lab-week-2-straighterline-anatomy-i-lab-2.doc

lOMoARcPSD| Lab week 2 Straighterline Anatomy I lab 2 Anatomy and Physiology I lOMoARcPSD| Lab 2 Cell Structure and Function BIO201L Pre-Lab Questions: 1. Identify the major similarities and differences between prokaryotic and eukaryotic cells. Similarities between prokaryotic and eukaryotic cells: Both cells have 1. Cell membrane 2. Genetic material (deoxyribonucleic acid (DNA)) 3. Ribosome 4. cytoplasm Differences between prokaryotic and eukaryotic cells 1. Prokaryotic cells are cells without a nucleus, while eukaryotic cells have double nucleus membranes. 2. Prokaryotic cells have no bounded organelle membrane, while eukaryotic are organelle membranes bonded together. 2. Where is the DNA housed in a prokaryotic cell? Where is it housed in a eukaryotic cell? In a prokaryotic cell, the DNA is housed in the nucleoid region of the cytoplasm, while in most eukaryotic cells, the DNA is in the nucleus, except for mitochondria DNA which has the DNA in the cellular mitochondria. 3. Name three structures which provide support and protection in a eukaryotic cell. 1. Cell membrane. 2. Cell walls. 3. Cytoskeleton Experiment 1: Cell Structure and Function Post-Lab Questions 1. Label each of the arrows in the following slide images: A- Chromosomes B- Nucleus lOMoARcPSD| Lab 2 Cell Structure and Function BIO201L Onion root tip, 1000x. 2. What is the difference between the rough and smooth endoplasmic reticulum? 1. The rough endoplasmic reticulum has ribosomes, while the smooth endoplasmic reticulum does not have ribosomes. 2. The function of the rough endoplasmic reticulum is to synthesize protein and transport the protein to the Golgi apparatus while the smooth endoplasmic reticulum function is to produce lipid and transport it to the cell. 3. Would an animal cell be able to survive without a mitochondria? Why or why not? The animal cell will not be able to survive without mitochondria because mitochondria are the powerhouse that metabolizes glucose to produce energy (ATP) needed during aerobic respiration. 4. What is the function of a lysosome? The function of lysosome is to produce digestive enzymes that help break down foreign or damaged materials in the cell. Experiment 2: Exploring Cell Size Record the cell radius and time to center of the cell in Table 2. Then, calculate the surface area, volume, and surface area: volume ratio (equations provided below). lOMoARcPSD| Lab 2 Cell Structure and Function BIO201L Table 2: Surface Area and Volume in Relation to Cell Size Radius (µm) Surface Area Surface Area: (µm2) Volume Volume Ratio Let R=radius and Let pi=22/7 (µm3) Let M be the ratio of d=diameter. Let A= Let pi=22/7 surface are to volume R= d/2 surface area V= Volume M= A/V (µm2/ µm3) Time to Center of the Cell (s) Let t= time to center of the cell. D=30 R=15 D=60 R=30 D=90 R=45 D=120 R=60 D=150 R=75 D=180 R=90 D=210 R=105 D=240 R=120 D=270 R=135 D=300 R=150 2828.57 T= 0.075 14142.86 0.19999 = 0.20 11314.29 .86 0.1000 T=0.15 25457.14 .14 0.0667 T=0.225 T=0.3 45257.14 .86 0.0499 = 0.05 70714.29 .14 0.0400 T=0.375 .57 .14 0.0333 T=0.45 0.0286 T=0.525 .57 .86 0.02499 = 0.0250 T=0.6 .29 .86 0.0222 T=0.675 .14 .14 0.01999 = 0.02 T=0.75 lOMoARcPSD| Lab 2 Cell Structure and Function BIO201L Post-Lab Questions 1. Create a graph of the time to the center of the cell as a function of the surface area: volume ratio. 2. How did the surface area affect diffusion to the center of the cell? What about volume? What about the surface area-to-volume ratio? As the surface area increases, the time to the center of the cell increases. This means that as the surface area increases, the rate of diffusion to the center of the cell also increases, likewise is the volume. As the volume increases, the time to the center of the cell increases. This means that as the volume increases, the rate of diffusion to the center of the cell increases. As the surface area lOMoARcPSD| Lab 2 Cell Structure and Function BIO201L to the volume ratio decreases, the time to the center of the cell increases. This means that as the surface area to the volume decreases, the rate of diffusion to the cell decreases or slows down. 3. How does this experiment demonstrate the need for larger cells to divide? Essential materials such as food, oxygen, and nutrients need to be transported inside the cell by the cell membrane, which leads to the internal cell growing bigger. As the internal cell increases, the surface area inside the cell becomes smaller. For the external cell membrane to meet the demand or need of the cell, the larger cell needs to divide. When a larger cell divides into smaller cells, it creates a larger surface area that allows materials, such as nutrients or oxygen to easily diffuse or cross into the cell.