3. C) Chromosomal clustering at opposite spindle poles

   After chromosomes move to the poles, they cluster together during Telophase I (or Telophase II, but here the context is meiosis I). The nuclear envelope may reform. So C matches I) Telophase.

4. D) Disc shaped part of centromere

   The kinetochore is the disc-shaped protein structure on the centromere where spindle fibers attach. So D matches II) Kinetochore.

Thus, the correct pairing is:

A → III, B → IV, C → I, D → II.

The synaptonemal complex is a protein structure that forms between homologous chromosomes during meiosis I. Its job is to hold the paired homologs together tightly so that crossing over can happen. Without it, the precise exchange of genetic material between non-sister chromatids would be impossible.

The key to this question is knowing the sequence of prophase I stages: leptotene → zygotene → pachytene → diplotene → diakinesis. Each stage has a distinct chromosomal event, and the synaptonemal complex appears and disappears at specific points.

1. Leptotene — Chromosomes begin to condense and become visible as thin threads. Homologs are still unpaired. No synaptonemal complex is present yet. So option (A) is wrong.

2. Zygotene — Homologous chromosomes start to pair up in a process called synapsis. As they align side by side, the synaptonemal complex begins to assemble between them. This is the stage where the complex forms, but it is not yet fully mature or at its most visible.

3. Pachytene — Synapsis is complete. The synaptonemal complex is now fully formed and holds the homologs together along their entire length. This is the stage where the complex is best visible under a microscope. Crossing over occurs here, facilitated by the complex.

4. Diplotene — The synaptonemal complex disassembles and the homologs begin to separate, though they remain attached at chiasmata (the sites of crossing over). The complex is no longer visible.

So the complex forms in zygotene and is best visible in pachytene. That means it is present and observable in both stages, making option (C) the correct choice.

Mendel's experiments with pea plants revealed the fundamental principles of inheritance. He worked with contrasting traits—tall versus dwarf, round versus wrinkled seeds—and discovered that these traits are controlled by discrete "factors" (what we now call genes) that exist in pairs. When these paired factors differ, one may mask the other; the visible trait comes from the dominant factor, while the recessive factor remains hidden but can reappear in later generations.

Let's evaluate each statement against Mendel's actual findings and the conventions that arose from his work:

Evaluating the statements

1. Statement A: "In self-pollinated heterozygous tall F₁ plants, some offspring of F₂ were dwarf."

   This is the cornerstone of Mendel's monohybrid cross. When he crossed pure-breeding tall plants ($TT$) with pure-breeding dwarf plants ($tt$), all F₁ offspring were tall ($Tt$—heterozygous). Self-pollinating these F₁ plants produced an F₂ generation with a 3:1 ratio—three tall to one dwarf. The dwarf plants ($tt$) reappeared because each parent contributed a recessive allele. This reappearance of the recessive phenotype in F₂ was Mendel's evidence that factors do not blend but remain discrete.

   Statement A is correct.

2. Statement B: "The factors to express specific characters are present in the genes of an organism."

   Mendel called them "factors"; we call them genes. Each gene carries the information for a specific character (height, seed shape, etc.). The statement correctly captures that these hereditary units—genes—are what determine an organism's traits. While Mendel did not know the physical nature of genes (DNA, chromosomes), he correctly inferred their existence and behavior.

   Statement B is correct.

3. Statement C: "The symbol for dominance tall will be 'T' and recessive is 'd' dwarf."

   The standard genetic notation uses the same letter for both alleles of a gene, with uppercase for dominant and lowercase for recessive. For the height gene, if we use $T$ for the dominant (tall) allele, the recessive (dwarf) allele must be $t$, not $d$. Using different letters ($T$ and $d$) would incorrectly suggest two unrelated genes rather than two forms of the same gene.

   Statement C is incorrect due to faulty notation.

4. Statement D: "Genes that code for a pair of contrasting traits are known as alleles."

   This is the definition of alleles. A gene is a locus (position) on a chromosome; alleles are the alternative forms of that gene. For the height gene, $T$ (tall) and $t$ (dwarf) are alleles—different versions coding for contrasting expressions of the same trait. Mendel's insight was that these alternatives segregate during gamete formation, so each gamete carries only one allele of each gene.

   Statement D is correct.

Concept & Intuition

This question is about the core principles of Mendelian genetics. Mendel’s first law (segregation) says that during gamete formation, the two alleles of a gene separate so each gamete gets only one. His second law (independent assortment) says that genes on different chromosomes sort independently. Statements C and D are false: genes can be altered by mutation, and starch synthesis in pea seeds is indeed controlled by a single gene (the r gene, where wrinkled peas have a defective starch-branching enzyme). So only A and B are correct.

Step-by-step reasoning

1. Statement A: “Segregate at gamete formation, only one of each pair is transmitted to a gamete”

   This is a direct restatement of Mendel’s law of segregation. During meiosis, homologous chromosomes (and thus the alleles on them) separate, so each gamete receives exactly one allele from each gene pair. True.

2. Statement B: “Independent pairs segregate independently”

   This is Mendel’s law of independent assortment. It applies to genes located on different chromosomes (or far apart on the same chromosome). For such genes, the allele a gamete receives for one gene does not affect the allele it receives for another. True.

3. Statement C: “Genes will not be modified at any time”

   This is false. Genes can be modified by mutations (point mutations, insertions, deletions, etc.), epigenetic changes, or recombination. The constancy of genes is not a principle of genetics. False.

4. Statement D: “Starch synthesis in pea seeds is not controlled by one gene”

   This is also false. In Mendel’s classic pea experiments, the round vs. wrinkled seed trait is controlled by a single gene (the r gene). The dominant allele (R) produces a functional starch-branching enzyme, leading to round seeds; the recessive allele (r) produces a defective enzyme, leading to wrinkled seeds. So starch synthesis is controlled by one gene. False.

Thus, only statements A and B are correct.