Genetics

(a) (i) - Karyotype: The karyotype of this individual will be XY, as the egg provides the X chromosome and the sperm provides the Y chromosome.
- Genetic Disorder: The individual may have Klinefelter syndrome, a disorder caused by the presence of an extra X chromosome (XXY), leading to male infertility and other physical and developmental features.
(ii) Two common symptoms of Klinefelter syndrome are: 1. Infertility: Due to underdeveloped testes and reduced sperm production. 2. Tall stature and long limbs: Affected individuals are typically taller than average, with longer arms and legs.
(iii) The formation of an XX egg can occur due to a failure in meiosis, specifically during oogenesis. This error, known as non-disjunction, leads to an egg with two X chromosomes instead of just one.
OR
(b) The DNA in a small sample of tissue or blood contains unique genetic markers (alleles) inherited from both parents. By comparing the DNA of the child with the DNA of the alleged father, we can identify matching markers that confirm or rule out paternity. This process is called DNA fingerprinting.

The two differences in inheritance patterns between Hemophilia and Sickle Cell Anemia are:
1. Hemophilia is X-linked recessive, meaning it is carried on the X chromosome and primarily affects males, while Sickle Cell Anemia is autosomal recessive, meaning it is inherited through non-sex chromosomes.
2. In Hemophilia, a single X chromosome carrying the defective gene is enough to express the disorder in males, but in Sickle Cell Anemia, both alleles must be defective (homozygous) for the disorder to be expressed in an individual.

Plasmids are typically circular and extra-chromosomal, meaning they are not part of the chromosome and can replicate independently. They are also usually double-stranded, not single-stranded.

Since the woman has a colour-blind father, she is a carrier (XNXc). Her husband, having normal vision, has the genotype XNY. Their offspring will have the following probabilities:
- Sons (XY) will inherit their father’s Y chromosome and will not be colour-blind.
- Daughters (XNXc) will inherit one X chromosome from the mother and will be carriers (not colour-blind).
Thus, there is a 25 percent chance that a daughter could inherit the colour-blind allele, making her a carrier.

According to Hardy-Weinberg equilibrium, the frequency of heterozygotes (Aa) is given by 2pq, where:
- p is the frequency of allele A (p = 0.6)
- q is the frequency of allele a (q = 1−p=0.4)
The frequency of genotype Aa is:
2pq = 2(0.6)(0.4) = 0.48

Mendelian monohybrid cross (1): This cross results in a phenotypic ratio of 3:1 in the F2 generation.
- Mendelian dihybrid cross (2): The phenotypic ratio in the F2 generation is 9:3:3:1.
- Incomplete dominance (3): In this case, the phenotypic ratio is 1:2:1 because of the blending of traits in the heterozygote.
- Test cross (4): A test cross between a heterozygous and a homozygous recessive plant results in a phenotypic ratio of 1:1 in the F2 generation.

Analyze the inheritance of Haemophilia.
- Haemophilia is a sex-linked recessive disorder, predominantly affecting males who have only one X chromosome.
Analyze the inheritance of sickle cell anemia.
- Sickle cell anemia is an autosomal recessive disorder, meaning both males and females are equally affected because it is inherited through autosomes, not sex chromosomes.
1. Haemophilia is a sex-linked recessive disorder, while sickle cell anemia is an autosomal recessive disorder.
2. Haemophilia predominantly affects males, whereas sickle cell anemia affects both sexes equally.

Mice are the transgenic animals with the largest numbers among all existing transgenic animals. They are widely used due to their short generation time and genetic similarity to humans.

Transgenic animals are animals whose genomes have been altered through the insertion of a foreign gene. This enables them to express specific traits, such as producing therapeutic proteins, serving as model organisms for disease research, or improving agricultural products.

The contraceptives and their corresponding modes of action are: A. The pill – III. Inhibits ovulation. B. Condom – I. Prevents sperm from reaching the cervix. C. Vasectomy – IV. Semen contains no sperm. D. Copper-T – III. Prevents implantation.

Offspring 1: Turner’s Syndrome (45, XO)
Caused due to absence of one X chromosome (Monosomy).
Genotype: 44 + XO (total 45 chromosomes).
Phenotype: Female with underdeveloped secondary sexual characters, short stature, and sterility. Offspring 2: Klinefelter’s Syndrome (47, XXY)
Caused due to the presence of an extra X chromosome (Trisomy).
Genotype: 44 + XXY (total 47 chromosomes).
Phenotype: Male with feminine body features, gynecomastia, and sterility.

Parental combinations dominate when genes are on the same chromosome due to **linkage**. Crossing over promotes new (non-parental) combinations. So, the assertion is true, but the reason is false.

While both the assertion and the reason are true, the reason is not a direct explanation for the assertion. Stirring-type bioreactors are common, but they are only one design feature among many for maintaining optimum growth conditions.

Degeneracy of the genetic code means that more than one codon can code for the same amino acid, which makes both the assertion and reason true, and the reason is the correct explanation.

Dominance occurs when one allele masks the other in a heterozygote (a-ii). Codominance is when both alleles are expressed fully in a heterozygote (b-iii). Pleiotropy refers to one gene affecting multiple traits (c-iv). Polygenic inheritance involves multiple genes influencing a single trait (d-i).

(a) The two closely linked genes that control -thalassemia are and , which are present on chromosome 16.
(b)

• Thalassemia: Caused by deletion or mutation of globin genes, leading to reduced or absent synthesis of one of the globin chains. The structure of the globin molecule is normal, but the quantity is insufficient.
• Sickle Cell Anaemia: Caused by a point mutation in the -globin gene, resulting in abnormal hemoglobin (HbS). The globin molecule becomes deformed under low oxygen conditions, causing sickling of RBCs.

Step 1: For an organism heterozygous at loci, the number of gamete types = .
Step 2: Here, , so number of gametes = .

Offspring 1: Turner’s Syndrome (45, XO) \begin{itemize}
Caused due to absence of one X chromosome (Monosomy).
Genotype: 44 + XO (total 45 chromosomes).
Phenotype: Female with underdeveloped secondary sexual characters, short stature, and sterility. \end{itemize} Offspring 2: Klinefelter’s Syndrome (47, XXY) \begin{itemize}
Caused due to the presence of an extra X chromosome (Trisomy).
Genotype: 44 + XXY (total 47 chromosomes).
Phenotype: Male with feminine body features, gynecomastia, and sterility. \end{itemize}

Parental combinations dominate when genes are on the same chromosome due to **linkage**. Crossing over promotes new (non-parental) combinations. So, the assertion is true, but the reason is false.

Degeneracy of the genetic code means that more than one codon can code for the same amino acid, which makes both the assertion and reason true, and the reason is the correct explanation.

Step 1: Understanding the assertion.
Repetitive sequences make up a significant portion of the human genome. These include satellite DNA, transposons, and other repetitive regions that do not code for proteins.

Step 2: Verifying the reason.
The reason is also correct. While repetitive sequences are abundant in the human genome, they do not directly code for proteins. Instead, they often play structural or regulatory roles in the genome.

Step 3: Conclusion.
Therefore, both the assertion and reason are true, and the reason correctly explains the assertion. The correct answer is (A).
Final Answer: (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).

Step 1: Understanding the question.
This question asks how many adenine bases are present in a DNA molecule given the percentage of guanine.
Step 2: Base pair rule.
In a DNA molecule, guanine pairs with cytosine (G-C), and adenine pairs with thymine (A-T). If 30% of the bases are guanine, 30% must also be cytosine. This leaves 40% for adenine and thymine combined.
Step 3: Calculating adenine bases.
Since adenine and thymine pair in equal amounts, half of the remaining 40% will be adenine. Therefore, the percentage of adenine is 20%. To find the number of adenine bases: Thus, the correct answer is 64 as per the given options.
Final Answer: 64.

Step 1: Chemical Transformation.
Bacteria can be made competent by treating them with a solution of calcium chloride (CaCl₂), which makes their cell membrane more permeable to DNA.
Step 2: Heat Shock Treatment.
After incubation with calcium chloride, the bacteria are subjected to a heat shock at 42°C for a brief period. This creates a temperature gradient that drives the DNA into the bacterial cells.
Step 3: Recovery and Selection.
After the heat shock, the bacteria are allowed to recover in a nutrient medium. Then, they can be grown on selective media to isolate those that have successfully taken up the recombinant DNA.

Step 1: Explanation of the replicating fork.
A replicating fork is formed during DNA replication when the double helix is unwound, creating two single-stranded DNA templates. The replication fork is the region where the DNA strands are being separated, and new complementary strands are synthesized.
Step 2: Diagram of a replicating fork.
(Note: Include a diagram of the replicating fork showing two separated DNA strands, with the leading strand and lagging strand, along with arrows showing the direction of replication.)
Step 3: Why replication occurs in the fork.
DNA replication occurs in the fork because the single-stranded DNA templates are exposed, allowing DNA polymerase to synthesize new complementary strands. The direction of replication is influenced by the polarity of the strands. Replication occurs in the 5' to 3' direction on the leading strand, and the lagging strand is synthesized in short segments (Okazaki fragments) due to its opposite polarity.

Step 1: Define Amniocentesis.
Amniocentesis is a medical procedure used during pregnancy to obtain a sample of amniotic fluid from the sac surrounding the fetus. This fluid contains fetal cells, which can be analyzed to detect genetic conditions, birth defects, or other health issues.
Step 2: Reason for statutory ban.
The government has imposed a statutory ban on amniocentesis due to its misuse for sex determination of the fetus, leading to female foeticide. While amniocentesis is important for detecting genetic disorders, its use for gender selection was deemed unethical and illegal under the Pre-conception and Pre-natal Diagnostic Techniques (PCPNDT) Act, 1994.

Step 1: Understanding transgenic animals.
Transgenic animals are genetically modified organisms that have foreign DNA introduced into their cells. This foreign genetic material can be integrated into their genome, leading to the expression of new traits.

Step 2: Analysis of the options.
- (A) Foreign DNA and RNA are present in their cells: Incorrect. While foreign DNA is integrated into the cells of transgenic animals, the presence of foreign RNA is not guaranteed in all cases, as RNA is typically transcribed from the foreign DNA but not always present in all cells.
- (B) Foreign RNA is present in all of their cells: Incorrect. While foreign RNA may be expressed, it is not necessarily present in all cells, as it is usually transcribed from the foreign DNA when required.
- (C) Foreign DNA is present in all of its cells: Correct. Transgenic animals typically have foreign DNA integrated into all of their cells, as the foreign DNA is integrated into the genome and inherited by all cells in the organism.
- (D) Foreign RNA is present in some of its cells: Incorrect. While foreign RNA might be present in certain cells, it is not a characteristic of all transgenic animals, as foreign RNA is typically present in the cells where the transgene is expressed.

Step 3: Conclusion.
Therefore, the correct answer is (C) Foreign DNA is present in all of its cells, as transgenic animals inherit the foreign DNA in all their cells.

Final Answer: Foreign DNA is present in all of its cells.

Step 1: Understanding the restriction site.
Pst I is a restriction enzyme that recognizes and cuts DNA at the sequence 5' C - T - G - C - A 3', with a sticky end created at the cleavage site.
Step 2: Analyzing the cleavage.
The enzyme cuts between the G and C, resulting in two fragments: 1. 5' C - T - G - C - A \, C - A - G 3' 2. 3' G - A - C - G - T \, C 5'
Step 3: Conclusion.
The correct resultant fragments are option (A) 5' C - T - G - C - A \, C - A - G 3' and 3' G - A - C - G - T \, C 5'.
Final Answer: (A) 5' C - T - G - C - A \, C - A - G 3'
3' G - A - C - G - T \, C 5'.

Step 1: Understanding the genetic cross.
In this question, we are dealing with a cross between two plants with genotypes "VV" (homozygous violet) and "vv" (homozygous white).
Step 2: Punnett square analysis.
The cross between "VV" and "vv" will produce all heterozygous "Vv" offspring, since each parent contributes one allele. The "V" allele (violet) is dominant, so all offspring will have violet flowers.
Step 3: Conclusion.
Thus, 100% of the offspring from the cross "VV × vv" will have violet flowers.
Final Answer: 100%.

Monohybrid Cross in Pea Plants (Mendel's experiment):
For pea plants, let us assume that the contrasting traits are: Purple flower (P) is dominant, and White flower (p) is recessive.
- Parental Cross (P generation):
- \text{Purple flower} (P) x \text{White flower} (p)
- F1 x F1 Cross:
- - The offspring (F2 generation) will have the following genotypic and phenotypic ratio:
- Genotypic ratio: 1 PP : 2 Pp : 1 pp
- Phenotypic ratio: 3 Purple : 1 White

Monohybrid Cross in Antirrhinum Plants (Snapdragon):
For Antirrhinum plants, let us assume that the contrasting traits are: Red flower (R) is dominant, and White flower (r) is recessive.
- Parental Cross (P generation):
- \text{Red flower} (RR) x \text{White flower} (rr)
- F1 x F1 Cross:
- - The offspring (F2 generation) will have the following genotypic and phenotypic ratio:
- Genotypic ratio: 1 RR : 2 Rr : 1 rr
- Phenotypic ratio: 3 Red : 1 White

Conclusion:
- In both crosses, the dominant trait appears in the F1 generation.
- In the F2 generation, both the dominant and recessive traits are visible.
- The Mendelian inheritance pattern follows a 3:1 phenotypic ratio and a 1:2:1 genotypic ratio in both cases.

Step 1: Phenomenon Leading to XO Abnormality.
The phenomenon that leads to an XO chromosomal abnormality is Turner Syndrome. This occurs when there is a missing or incomplete second sex chromosome (only one X chromosome is present instead of two).
Step 2: Symptoms of XO Chromosomal Abnormality.
Individuals with Turner syndrome typically have short stature, infertility, and may experience heart defects, kidney problems, and learning difficulties. They also typically lack the development of secondary sexual characteristics such as breast development.
Step 3: Karyotype.
The karyotype of individuals with Turner syndrome is 45,X, which means they have a total of 45 chromosomes instead of the normal 46, with only one X chromosome instead of two sex chromosomes.

Step 1: Chromosomal Disorder.
Chromosomal disorders are caused by abnormalities in the number or structure of chromosomes. An example is Down syndrome, which occurs due to an extra copy of chromosome 21 (trisomy 21).
Step 2: Mendelian Disorder.
Mendelian disorders are caused by mutations in a single gene, following Mendelian inheritance patterns. An example is Cystic Fibrosis, which is caused by a mutation in the CFTR gene and follows autosomal recessive inheritance.