Genetic Techniques: Selective Breeding vs. Genetic Engineering assessment Answer

Question 1: Taking into Account What you have been Taught, What Two Approaches Could you take?

Answer 1:

1. The first approach is selective breeding another approach is genetic engineering.

Question 2: Explain your two Approaches(techniques) in detail.

Answer 2:

• Selective breeding is a natural and traditional method of approach novel plant varieties with desired traits. In this case, one need to cross-pollinate the local "Juliette" strawberries (Fragaria ananassa) with either the Alpine (Fragaria vesca) or Beach (Fragaria chiloensis) varieties, both of which develop white strawberries. The objective is to combine the desirable traits of "Juliette" (sweet flavor, large size) with the white color from either the Beach or Alpine strawberries.
• By cross-pollinating "Juliette" (SSRR) with either Alpine (ssrr) or Beach (ssrr). The result will be F1 plants with the genotype SsRr, which should have the desired large, sweet, white strawberries. Best performing F1 plants with the desired traits for further breeding.

Genetic Engineering 

Genetic engineering enables specific gene’ precise modification to attain desired traits. In this approach, CRISPR-Cas9 technology can be used to directly edit the genes responsible for fruit size (S) and color (R) in "Juliette." these genes can be modified to develop a plant with the desired characteristics. In this approach, the first step is recognising the specific genes responsible for color (R) and fruit size (S) in "Juliette" strawberries. Then, CRISPR-Cas9 can be used to edit these genes, introducing the small fruit size (ss) and white color (rr) alleles. Thereafter, the plants will be regenerated from the edited cells and select those with the desired traits. Lastly, edited genes will be confirmed through DNA sequencing.

Question 3: As we do not know if the "Juliette" strawberry plant is Homozygous or Heterozygous, what combination will result in "Juliette's" visible traits?

Answer 3: Allele Combinations for "Juliette" Visible Traits:

1. If "Juliette" is homozygous for both traits, it would have the genotype SSRR, it will result in large red fruit.
2. If "Juliette" is heterozygous for both traits, it would have the genotype SsRr (large red fruit).

Question 4: Show all possible dihybrid crosses. Point out the desired phenotype.

Answer 4: To achieve the required white, large fruit, cross "Juliette" (SsRr) is made with either Alpine (ssrr) or Beach (ssrr)

Scenario 2:

1. The white-leaved plant may not have a specific set of genes or gene required for proper leaf growth and development, even though light and water conditions are appropriate.
2. Cellular respiration is the procedure by which cells produce ATP, the energy currency of the cell. It encompasses 3 main phases: glycolysis, the citric acid cycle/ Krebs cycle, and oxidative phosphorylation. During glycolysis, glucose is degraded into pyruvate, developing a small amount of NADH and ATP. Thereafter, pyruvate enters the Krebs cycle, where it is further oxidized, releasing CO2 and developing more FADH2 and NADH. The electrons contained by FADH2 and NADH are employed in the third stage, oxidative phosphorylation, to develop a proton gradient across the inner mitochondrial membrane. This gradient results in ATP synthesis.
3. The Sundew gene may not have been sufficient because it only addressed one component of the plant's nutrition. Although the sticky leaves would assist in trapping and digesting prey, the plant may still not have essential nutrients that it cannot get solely through this approach. The integrated gene should assist the plant absorbing essential nutrients or minerals from the soil.
4. To ascertain that the inserted gene is expressed appropriately, it should be integrated into a region of the genome of the plant where it can be effectively transcribed and translated. This could involve placing it near a strong promoter region or within a regulatory sequence that is active in the tissues where the protein is needed.
5. DNA replication is the process by which a cell makes an DNA’s identical copy. It happens in the eukaryotic cell’s nucleus and encompasses numerous steps:

1. Helicase enzyme unwinds and divides the Ds DNA molecule. 
2. DNA polymerase enzyme adds complementary nucleotides to each of the original strands, resulting in two identical double-stranded DNA molecules. 
3. Primase develops a short RNA primer on every template strand, developing a starting point for DNA polymerase to elongate. 
4. DNA polymerase elongates the RNA primers with DNA nucleotides, developing Okazaki fragments on the lagging strand. 
5. The enzyme “DNA ligase” binds the Okazaki fragments together on the lagging strand. 
6. The result is two identical and complete molecules of and double-stranded.

1. To produce a protein from a given gene, the following steps occur:

1. Transcription: In the nucleus, RNA polymerase binds to the DNA at the promoter region. It then synthesizes a complementary RNA strand using the DNA as a template. This forms a single-stranded mRNA molecule.
2. mRNA processing: The newly synthesized mRNA undergoes processing, including the removal of introns (non-coding regions) and the addition of a 5' cap and a poly-A tail. This mature mRNA molecule is then transported to the cytoplasm.
3. Translation: In the cytoplasm, the mRNA attached to a ribosome. tRNA molecules, transport specific amino acids, attached to the codons through complementary base pairing on the mRNA. This develops a polypeptide chain as the ribosome shifts along the mRNA.
4. Protein folding and post-translational modifications: The newly develops polypeptide undergoes folding and may undergo numerous modifications, like glycosylation or phosphorylation, to develop a functional protein.

1. Before mRNA can be translated, it must undergo numerous steps:

1. Transcription: The DNA sequence containing the gene is used as a template to synthesize a complementary mRNA strand by RNA polymerase.
2. mRNA processing: In eukaryotic cells, the initial transcript (pre-mRNA) undergoes modification, including the removal of introns and the addition of a 5' cap and a poly-A tail. This creates a mature mRNA molecule.
3. nuclear export: The mature mRNA molecule is transported from the nucleus to the cytoplasm through nuclear pores.
4. Initiation of translation: In the cytoplasm, the mRNA binds to a ribosome, and the translation initiation complex is formed with the help of initiation factors and initiator tRNA. This marks the beginning of protein synthesis

8.

This process is lysogenic cycle. This is an alternative process of viral replication in which the bacteriophage attaches its genetic material into DNA of the host cell and remains dormant for an extent ended period. The first process is bacteriophage attachment and penetration. The second step is integration. In this step, the viral genetic material integrated into the DNA of the host cell. This integrated DNA of virus is called prophage. The third step is replication. In this step, the host cell replicates as it usually would, and the genetic material of virus is replicated along the DNA of the host. This implies that every time the host cell splits, the viral DNA is replicated and transmitted to the daughter cells. The fourth step is induction. In this step, the integrated virus DNA can be induced to enter the lytic cycle.

9.

This process is lytic cycle. This is an alternative method of replication of virus in which virus infects a host cell and completely hijacks the cellular machinery of host to develop new virus cell, eventually predisposing to the host cell’s destruction. The first step is bacteriophage attachment. In the first step, the virus attaches to the host cell’s specific receptor. This attachment is highly specific. The second step is penetration. Post attachment, the bacteriophage inserts its genome into the host cell. The third step is biosynthesis. Once the viral genetic material controls the cellular machinery of the host. It employs the resources of host to duplicate its own genetic material and develops viral proteins. The fourth phase is maturation. The duplicated viral components are assembled into new virus particles inside the host cell. The last phase is release. Ultimately, the host cell bursts open (lyses), releasing the newly formed virus particles into the surrounding environment. These viruses’ particles can then infect other host cells and continue the lytic cycle.

10: 

Scenario 4

Part 1

Forested Sequence:

Original DNA Sequence:

Corresponding mRNA Sequence:

Amino Acid Sequence (using the Genetic Code Chart):

Urban Sequence:

Modified DNA Sequence in Urban Population: 

mRNA: 

Amino Acid: Val Met Ser Leu Stop Stop Leu Pro Asp

Based on the sequences analysis, it is evident that a mutation has happened in the urban population that leads to a stop codon (UAA) in the mRNA sequence. This “stop” codon prematurely terminates the translation procedure, developing a shorter protein. The occurrence of stop codon perturbs the usual amino acids sequence, resulting in the shift from grey to black coloration in the moths.

PART B

1. The Urban Sequence has one mutation compared to the Forested Sequence.
2. Silent Mutation: This form of mutation does not alter the sequence of amino acid of the protein. It happens when a nucleotide substitution produces codon that still codes for the same amino acid. These mutations are usually inconsequential for protein function.
• Missense Mutation: This mutation leads to alteration in the amino acid sequence of the protein. It happens when a nucleotide substitution predisposes to a codon that codes for a distinct amino acid. Missense mutations can have different impact on protein function, based on the specific amino acid change.
• Nonsense Mutation: This mutation adds a premature stop codon into the mRNA sequence, predisposing to the protein synthesis termination. Nonsense mutations result in a nonfunctional and usually truncated protein.
3. In the Urban Sequence, there is one mutation in contrast to the Forested Sequence. This mutation is a missense mutation because it alters a codon from "CUI" to "UAA," which introduces a stop codon, predisposing to premature termination. So there is 1 missense mutation.
4. The black coloration in urban moths may be due to of natural selection casued by the changing environment. In highly industrialized and polluted urban areas, the presence of darker surfaces such as soot-covered buildings and trees may provide a selective advantage to moths with black coloration. This advantage could be due to increased camouflage against the dark and sooty backgrounds, which makes them less visible to predators. Over time, moths with black coloration would have a higher likelihood of surviving and passing on their genes to the next generation, leading to the prevalence of the black color morph.
5. If pollution worsens and surfaces become darker, it is likely that the prevalence of black urban moths would increase. Natural selection would favor individuals with black coloration as they would have a better chance of survival due to improved camouflage against the darker background. Over successive generations, the allele for black coloration would become more common in the population.
6. The mutation in the DNA sequence from ATG to ATC affects the translated protein because it changes the codon for the amino acid methionine (AUG) to a codon for the amino acid isoleucine (AUC). Methionine is the typical start codon for protein synthesis, indicating the beginning of the protein sequence. By changing it to isoleucine, the mutation may lead to the incorporation of isoleucine at the beginning of the protein instead of methionine. This could potentially affect the protein's structure and function, as the choice of the starting amino acid is critical for proper protein function. Depending on the specific protein and its role, this mutation could have various consequences

Scenario 5

1. 
   In forensic DNA analysis, genetic markers like short tandem repeats (STRs) are commonly used to identify individuals. These markers consist of repeated sequences of DNA that vary in the number of repeats between individuals. By analyzing the number of repeats at specific STR loci, forensic geneticists can differentiate between individuals. Each person has a unique combination of STR allele lengths at multiple loci. This information helps in distinguishing between two suspects by comparing their STR profiles.
2. In the given sequence, the repeat sequence is "GGCGGGCGGGCGGGCGGGCGGGCGG." 
3.  to amplify the provided region, a pair of primers should be dsigned to flank the repeat sequence. For example, the forward primer might be based on the sequence "TAATGTTAA" (starting from the beginning of the provided sequence), and the reverse primer might be based on the sequence "TCTGACCGTT" (at the end of the provided sequence). These primers will help amplify the region of interest.
4. In the programmed thermocycler, the different steps represent different stages of the polymerase chain reaction (PCR) process.

• Step 1 (54°C for 2:00): This is the annealing step, where the primers bind to their complementary sequences on the target DNA. The lower temperature facilitates primer binding.
• Step 2 (34°C for 0:15): This step seems unusual, as it's much lower than the typical annealing temperature. It's possible that this step has a specific purpose not mentioned. It's not a standard part of PCR.
• Step 3 (72°C for 5:00): This is the extension step, where the DNA polymerase synthesizes new DNA strands using the primers as a template. The high temperature helps the enzyme work efficiently.
• The -65°C for 0:20 is also not a typical step in a PCR cycle.

The PCR process typically consists of denaturation, annealing, and extension steps, but the provided program has some unusual conditions. The number of copies of the target DNA sequence produced after 40 cycles depends on various factors, including the initial DNA concentration and the efficiency of the PCR. In a typical PCR, the number of target DNA copies would double with each cycle. However, the provided program lacks some standard steps and includes unusual conditions, so it's challenging to estimate the number of copies without additional information.

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