Pre AP Biology -GEN 6 Biotechnology- MCQ Exam Style Questions -New Syllabus 2025-2026
Pre AP Biology -GEN 6 Biotechnology- MCQ Exam Style Questions – New Syllabus 2025-2026
Pre AP Biology -GEN 6 Biotechnology- MCQ Exam Style Questions – Pre AP Biology – per latest Pre AP Biology Syllabus.
Question
▶️ Answer/Explanation
Correct Option: c
Plasmids are extrachromosomal genetic elements found primarily in bacteria.
They consist of small, circular double-stranded \(\text{DNA}\) molecules.
Unlike the main chromosome, they replicate independently within the cell.
They typically carry accessory genes, such as those for antibiotic resistance.
Options a and b are incorrect because plasmids are made of \(\text{DNA}\), not \(\text{RNA}\).
Option d is incorrect because it describes transformation rather than the definition of a plasmid itself.
Question
b. It sequences more DNA fragments at the same time.
c. It does not require amplification of DNA fragments before sequencing.
d. It does not require the use of computer algorithms to find places where sequence fragments overlap.
▶️ Answer/Explanation
The Illumina method utilizes massive parallel sequencing, allowing millions of fragments to be processed simultaneously.
In contrast, the Sanger method is relatively low-throughput, typically sequencing only one DNA fragment per reaction.
By sequencing many fragments at once, the time per base pair is significantly reduced.
This high-throughput capability drastically lowers the cost per gigabase (Gb) of data generated.
While Sanger reads are longer ($800-1000$ bp), Illumina’s speed comes from the sheer volume of short reads ($50-300$ bp).
Therefore, the ability to sequence more fragments at the same time is the primary reason for its efficiency.
Question
b. transgenic technique
c. DNA fingerprinting technique
d. Southern blot technique
▶️ Answer/Explanation
The correct option is d. Southern blot technique.
Sickle cell disease is caused by a point mutation in the \(\beta\)-globin gene.
This mutation often eliminates a specific restriction enzyme recognition site.
Southern blotting allows researchers to detect these Restriction Fragment Length Polymorphisms (RFLPs).
By comparing fragment sizes, one can distinguish between homozygous dominant, heterozygous (trait), and homozygous recessive (disease) genotypes.
Cloning and transgenic techniques are used for gene manipulation, not primary diagnosis.
DNA fingerprinting is typically used for forensics or paternity rather than specific single-gene mutations.
Question
▶️ Answer/Explanation
The correct option is (a).
DNA fragments possess a uniform negative charge due to their phosphate backbone.
In an electric field, all fragments migrate toward the positive anode.
The agarose gel matrix acts as a molecular sieve.
Smaller DNA fragments move through the pores with less resistance.
Consequently, smaller fragments travel faster and further than larger ones.
Both small and large fragments are visible under UV light if a fluorescent dye is used.
Question
b. find promoters.
c. determine relative expression of genes.
d. create libraries that can maintain themselves on their own.
▶️ Answer/Explanation
The correct option is c.
$\text{cDNA}$ is synthesized from $\text{mRNA}$ using the enzyme reverse transcriptase.
Unlike genomic DNA, $\text{cDNA}$ contains only the exons (expressed sequences).
Because $\text{mRNA}$ levels reflect gene activity, $\text{cDNA}$ libraries show which genes are active.
Introns and promoters are non-coding regions found only in genomic libraries.
Therefore, $\text{cDNA}$ libraries are ideal for comparing relative gene expression across tissues.
This allows researchers to identify proteins being produced in specific cellular conditions.
Question
▶️ Answer/Explanation
DNA fingerprinting identifies unique genetic sequences present in all living organisms.
The process utilizes non-coding regions of $\text{DNA}$ that vary significantly between individuals.
Since humans, animals, and plants all possess unique $\text{DNA}$ profiles, the technique is universal.
In humans, it is a primary tool for forensics and determining biological parentage.
In animals, it is used for tracking wildlife diversity and verifying pedigrees.
In plants, it facilitates cultivar identification and ensures agricultural consistency.
Therefore, the value of $\text{DNA}$ fingerprinting extends across all three biological groups.
The correct option is c.
Question
b. Recombinant vectors make their hosts more antibiotic resistant. Colonies are larger.
c. Recombinant vectors make their hosts replicate more slowly. Colonies are smaller.
d. Recombinant vectors have more restriction enzymes to degrade X-gal. Colonies are blue.
▶️ Answer/Explanation
The correct option is a.
In blue-white screening, the vector contains a $lacZ$ gene encoding $\beta$-galactosidase.
Insertion of foreign DNA into the Multiple Cloning Site (MCS) causes insertional inactivation of $lacZ$.
Recombinant colonies cannot process X-gal and appear white.
Nonrecombinant colonies have a functional $lacZ$ gene, process X-gal, and appear blue.
Therefore, white colonies indicate successful insertion, while blue colonies identify nonrecombinants.
Antibiotic resistance genes are typically used for selection of transformants, not for distinguishing recombinants.
Question
b. stem cells
c. bone cells
d. muscle cells
▶️ Answer/Explanation
The correct answer is b. stem cells.
To create a knockout mouse, specific genes are disrupted in embryonic stem (ES) cells.
These modified $ES$ cells are pluripotent, meaning they can differentiate into any cell type.
The cells are injected into a blastocyst to create a chimeric mouse.
Through breeding, the genetic “knockout” is eventually passed into the germline.
Somatic cells like skin, bone, or muscle cells cannot be used as they are already differentiated.
Only $ES$ cells allow the altered DNA to contribute to the entire offspring’s development.
Question
b. It carries antibiotic resistance.
c. Its natural host cells are plant cells.
d. It has sites that are cut by restriction endonucleases.
▶️ Answer/Explanation
The correct option is c. Its natural host cells are plant cells.
The $\text{Ti}$ (Tumor-inducing) plasmid is derived from the bacterium Agrobacterium tumefaciens.
It is uniquely useful because it can naturally transfer a segment of its $\text{DNA}$ ($\text{T-DNA}$) into the plant genome.
While characteristics like being circular or having restriction sites are common to many vectors, the ability to infect plants is specific to $\text{Ti}$.
Scientists “disarm” the plasmid by removing disease-causing genes and replacing them with desired foreign genes.
This makes it the most effective tool for creating transgenic plants or Genetically Modified Organisms ($\text{GMOs}$).
Question
▶️ Answer/Explanation
[Image of ex vivo gene therapy process] The correct option is d.
Somatic cell gene therapy targets non-germline cells, so genetic changes are not inherited by offspring.
Mature red blood cells lack a nucleus, making them unsuitable for direct gene insertion.
The therapy is primarily effective for monogenic disorders rather than complex polygenic diseases.
In ex vivo techniques, patient cells are harvested and cultured outside the body.
Desired DNA is then integrated into these cultured cells using viral or non-viral vectors.
The modified, functional cells are then re-infused back into the patient’s body.
Question
▶️ Answer/Explanation
The correct answer is a.
Dolly was created using Somatic Cell Nuclear Transfer ($SCNT$).
A nucleus from a $2n$ (diploid) mammary gland cell was extracted.
This somatic nucleus was fused with an enucleated egg cell.
The egg’s cytoplasm “reprogrammed” the somatic DNA to an embryonic state.
The resulting embryo was genetically identical to the original somatic cell donor.
This process bypasses traditional fertilization between two haploid ($n$) gametes.
Question

B. Turner’s syndrome (mono X) – the individual is a female that has only one $X$ chromosome.
C. Klinefelter’s syndrome ($XXY$) – the individual is a male that has an extra $X$ chromosome.
D. Cystic Fibrosis Disease – the individual has a mutation that created a mutant recessive allele.
▶️ Answer/Explanation
The correct option is C.
By examining the sex chromosomes in the bottom right, there are two $X$ chromosomes and one $Y$ chromosome.
This $XXY$ configuration is the hallmark of Klinefelter’s syndrome.
The individual is phenotypically male due to the presence of the $Y$ chromosome.
The total chromosome count for this individual is $47$ instead of the typical $46$.
All other autosomal pairs, including pair $21$, consist of only two chromosomes.
Cystic Fibrosis is a gene mutation and would not be visible on a standard karyotype.
Question

B. Trisomy $18$, in which the individual has a third or extra chromosome number $18$.
C. Trisomy $21$ (Down syndrome), in which the individual has a third or extra chromosome number $21$.
D. Turner’s syndrome, in which a person has only one sex chromosome, such as only one $X$.
▶️ Answer/Explanation
The karyotype displays $22$ pairs of autosomes and one pair of sex chromosomes.
By inspecting chromosome pair $21$, there are three chromosomes instead of the usual two.
This condition of having an extra chromosome at position $21$ is known as Trisomy $21$.
Trisomy $21$ is the genetic cause of Down syndrome.
The sex chromosomes show both $XX$ and $XY$ options, indicating the disorder affects both genders.
Other pairs, such as $13$ and $18$, correctly show only two chromosomes.
Therefore, the correct option is C.
Question

▶️ Answer/Explanation
The correct option is (C).
Gel electrophoresis is a technique used to separate macromolecules like $\text{DNA}$ or proteins.
$\text{DNA}$ fragments are loaded into a gel and an electric current is applied.
Since $\text{DNA}$ is negatively charged, it migrates toward the positive electrode.
Fragments are separated based on their size and charge.
Smaller fragments move faster and further through the gel matrix than larger ones.
This allows scientists to identify the presence of specific resistance gene fragments.
Question

(B) transgenic
(C) polyploid
(D) vectors
▶️ Answer/Explanation
The correct answer is (B) transgenic.
A transgenic organism contains genetic material into which DNA from an unrelated organism has been artificially introduced.
In this case, genes from rye or wild grasses were moved into the wheat genome.
Option (A) recombinant usually refers to the DNA molecule itself rather than the whole organism.
Option (C) polyploid refers to organisms with more than two complete sets of chromosomes.
Option (D) vectors are the vehicles (like plasmids or viruses) used to carry the DNA into the host cell.
Therefore, the modified wheat plant is specifically described as being transgenic.
Question

(B) identical to Cell B.
(C) very similar to Cell A, with some DNA from Cell B.
(D) very similar to Cell B, with some DNA from Cell A.
▶️ Answer/Explanation
The correct option is (C).
In SCNT, the nucleus containing the nuclear DNA is taken from Cell A.
This nucleus is inserted into Cell B, which has had its own nucleus removed (enucleated).
The resulting embryo, Cells C, inherits the nuclear DNA exclusively from Cell A.
However, Cell B (the egg cell) still contains mitochondrial DNA in its cytoplasm.
Therefore, the offspring is genetically very similar to Cell A but not 100% identical.
It contains a small fraction of DNA (mitochondrial) from Cell B.
Question

▶️ Answer/Explanation
The correct option is (C).
Process $Y$ illustrates the enucleation of Cell $B$, which is an unfertilized egg cell.
The arrow pointing away from the cell indicates the removal of the original nucleus.
This ensures the egg contains no genetic material before the somatic nucleus from Cell $A$ is inserted.
In $\text{SCNT}$, the egg serves only as a cytoplasmic environment to support development.
Therefore, Cell $B$ is having its nuclear $\text{DNA}$ removed to make room for the donor nucleus.
This step is essential to ensure the clone has the exact $\text{DNA}$ of the donor organism.
Question


▶️ Answer/Explanation
The correct option is (B).
In SCNT, Cell $A$ is the Somatic cell which provides the diploid ($2n$) nucleus containing the donor’s DNA.
Cell $B$ is the Donor egg (oocyte) used as the host for the transferred nucleus.
Process $Y$ represents enucleation, where the original haploid ($n$) nucleus is removed from the egg.
The nucleus from Cell $A$ is then inserted into the enucleated egg from Cell $B$.
The resulting cell is stimulated to divide, forming the cluster of Cells $C$ (the embryo).
The offspring will be genetically identical to the individual that provided the somatic Cell $A$.
Therefore, Cell $A$ must be the somatic cell and Cell $B$ must be the egg cell.
