Genetics

Step 1: Recall the concept.
Dosage compensation ensures that the expression of X-linked genes is balanced between males (XY) and females (XX).
Step 2: Mechanism in mammals.
In female mammals, one X chromosome is randomly inactivated during early embryonic development (Lyonization), forming a Barr body.
Step 3: Conclusion.
Hence, equal expression is achieved through dosage compensation.

Step 1: Analyze the phenotypic ratio.
The observed distribution shows:
Violet-green (2340), Violet-yellow (47), White-green (43), White-yellow (770).
Step 2: Compare with expected Mendelian ratio.
If genes were independently assorting, a dihybrid cross (VVGG × vvgg) would produce F phenotypes in a 9:3:3:1 ratio.
However, the numbers here are highly skewed — most offspring show parental combinations (violet-green and white-yellow), while recombinants (violet-yellow and white-green) are rare.
Step 3: Interpret the deviation.
Low frequency of recombinant phenotypes (47 and 43) suggests that the genes for flower color and seed color are linked — i.e., located close together on the same chromosome.
Crossing over between them is infrequent, leading to a majority of parental-type combinations.
Step 4: Conclusion.
Thus, the distribution arises due to genetic linkage, meaning both genes are on the same chromosome.

Step 1: Determine genotypes.
Mother (AB) → genotype I I
Father (O) → genotype ii
Step 2: Write possible combinations.
Cross I I × ii → children genotypes: I i and I i
Step 3: Interpret phenotypes.
I i → blood group A
I i → blood group B
Step 4: Conclusion.
Possible blood groups = A or B.

Step 1: Recall Hardy–Weinberg equation.
where
= frequency of allele A,
= frequency of allele a.
Step 2: Given.
Step 3: Compute heterozygote frequency.
Step 4: Conclusion.

Step 1: Determine number of heterozygous gene pairs.
Genotype = Aa Bb Cc Dd → 4 heterozygous loci.
Step 2: Formula for gamete types.
Each heterozygous gene pair produces two possible alleles (A or a, B or b, etc.). Thus, the total possible gametes = , where = number of heterozygous pairs.
Step 3: Apply formula.
Step 4: Conclusion.
Hence, the maximum number of genotypes possible in gametes = 16.

Step 1: Concept.
Chromosome 4 is an autosome, hence both males and females have 2 copies of it. Step 2: Determine total alleles.
Total individuals = 5 males + 10 females = 15 individuals.
Each has 2 alleles for the autosomal gene: Step 3: Conclusion.
Maximum number of alleles = 30.

Step 1: Recall group definitions.
- Monophyletic group: Includes a common ancestor and all descendants.
- Paraphyletic group: Includes a common ancestor but only some descendants (not all).
- Polyphyletic group: Includes members without their most recent common ancestor.
Step 2: Analyze tree.
- A, B, C: These form a monophyletic group (all descendants included).
- D, E, F: Share a common ancestor, but G (a descendant of the same ancestor) is excluded → Paraphyletic.
- B, C, D: Members from two different clades → polyphyletic.
- C, D, E: Not valid clade grouping.
Step 3: Conclusion.
The paraphyletic group is D, E, F. Final Answer:

Step 1: Understanding the problem. We are given a cDNA of 614 base pairs (bp) that includes 125 bp 5'-UTR and 120 bp 3'-UTR. The coding region (which codes for the polypeptide) will be the remaining part of the cDNA after subtracting the UTR regions. Step 2: Converting the coding region length to amino acids. Each codon consists of 3 base pairs, and each codon codes for one amino acid. To find the number of amino acids, divide the length of the coding region by 3: Final Answer:

Step 1: Understanding Hardy-Weinberg equilibrium.
According to Hardy-Weinberg equilibrium, the frequency of the homozygous recessive genotype ( ) is given by , where is the frequency of the recessive allele . The given information tells us that .
Step 2: Finding the frequency of allele .
To find , take the square root of : Step 3: Finding the frequency of the dominant allele.
Since , where is the frequency of the dominant allele , we can calculate : Step 4: Finding the carrier frequency.
The frequency of the carriers (heterozygous ) is : So, the percentage of carriers is: Final Answer:

Step 1: Identifying the inheritance pattern.
We are dealing with a tetra-hybrid cross, which involves four traits (A, B, D, and E) with two alleles each. The genotypes of the parents are AaBbDdEe. The desired phenotypes for each trait are:
- = dominant phenotype
- = recessive phenotype
- = dominant phenotype
- = recessive phenotype
Step 2: Calculating probabilities for each trait.
- For the A trait: cross gives a probability of for offspring with dominant .
- For the b trait: cross gives a probability of for offspring with recessive .
- For the D trait: cross gives a probability of for offspring with dominant .
- For the e trait: cross gives a probability of for offspring with recessive .
Step 3: Calculating the overall probability.
The overall probability of obtaining the desired phenotypes is the product of the probabilities for each individual trait: Final Answer:

Step 1: Understanding the problem The genes are linked, which means the two loci are on the same chromosome and do not assort independently. The distance between the two genes is given as 20 cM, which corresponds to 20% recombinant offspring. This means that 20% of the gametes will be recombinant, and 80% will be parental.
Step 2: Calculating recombinant and parental gametes
- The parental gametes are and , which make up 80% of the gametes.
- The recombinant gametes are and , which make up 20% of the gametes.
Step 3: Genotype of F1 and test cross
The F1 plants are . When test crossed with , the possible gametes from the F1 plant will be:
- Parental: and (80%)
- Recombinant: and (20%)
The test cross will produce the following combinations:
- (from and )
- (from and )
- (from and )
- (from and )
Step 4: Percentage of genotype
The genotype is produced by the combination of two gametes. The percentage of this genotype is: Final Answer:

Step 1: Understanding the lac repressor mechanism.
The lac repressor is a protein expressed from the lacI gene that regulates the lac operon, which controls the metabolism of lactose in bacteria. It works by binding to the lac operator, preventing transcription of the lac genes unless an inducer (e.g., allolactose) is present to relieve repression.
Step 2: Analyzing the options.
- (A) Correct: The lac repressor is indeed allosterically controlled. It binds to the lac operator region on the DNA and prevents the transcription of the lac operon unless an inducer like allolactose is present. The binding of the inducer changes the shape of the repressor, reducing its affinity for the operator and allowing transcription.
- (B) Incorrect: The lacI gene is in the 'trans' configuration with respect to the lac operon. The lac repressor is a diffusible molecule that can act on any lac operon in the cell, regardless of whether it is on the same DNA molecule as the lacI gene. It does not act 'cis' (on the same DNA molecule).
- (C) Incorrect: The presence of glucose actually weakens the effect of the lac repressor, but through a different mechanism, involving catabolite repression. When glucose levels are high, the levels of cyclic AMP (cAMP) are low, leading to reduced activation of the lac operon. However, this is separate from the direct action of the lac repressor.
- (D) Correct: The lac repressor regulates gene expression 'in trans', meaning it can act on any lac operon in the cell, even if the lacI gene and the lac operon are on different DNA molecules. This allows the lac repressor to regulate genes that are not on the same chromosome.
Final Answer:

Step 1: Understanding the ABC model of flower development.
The ABC model describes how three classes of genes (A, B, and C) control flower organ identity during development. These genes work in overlapping combinations to specify the identity of each floral organ.
Step 2: Analyzing the options.
- (A) Correct: ‘Class A’ genes are responsible for the development of sepals. Sepals form the outermost part of the flower, and their identity is regulated by Class A genes such as APETALA1 (AP1).
- (B) Correct: ‘Class B’ genes are responsible for the development of petals and stamens. Petals, which are the second layer of the flower, are regulated by Class B genes like APETALA3 (AP3) and PISTILLATA (PI).
- (C) Correct: ‘Class C’ genes are responsible for the development of stamens and carpels. Class C genes such as AGAMOUS (AG) are critical for the formation of the male and female reproductive organs.
- (D) Incorrect: Class C genes do not solely determine carpel identity. Although Class C genes contribute to carpel development, the carpel is also influenced by the interaction of Class A and Class B genes.
Final Answer:

Step 1: Understanding the methods.
- Agarose gel electrophoresis is the most commonly used method to separate and visualize DNA fragments. It allows DNA of various sizes to be separated by size, with shorter fragments traveling faster than longer ones. A DNA fragment of 150 base pairs can be easily separated using this method.
- Northern blotting involves the detection of RNA, not DNA, and thus is not suitable for detecting a DNA fragment.
- Polyacrylamide gel electrophoresis is typically used for separating smaller fragments, such as proteins or smaller nucleic acid sequences, but for 150 base pairs, agarose gel electrophoresis is more suitable.
- Western blotting is used to detect proteins, not nucleic acids, so it is not applicable for detecting DNA fragments.
Step 2: Conclusion.
The correct method to detect a DNA fragment of 150 base pairs is agarose gel electrophoresis.
Final Answer:

Step 1: Definition.
Down syndrome is a genetic disorder caused by an extra copy of chromosome 21 (also called trisomy 21).
Step 2: Exclude other options.
- Option (A) Trisomy 16: Causes miscarriage, not Down syndrome.
- Option (B) Trisomy 18: Causes Edwards syndrome.
- Option (C) Trisomy 21: Correct – leads to Down syndrome.
- Option (D) Trisomy 22: Causes severe abnormalities but not Down syndrome.
Step 3: Clinical features.
Common features of Down syndrome include intellectual disability, characteristic facial features, and congenital heart defects.
Final Answer: