Principles Of Inheritance And Variation

Colour blindness is an X-linked recessive disorder. Males inherit the X chromosome from their mother, while females inherit one X chromosome from each parent. The daughter inherits the Xc chromosome from her father and XC (normal) from her mother, making her a carrier. The son cannot inherit Xc from the father as he receives only the Y chromosome from him. Cross:

Father (XcY) ×Mother (XCXC)
Gametes: Xc, Y (Father) and XC, XC (Mother)
Offspring:Daughter (XCXc) (Carrier)
Son (XCY) (Normal).

- Hence, the son and daughter cannot inherit colour blindness from the father as it is X-linked.

Step 1 (a): The type of sex determination system observed in honey bees is called the Haplodiploid sex determination system.
Step 2 (b):

• [(i)] Mitosis — Males are haploid and produce sperm by mitosis.
• [(ii)] n — Since males are haploid, their gametes have n chromosomes.
• [(iii)] 2n — Fertilized eggs (diploid) form female progeny with 2n chromosomes.

Step 3 (c): Unfertilized eggs develop into haploid males (drones) with n chromosomes. The sex will be male.

Cross: Gg × gg

Phenotypic ratio = 1 Green : 1 Yellow

Step 1: The genetic disorder caused by monosomy of a sex chromosome is called Turner’s syndrome. It occurs due to the presence of only one X chromosome (45, ).
Step 2: Two symptoms of Turner’s syndrome are:

• Underdeveloped secondary sexual characters.
• Short stature and webbed neck.

(a)

• [(i)] A test cross is a cross between an individual of unknown genotype with a homozygous recessive individual to determine the genotype (zygosity). If the progeny shows all dominant traits, the test plant is homozygous dominant; if a 1:1 ratio of dominant and recessive traits appears, it is heterozygous.
• [(ii)] Human females are rarely haemophilic because the disorder is X-linked recessive. For a female to be haemophilic, she must inherit the defective gene from both parents (i.e., mother must be carrier and father must be haemophilic). Haemophilic individuals suffer from defective blood clotting due to absence of clotting factors, resulting in prolonged bleeding.

OR
(b)

• Event a: Transcription
  Event c: Translation
• [(i)] These events take place in the nucleus (transcription and splicing) and cytoplasm (translation).
• [(ii)] Event b is RNA splicing, where introns (non-coding sequences) are removed and exons (coding sequences) are joined. This process is necessary to produce a functional mRNA transcript that can be translated into a proper protein.

1. The scientific name of the fruit fly is Drosophila melanogaster.
2. Morgan preferred fruit flies for the following reasons:

1. They have a short life cycle and reproduce quickly.
2. They produce a large number of offspring and show clear genetic variations.

(a) Step 1: Reasons for using Drosophila melanogaster:

• Easy to rear in laboratory.
• Short life cycle of about 2 weeks.
• Produces large number of progeny.
• Clear and distinguishable sexual dimorphism and genetic traits.

(b) Step 2: Morgan and his group observed that genes located on the same chromosome tend to be inherited together.
Step 3: The frequency of recombination between two genes is directly proportional to the physical distance between them on a chromosome.
Step 4: This led to the concept of gene linkage and genetic mapping based on recombination frequency.

Step 1 (A):

1. Haemoph ilia and red-green colourblindness are X-linked recessive disorders. Since men have only one X chromosome, the presence of a single defective gene leads to expression of the disorder. Women, having two X chromosomes, must inherit two defective copies to express the condition, which is less likely.
2. For equal production of haemophilic daughters and sons, the cross must be between a carrier female ( ) and a haemophilic male ( ):
   
   

1. This cross results in 1 haemophilic daughter and 1 haemophilic son out of 4 children.

Step 2 (B):

1. In bacteria, both transcription and translation occur in the cytoplasm. In eukaryotes, transcription occurs in the nucleus, and translation occurs in the cytoplasm.
2. 
3. DNA base pairing:
   • Adenine (A) pairs with Thymine (T)
   • Guanine (G) pairs with Cytosine (C)

The Lac Operon in E. coli

The lac operon in E. coli is a classic example of gene regulation in bacteria. It controls the expression of genes that code for enzymes involved in the metabolism of lactose. The regulation of the lac operon depends on the presence or absence of lactose (specifically, its isomer allolactose).

In the Absence of Lactose:

• The repressor protein, encoded by the i gene (regulatory gene), is synthesized in its active form.
• This active repressor protein binds to the operator region of the lac operon.
• The binding of the repressor to the operator physically blocks the RNA polymerase from binding to the promoter and transcribing the structural genes (lacZ, lacY, lacA) of the operon.
• As a result, the expression of the lac operon is switched off, and the enzymes for lactose metabolism are not produced.

In the Presence of Lactose:

• When lactose is added to the growth medium, it is transported into the bacterial cell.
• Inside the cell, lactose is converted into its isomer allolactose by a small amount of -galactosidase (the enzyme encoded by lacZ is present at very low basal levels).
• Allolactose acts as an inducer. It binds to the repressor protein.
• The binding of allolactose to the repressor induces a conformational change in the repressor protein, causing it to become inactive.
• The inactive repressor loses its ability to bind to the operator region of the lac operon.
• With the repressor detached from the operator, RNA polymerase can now bind to the promoter and transcribe the structural genes (lacZ, lacY, lacA).
• These genes are then translated into the corresponding enzymes:
  • ( )-galactosidase (breaks down lactose into glucose and galactose),
  • lactose permease (facilitates the transport of lactose into the cell),
  • transacetylase (its exact role in lactose metabolism is not fully understood but is thought to detoxify other sugars that might be transported by permease).
• Thus, the addition of lactose (through allolactose) switches on the lac operon, allowing the bacteria to utilize lactose as a carbon and energy source.
• Once the lactose is metabolized and its concentration decreases, the repressor becomes active again and switches off the operon.

Correct Answer:

Lactose addition leads to the production of allolactose, which binds to the repressor protein, making it inactive and detaching it from the lac operon's operator region. This allows RNA polymerase to transcribe the structural genes, switching on the operon and enabling lactose metabolism.

Assertion (A) states that in dihybrid crosses involving sex-linked genes in , the generation shows non-parental gene combinations. This is true. Sex-linked genes are located on the X chromosome. During meiosis in the generation (which are heterozygous for these genes), recombination (crossing over) can occur between the sex chromosomes, leading to the formation of gametes with non-parental combinations of the linked genes. These non-parental combinations will be observed in the generation.
Reason (R) states that two genes present on different chromosomes show linkage and recombination in . This is false. Genes located on different chromosomes assort independently according to Mendel's law of independent assortment. Linkage occurs when genes are located on the same chromosome and tend to be inherited together. Recombination occurs due to crossing over between homologous chromosomes during meiosis, which can separate linked genes, but it doesn't apply to genes on different chromosomes in the context of linkage. Genes on different chromosomes exhibit independent assortment, resulting in non-parental combinations due to the random segregation of chromosomes. Therefore, Assertion (A) is true, but Reason (R) is false.