CUET UG Biology Booster Test 3 Genetic Disorders

Pedigree analysis evaluates heritable variations systematically across generations. Statement I is correct because it traces physiological abnormalities or clinical diseases down a family lineage. Statement II is correct because it assumes that DNA is transmitted faithfully from parent to offspring without constant random fluctuations, providing a baseline for tracking traits. Statement III is correct because sudden variations or unique inheritance lines point to permanent alterations (mutations) in the genetic material. Therefore, all three statements are conceptually sound and true according to NCERT.

When two unaffected parents produce a child who shows a genetic disorder, the trait cannot be dominant. Dominant traits express themselves in every individual who carries the allele, meaning at least one parent would have to show the condition. Instead, this pattern indicates that both parents are healthy, heterozygous carriers (AA) for an autosomal recessive trait. These carriers pass on their hidden mutant alleles (A) at the same time, giving their child a 25% chance of inheriting the homozygous recessive genotype (aA) and expressing the disease.

A pedigree chart is structured top-to-bottom to show the chronological flow of generations: 1. At the top of a generational layer, you place the Parents (2). 2. These parents are linked directly by a horizontal Mating line (4). 3. From this horizontal line, a vertical Line of descent drops downward (3). 4. At the bottom of this line, the Offspring are displayed from left to right in order of birth (1). This arrangement yields the sequence 2-4-3-1.

NCERT provides a standardized chart of pedigree symbols to avoid confusion. In this system, an open diamond represents an unspecified sex, a solid filled square represents an affected male, and a diagonal line across a shape indicates a deceased individual. However, a circle with a cross is not an NCERT standard notation for a carrier female. Carriers are either left unshaded because they are phenotypically normal, or they are shown using a half-filled shape or a central dot. Therefore, Option B is the correct answer to this "NOT" question.

The inheritance pattern of any Mendelian disorder depends on two main factors: whether the trait is dominant or recessive, and whether the mutated gene resides on an autosome or a sex chromosome. Autosomal disorders (like sickle-cell anemi A) affect males and females at equal rates. Sex-linked disorders (like haemophili A) are tied to the sex chromosomes, meaning they show different inheritance frequencies in males and females and often pass from carrier mothers to their sons.

In pedigree analysis, autosomal dominant traits are expressed whenever a dominant allele is present and therefore usually appear in every generation. Autosomal recessive traits may remain hidden in heterozygous carriers, causing the trait to skip generations. A carrier possesses one normal and one recessive allele without showing the disease phenotype. Pedigree analysis is a standard genetic tool used to study the inheritance of traits through successive generations of a family.

Color blindness is an X-linked recessive trait, meaning the mutant gene resides on the X chromosome, not the Y chromosome. Females (XX) have two X chromosomes, so a single normal allele will mask a recessive mutation on the other chromosome. For a female to be color blind, she must inherit two mutant copies (XcX C), which requires a color-blind father and at least a carrier mother. The claim that the gene is found only on the Y-chromosome is entirely false, making Option D the correct choice for this "NOT" question.

Let's map out the genetic cross for this couple: The color-blind man has the genotype XcY. The homozygous normal woman has the genotype XCXC. When they have a son, the child inherits the Y chromosome from his father and one of the normal XC chromosomes from his mother, resulting in an XCY genotype. Because the son receives his X chromosome entirely from his mother, the father's color blindness cannot be passed down to him. This means the probability of their son being color blind is 0%. Maternal Gametes: XC XC Paternal Gametes: Xc XCXc XCXc (Carrier Daughters) Y XCY XCY (Normal Sons)

Queen Victoria was a heterozygous carrier (XHXh) for haemophilia, carrying a mutation in a blood clotting factor gene on one of her X chromosomes. She passed this mutant X chromosome (Xh) down to several of her children. Her male descendants who inherited the mutant allele (XhY) expressed the full disease because they lacked a second X chromosome to mask the recessive trait. This classic transmission pattern allowed the disease to spread through European royal lineages.

Let's map out the genetic cross for this couple: The carrier female has the genotype XHXh. The normal male has the genotype XHY. Crossing these genotypes yields four distinct outcomes among their offspring: 1. XHXH: A clinically normal female. 2. XHXh: An unaffected carrier female. 3. XHY: A clinically normal male. 4. XhY: An affected haemophilic male. This distribution creates an equal phenotypic ratio of 1 normal female : 1 carrier female : 1 normal male : 1 haemophilic male, matching Option B. Maternal Gametes: XH Xh Paternal Gametes: XH XHXH XHXh (Normal / Carrier Females) Y XHY XhY (Normal / Haemophilic Sons)

Sickle-cell anaemia is an autosomal recessive disorder controlled by a single pair of alleles, $Hb^A$ and $Hb^S$. In a heterozygous individual ($Hb^A Hb^S$), the normal $Hb^A$ gene behaves as a dominant allele over the mutant $Hb^S$ allele. This single normal gene ensures the synthesis of enough fully functional, adult hemoglobin to maintain normal, biconcave red blood cell morphology under normal oxygen tensions. Consequently, these individuals remain healthy carriers who appear apparently unaffected unless they experience severe oxygen deprivation.

At the molecular level, a point mutation in the gene encoding the $\beta$-globin chain changes the DNA sequence, which alters the resulting mRNA codon at the sixth position from GAG to GUG. The original codon (GAG) codes for Glutamic acid, while the mutant codon (GUG) codes for Valine. This single alteration substitutes a hydrophilic amino acid for a hydrophobic one, causing the modified hemoglobin protein to polymerize into long fibers under low oxygen conditions.

Alpha Thalassemia is an autosome-linked recessive condition regulated by two closely linked genes, HBA1 and HBA2, located on chromosome 16. Since every individual carries two copies of each gene, there are four total alpha alleles in a diploid human genome. The clinical severity of Alpha Thalassemia depends directly on how many of these four alleles are mutated or deleted. Losing a single allele causes minimal symptoms, whereas losing all four causes severe anemia or hydrops fetalis. This clear dosage effect makes clinical expression directly proportional to the number of affected alpha genes.

Thalassemia is an autosome-linked recessive blood disorder categorized by a reduced synthesis rate of normal globin chains (a quantitative defect). Alpha Thalassemia is mapped to two genes on Chromosome 16, and Beta Thalassemia is tied to a single gene on Chromosome 11. However, the substitution of Valine at the 6th position of the globin chain is a structural, qualitative mutation that belongs to sickle-cell anaemia. Because this feature describes a different genetic disorder, statement D is false, making it the correct choice for this "NOT" question.

During mitotic or meiotic nuclear division, spindle fibers normally pull sister chromatids apart toward opposite poles during anaphase. If these chromatids fail to separate (non-disjunction), the distribution becomes uneven. One resulting cell gains an extra individual chromosome ($2n + 1$) while the companion cell loses one ($2n - 1$). This change in the number of individual chromosomes within a cell is defined as Aneuploidy, distinguishing it from structural mutations or whole-set polyploidy.

The term monosomy describes a form of aneuploidy where an individual is missing one chromosome from a standard pair, bringing the total count down from 46 to 45 ($2n - 1$). A karyotype of 45, X0 means the individual has inherited only a single sex chromosome (one X chromosome) instead of a normal pair ($XX$ or $XY$). This specific numerical reduction represents a monosomy of the sex chromosomes, which clinically manifests as Turner's syndrome.

A normal human somatic cell contains 46 chromosomes arranged in 23 pairs. Down's syndrome is caused by an error in cell division called trisomy 21, where an individual inherits an extra copy of chromosome 21. Because this chromosome pair gains an extra entity, the equation changes to $2n + 1$. This brings the total somatic chromosome count to 47, making Option C correct.

Let's match each physical symptom to its corresponding chromosomal disorder: Big and wrinkled tongue is a physical diagnostic sign of Down's Syndrome (Trisomy 21) (1-ii). Gynaecomastia refers to the development of breast tissue in biological males, which occurs in Klinefelter's Syndrome ($47, XXY$) (2-iii). Rudimentary ovaries are underdeveloped, non-functional ovaries characteristic of females with Turner's Syndrome ($45, X0$) (3-i). This complete matching sequence corresponds to Option A.

The provided reading passage states that Thalassemia is a quantitative problem because it synthesizes too few globin molecules. In contrast, sickle-cell anaemia is a qualitative problem because a point mutation substitutes an amino acid, causing the body to synthesize an incorrectly structured, abnormal globin protein. Down's syndrome and Turner's syndrome are chromosomal number problems (aneuploidies) rather than single-gene traits, leaving sickle-cell anemia as the correct answer.

Turner's syndrome is a chromosomal disorder caused by the absence of one of the X chromosomes in females ($45, X0$). Because they lack the second X chromosome required for full ovarian development, their gonads remain rudimentary and fibrous. Without functional ovaries, these individuals cannot produce mature ova or normal levels of female sex hormones, which leaves them sterile and unable to produce offspring. This confirms Option B.