BMS415 Cell Biology And Genetics UITM Assignment Answer Malaysia

BMS415 Cell Biology and Genetics at UITM, Malaysia, is an introductory course that covers fundamental concepts in cell biology and genetics. Students will learn about cell structures and functions, transport mechanisms, and cell divisions. The genetic topics include Mendelism, linkage, crossing over, and chromosome mapping, as well as sex chromosome and sex linkage. Additionally, gene interactions, mutations, and genetic population studies are explored. This BMS415  course provides a solid foundation in the essential aspects of cell biology and genetics, preparing students for further studies in the field.

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Assignment Task 1 : Describe the various cell structures and differences in prokaryotic and eukaryotic cell

Cells are the basic units of life and can be broadly classified into two main types: prokaryotic and eukaryotic cells. These cells exhibit distinct differences in their structures and functions.

Prokaryotic Cells:

Cell Structure:

• Prokaryotic cells are small and simple in structure.
• They lack a true nucleus, meaning their genetic material is not enclosed within a membrane-bound nucleus. Instead, their DNA is found in a region called the nucleoid.
• Prokaryotic cells have a cell wall, which provides structural support and protection.
• The cell membrane surrounds the cytoplasm, controlling the passage of materials into and out of the cell.
• Ribosomes, the cellular machinery for protein synthesis, are present in prokaryotic cells.

Differences from Eukaryotic Cells:

• No membrane-bound organelles: Prokaryotes lack membrane-bound organelles such as mitochondria, endoplasmic reticulum, and Golgi apparatus.
• Flagella structure: Prokaryotic flagella are simpler in structure compared to eukaryotic flagella.
• Single circular chromosome: Prokaryotic cells typically contain a single, circular chromosome, while eukaryotic cells have multiple linear chromosomes.

Eukaryotic Cells:

Cell Structure:

• Eukaryotic cells are larger and more complex compared to prokaryotic cells.
• They have a true nucleus, where the genetic material is enclosed within a double-membraned nuclear envelope.
• Eukaryotic cells contain various membrane-bound organelles, including mitochondria, endoplasmic reticulum, Golgi apparatus, lysosomes, and more.
• The cytoplasm in eukaryotic cells is organized into various compartments due to the presence of organelles.Eukaryotic cells have a cytoskeleton, providing structural support and aiding in cell movement.

Differences from Prokaryotic Cells:

• Membrane-bound organelles: Eukaryotic cells possess various membrane-bound organelles that perform specialized functions, which are absent in prokaryotic cells.
• More complex cell division: Eukaryotic cell division involves mitosis or meiosis, whereas prokaryotic cells divide through binary fission.
• Linear chromosomes: Eukaryotic cells have multiple linear chromosomes contained within the nucleus.

Assignment Task 2 : Explain the relation of structure to function of selected cellular organelles, the differences between prokaryotic and eukaryotic cells, the principles of cellular transport processes, cell division processes, DNA structure, flow of genetic information and population genetics

Relation of Structure to Function of Selected Cellular Organelles:

Each organelle in eukaryotic cells has a specific structure that correlates with its function. For example:

• Mitochondria have folded inner membranes (cristae) to increase the surface area for ATP production in cellular respiration.
• Endoplasmic reticulum (ER) is composed of rough ER, studded with ribosomes involved in protein synthesis, and smooth ER, which plays a role in lipid metabolism.

Golgi apparatus modifies, sorts, and packages proteins for transport within or outside the cell.

Principles of Cellular Transport Processes:

• Passive Transport: It does not require energy and includes diffusion, osmosis, and facilitated diffusion.
• Active Transport: It requires energy and moves substances against their concentration gradient, such as the sodium-potassium pump. 

Cell Division Processes:

Mitosis: The process of somatic cell division, leading to the formation of two genetically identical daughter cells.

Meiosis: The process of germ cell division, resulting in the formation of haploid gametes (sperm and egg cells) with genetic variation

DNA Structure and Flow of Genetic Information:

• DNA is a double-stranded helix composed of nucleotides containing sugar, phosphate, and nitrogenous bases (adenine, thymine, cytosine, guanine).
• Genetic information flows from DNA to RNA through transcription and then from RNA to proteins through translation.Population Genetics:Population genetics studies the distribution of alleles in a population and how they change over time through processes like natural selection, genetic drift, and gene flow.

Assignment Task 3 : Apply the principles of Mendelian genetics, mutation and genetic linkage to construct a genetic cross and predict its outcome

For this task, we will construct a genetic cross and predict its outcome using Mendelian genetics principles, considering mutation and genetic linkage. Let’s begin:

Genetic Cross: We will consider a cross between two fruit flies (Drosophila melanogaster) with different traits.

Parental Generation (P):

Female (F1): Homozygous dominant for eye color (Red eyes) – RR

Male (M1): Heterozygous for eye color (White eyes) – Rr

Step 1: Determine the gametes each parent can produce.

• Female (F1) can produce only one type of gamete – R.
• Male (M1) can produce two types of gametes – R and r.

Step 2: Set up the Punnett square to determine the possible genotypes of the offspring.

| R | r |

R | RR | Rr |

r | Rr | rr |

Step 3: Determine the phenotypes of the offspring based on the genotypes.

• RR: Red eyes
• Rr: Red eyes
• rr: White eyes

Step 4: Calculate the genotypic and phenotypic ratios of the offspring.

Genotypic Ratio:

• RR: 1/4 (25%)
• Rr: 2/4 (50%)
• rr: 1/4 (25%)

Phenotypic Ratio:

• Red eyes: 3/4 (75%)
• White eyes: 1/4 (25%)

Step 5: Consider Mutation:

Let’s assume that during the formation of gametes, a mutation occurs in 5% of the cases, resulting in a change of the R allele to a new allele, M.

• Female (F1): RR (95%) and RM (5%)
• Male (M1): R (50%) and r (50%)

Now, we need to modify the Punnett square:

 |   R   |   M   |   r   |

R | RR | RM | Rr |

r | RM | MM | Mr |

Step 6: Calculate the genotypic and phenotypic ratios with the mutation considered.

Genotypic Ratio:

• RR: 2/8 (25%)
• RM: 2/8 (25%)
• MM: 1/8 (12.5%)
• Mr: 2/8 (25%)
• rr: 1/8 (12.5%)

Phenotypic Ratio:

• Red eyes: 6/8 (75%)
• White eyes: 2/8 (25%)

Assignment Task 4 : Prepare slide specimens and manipulate a microscope to observe selected cells and cellular structures; perform Mendelian calculations when given genetic data and present a written report

For this task, you will need to:

• Prepare slide specimens of selected cells or cellular structures (e.g., onion cells for observing under the microscope).
• Observe the specimens under the microscope, noting their characteristics, such as cell shape, size, and any observable organelles.

Perform Mendelian calculations using provided genetic data or data from your genetic cross in Assignment Task 3.

Present a written report that includes the following sections:

1. Introduction: Briefly explain the purpose of the experiment and the significance of observing the selected cells and cellular structures.
2. Materials and Methods: Describe the materials used for preparing slide specimens and the microscope manipulation. Include the genetic data or genetic cross setup used for Mendelian calculations.
3. Results: Present your observations of the selected cells and cellular structures obtained through the microscope. Provide the genotypic and phenotypic ratios from the genetic cross and any calculations performed.
4. Discussion: Interpret the results, compare them with the expected outcomes based on Mendelian genetics, and explain any deviations observed. Discuss the implications of the mutation and its effect on the predicted ratios.
5. Conclusion: Summarize the findings of the experiment, including the key observations of the cellular structures and the predicted outcomes of the genetic cross.

Remember to use appropriate scientific terminology, data presentation methods (e.g., tables, graphs), and references if you used external sources. Good luck with your assignments!

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