2.2.1: A Green Kid's Guide to Watering Plants
Quiz by Adam Wilkinson
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My Chinese teacher and English teacher are ________ people.
My sister has six apples and I have six. We have an _____ number of apples.
I sit with my legs ______ the desk and the floor.
I will _________ a party on Saturday.
I am _________ the unit test will happen tomorrow morning.
The ______ air in the rainforest helps the plants grow.
My water bottle opened in my bag, so now my books are ______.
A library is the _______ to go if you need to borrow a book.
What is the correct plural of sandwich?
What is the correct plural of park?
What is the correct plural of bench?
What is the correct plural of beach?
What is the correct plural of fox?
What is the correct plural of dress?
What is the correct plural of loss?
What is the correct plural of paper?
Is the sentence 'singular' or 'plural':
The cats played in the garden.
Is the sentence 'singular' or 'plural':
This sentence ends with a period.
Is the sentence 'singular' or 'plural':
The book has a happy ending.
Is the sentence 'singular' or 'plural':
My friend has three sisters.
What is the genre of this text?
What color of gardener does the reading talk about?
According to the story, what is the correct way to plant?
What does 'conserve' water mean?
Which section will tell you if your plants need water?
What is the purpose of this reading?
Which statement is true.
Why should you plant your seeds in groups.
What is the contraction of 'can' and 'not'?
Which phoneme can be added to 'lip' to make a new word?
A BAD CASE OF THE STRIPES By David Shannon Parts(18): Camilla Narrator 1 Narrator 2 Narrator 3 Narrator 4 Mr. Harms Mother Father Dr. Bumble Old Woman Environmental Therapist Dr. Grop Dr. Gourd Dr. Sponge Mr. Mellon Dr. Cricket Dr. Young <><><><><><><><><><><><><><><><><><><><><><><><><><><><><><><><><><><><><><><><><><><><><><><><><> Narrator 1: A BAD CASE OF THE STRIPES By David Shannon Narrator 2: Camilla Cream loved lima beans. But she never ate them. Narrator 3: All of her friends hated lima beans, and she wanted to fit in. Camilla always worried about what other people thought of her. Narrator 4: Today she was fretting even more than usual. It was the very first day of school, and she couldn't decide what to wear. There were so many people to impress! Narrator 1: She tried on forty-two outfits, but none seemed quite right. She put on a pretty red dress and looked in the mirror. Then she screamed. Narrator 2: Her mother ran into the room, and she screamed, too. Mother: "Oh my heavens! You're completely covered with stripes!" Narrator 3: she cried. This was certainly true. Camilla was striped from head to toe. She looked like a rainbow. Narrator 4: Mrs. Cream felt Camilla's forehead. Mother: "Do you feel all right?" Narrator 1: she asked. Camilla: "I feel fine, but just look at me!" Narrator 2: Camilla answered. Mother: "You get back in bed this instant. You're not going to school today." Narrator 3: her mother ordered. Camilla was relieved. She didn't want to miss the first day of school, but she was afraid of what the other kids would say. And she had no idea what to wear with those crazy stripes. Narrator 4: That afternoon, Dr. Bumble came to examine Camilla. Dr. Bumble: "Most extraordinary! I've never seen anything like it! Are you having any coughing, sneezing, runny nose, aches, pains, chills, hot flashes, dizziness, drowsiness, shortness of breath, or uncontrollable twitching?" Narrator 1: he asked. Camilla: "No, I feel fine." Narrator 2: Camilla told him. Dr. Bumble: "Well then, I don't see any reason why she shouldn't go to school tomorrow. Here's some ointment that should help clear up those stripes in a few days. If it doesn't, you know where to reach me." Narrator 3: Dr. Bumble said, turning to Mrs. Cream. And off he went. Narrator 4: The next day was a disaster. Everyone at school laughed at Camilla. They called her "Camilla Crayon" and "Night of the Living Lollipop." Narrator 1: She tried her best to act as if everything were normal, but when the class said the Pledge of Allegiance, her stripes turned red, white, and blue, and she broke out in stars! Narrator 2: The other kids thought this was great. One yelled out, Narrator 3: "Let's see some purple polka dots!" Narrator 4: Sure enough, Camilla turned all purple polka-dotty. Someone else shouted, Narrator 1: "Checkerboard!" Narrator 4: and a pattern of squares covered her skin. Soon everyone was calling out different shapes and colors, and poor Camilla was changing faster than you can change channels on a T.V. Narrator 2: That night, Mr. Harms, the school principal, called. Mr. Harms: "I'm sorry, Mrs. Cream, I'm going to have to ask you to keep Camilla home from school. She's just too much of a distraction, and I've been getting phone calls from the other parents. They're afraid those stripes may be contagious." Narrator 3: he said. Camilla was so embarrassed. She couldn't believe that two days ago everyone liked her. Now, nobody wanted to be in the same room with her. Narrator 1: Her father tried to make her feel better. Father: "Is there anything I can get you, sweetheart?" Narrator 2: he asked. Camilla: "No, thank you," Narrator 3: sighed Camilla. What she really wanted was a nice plate of lima beans, but she had been laughed at enough for one day. Dr. Bumble: "Hmm, well, yes, I see. I think I'd better bring in the Specialists. We'll be right over.” Narrator 4: said Dr. Bumble to Mr. Cream on the phone. About an hour later, Dr. Bumble arrived with four people in long white coats. He introduced them to the Creams. Dr. Bumble: "This is Dr. Grop, Dr. Sponge, Dr. Cricket, and Dr. Young." Narrator 1: Then the Specialists went to work on Camilla. They squeezed and jabbed, tapped and tested. It was very uncomfortable. Dr. Grop: "Well, it's not the mumps." Dr. Sponge: "Or the measles." Dr. Cricket:"Definitely not chicken pox." Dr. Young: "Or sunburn." Narrator 2: replied the Specialists. Specialists:"Try these. Take one of each before bed." Narrator 4: said the specialists. They each handed her a bottle filled with different colored pills. Then they filed out the front door followed by Dr. Bumble. Narrator 1: That night, Camilla took her medicine. It was awful. Narrator 2: When she woke up the next morning, she did feel different, but when she got dressed, her clothes didn't fit right. She looked in the mirror, and there, staring back at her, was a giant, multi-colored pill with a face on it. Narrator 3: Dr. Bumble rushed over as soon as Mrs. Cream called. But this time, instead of the Specialists, he brought the Experts. Narrator 4: Dr. Gourd and Mr. Mellon were the finest scientific minds in the land. Once again, Camilla was poked and prodded, looked at and listened to. Narrator 1: The Experts wrote down lots of numbers. Then they huddled together and whispered. Dr. Gourd finally spoke. Dr. Gourd: "It might be a virus," Narrator 2: he announced with authority. Suddenly, fuzzy little virus balls appeared all over Camilla. Mr. Mellon: "Or possibly some form of bacteria," Narrator 3: said Mr. Mellon. Out popped squiggly little bacteria tails. Dr. Gourd: "Or it could be a fungus," Narrator 4: added Dr. Gourd. Instantly, Camilla was covered with different colored fungus blotches. The experts looked at Camilla, then each other. Experts: "We need to go over these numbers again back at the lab. We’ll call you when we know something," Narrator 1: said the Experts. But the Experts didn't have a clue, much less a cure. Narrator 2: By now, the T.V. news had found out about Camilla. Reporters from every channel were outside her house, telling the story of "The Bizarre Case of the Incredible Changing Kid." Narrator 3: Soon a huge crowd was camped out on the front lawn. Narrator 4: The Creams were swamped with all kinds of remedies from psychologists, allergists, herbalists, nutritionists, psychics, an old medicine man, a guru, and even a veterinarian. Narrator 1: Each so-called cure only added to poor Camilla's strange appearance until it was hard to even recognize her. She sprouted roots and berries and crystals and feathers and a long furry tail. But nothing worked. Narrator 2: One day, a woman who called herself an Environmental Therapist claimed she could cure Camilla. She said, Environmental Therapist: "Close your eyes, breathe deeply, and become one with your room." Camilla: "I wish you hadn't said that," Narrator 3: Camilla groaned. Slowly, she started to melt into the walls of her room. Her bed became her mouth, her nose was a dresser, and two paintings were her eyes. The therapist screamed and ran from the house. Mother: "What are we going to do? It just keeps getting worse and worse!" Narrator 4: cried Mrs. Cream. She began to sob. Narrator 1: At that moment, Mr. Cream heard a quiet little knock at the front door. He opened it, and there stood an old woman who was just as plump and sweet as a strawberry. Old Woman: "Excuse me, but I think I can help." Narrator 2: she said brightly. Narrator 3: She went into Camilla's room and looked around. Old Woman: "My goodness, what we have here is a bad case of the stripes. One of the worst I've ever seen!" Narrator 4: she said with a shake of her head. She pulled a container of small green beans from her bag. She said, Old Woman: "Here. These might do the trick." Mother: "Are those magic beans?" Narrator 1: asked Mrs. Cream. The old woman replied, Old Woman: "Oh my, no, there's no such thing. These are just plain old lima beans. I'll bet you'd like some, wouldn't you?" Narrator 2: she asked Camilla. Camilla wanted a big, heaping plateful of lima beans more than just about anything, but she was still afraid to admit it. She said, Camilla: "Yuck! No one likes lima beans, especially me!" Old Woman: "Oh, dear, I guess I was wrong about you." Narrator 3: said the old woman sadly. She put the beans back in her bag and started toward the door. Narrator 4: Camilla watched the old woman walk away. Those beans would taste so good. And being laughed at for eating them was nothing, compared to what she'd been going through. She finally couldn't stand it. Camilla: "Wait! The truth is...I really love lima beans." Narrator 1: she cried. The old woman smiled, popping a handful of beans into Camilla's mouth, and said, Old Woman: "I thought so." Camilla: "Mmmmmmm," Narrator 2: said Camilla. Suddenly the branches, feathers, and squiggly tails began to disappear.Then the whole room swirled around. When it stopped, there stood Camilla, and everything was back to normal. Camilla: "I'm cured!" Narrator 3: she shouted. The old woman said, Old Woman: "Yes, I knew the real you was in there somewhere." Narrator 4: She patted Camilla on the head and went outside and vanished into the crowd. Narrator 1: Afterward, Camilla wasn't quite the same. Narrator 2: Some of the kids at school said she was weird, but she didn't care a bit. Narrator 3: She ate all the lima beans she wanted, and she never had even a touch of stripes again.
好的,根據您提供的表格 [NEW SOURCE],以下是表格中例句的中文意思: * **1. all** * **全部的**:**All** my friends were here with me. (我所有的朋友當時都在這裡。) * **全部**:**All** of us enjoyed the movie. (我們所有人都很喜歡這部電影。) * **都**:He got **all** wet. (他全身都濕透了。) * **2. along** * **沿著**:We walked **along** the river yesterday evening. (我們昨天傍晚沿著河邊散步。) * **帶……一起**:When my mother goes out, she takes my little brother **along**. (我媽媽外出時,會帶著我的小弟弟一起去。) * **3. angle** * **觀點**:We should look at the problems from different **angles**. (我們應該從不同的觀點來看待這些問題。) * **角度**:The picture is hanging at an **angle** of 45°. (這張畫以 45 度的角度懸掛著。) * **4. answer** * **答案**:Do you know the **answer** to the question? (你知道這個問題的答案嗎?) * **回答;回應**:Could you **answer** the phone for me? (你可以幫我接一下電話嗎?) * **5. back** * **後面**:She wrote her cellphone number down on the **back** of the paper. (她把她的手機號碼寫在紙的背面。) * **後面的**:Open the **back** door, please. (請打開後面的門。) * **回原處**:It’s time to go **back** home. (該回家了。) * **6. bat** * **蝙蝠**:Did you ever see a **bat** flying quickly in the sky at night? (你曾經看過蝙蝠在夜空中快速飛行嗎?) * **球棒**:Swing the **bat** higher. (把球棒揮高一點。) * **擊**:It’s your turn to **bat**. (輪到你打擊了。) * **7. bite** * **一口的量**:Jane took a **bite** of the guava. (珍咬了一口芭樂。) * **咬**:The dog **bit** the woman’s leg. (那隻狗咬了那個女人的腿。) * **8. book** * **書**:I’ve just started reading a **book** by Stephen King. (我剛開始讀一本史蒂芬·金的書。) * **預訂;預約**:They **booked** two seats at the theater. (他們在劇院預訂了兩個座位。) * **9. block** * **街區**:Nancy and I live on the same **block**. (南希和我住在同一個街區。) * **阻擋**:Those heavy boxes **blocked** my way to the restroom. (那些沉重的箱子擋住了我去洗手間的路。) * **10. bow** * **蝴蝶結**:David chose a gray **bow** tie to go with his black suit. (大衛選擇了一個灰色蝴蝶領結來搭配他的黑色西裝。) * **鞠躬**:The actor **bowed** to everyone before he left the stage. (那位演員在離開舞台前向大家鞠躬。) * **11. break** * **暫停;休息**:I’m tired. Can we take a **break**? (我累了。我們可以休息一下嗎?) * **分解**:These plastic forks are hard to **break** down. (這些塑膠叉子很難分解。) * **打破**:The glass is very expensive. Don’t **break** it. (這個玻璃很貴。不要打破它。) * **12. bright** * **晴朗的**:It’s a **bright** morning. Why not take a walk along the river? (這是個晴朗的早晨。何不沿著河邊散步呢?) * **明亮的**:The room isn’t **bright** enough. Let’s not read here. (這個房間不夠明亮。我們不要在這裡閱讀。) * **13. call** * **打電話**:I got a **call** from my old friend last night. (我昨晚接到我老朋友的電話。) * **打電話**:Tina **called** me last night. We talked a lot about music. (蒂娜昨晚打電話給我。我們聊了很多關於音樂的事。) * **呼喊**:Listen! Is that a **call** for help? (聽!那是求救的呼喊嗎?) * **呼喊**:Why did you **call** my name then? (那你當時為什麼喊我的名字?) * **14. camp** * **營隊**:Patrick joined a science **camp** this summer. (派屈克今年夏天參加了一個科學營隊。) * **露營**:They **camped** by the river yesterday. (他們昨天在河邊露營。) * **15. case** * **箱;盒**:The kids drank the whole **case** of Coke. (孩子們喝掉了一整箱可樂。) * **實例;情況**:The number of new **cases** of Covid-19 is growing. (新冠肺炎的新增病例數正在增加。) * **16. catch** * **接球**:Nice **catch**! My good dog. (接得好!我的好狗狗。) * **罹患(病)**:My head hurts. I may **catch** a cold. (我頭痛。我可能感冒了。) * **抓住**:I didn’t **catch** the ball. (我沒有接到那個球。) * **17. change** * **零錢;找零**:I think you’ve given me the wrong **change**. (我想你找錯錢了。) * **改變;交換**:The leaves **change** (in color) from green to red in the fall. (秋天時,樹葉的顏色從綠色變成紅色。) * **18. clean** * **打掃;清理**:Tom **cleans** the toilet once a week. (湯姆一週打掃一次馬桶。) * **乾淨的**:The water isn’t **clean**. Don’t drink it. (這水不乾淨。不要喝。) * **19. close** * **關;闔**:**Close** your books, students. Let’s have a pop quiz. (同學們,把你們的書闔上。我們要進行隨堂測驗。) * **靠近地**:Jane sat **close** to her husband at the party. (在派對上,珍緊挨著她的丈夫坐著。) * **20. cold** * **感冒**:I had a **cold** a week ago. (我一個星期前感冒了。) * **寒冷的**:It was **cold** last night. (昨晚很冷。)
Lopez Family Quiz Questions Family History Who are the original ancestors of the Lopez clan? Agapito and Cemona Lopez Antonio and Carmen Lopez Alejandro and Cecilia Lopez Alfonso and Clara Lopez How many children did Agapito and Cemona Lopez have? 6 8 10 12 Which family branch is represented by the color Yellow? Jeremias Lopez Rufina Lopez-Solivio Samuel Lopez Marina Lopez-Tenizo Which family member hosted the 28th Lopez Family Reunion? Marina Lopez-Tenizo Jeremias A. Lopez Family David Lopez Rebecca Lopez-Diaz What Bible verse was featured in the 2019 reunion theme "My Family, My Home"? Luke 12:34 John 3:16 Psalm 133:1 Proverbs 22:6 Family Traditions What traditional Filipino dish is always served at Lopez family gatherings? [Insert correct dish] [Option 2] [Option 3] [Option 4] What activity traditionally closes Lopez family reunions? Group photo Prayer circle Talent show Raffle drawing Which color represents the 3rd Generation in the family's color coding system? Yellow Blue Green Pink Gold/Brown What was the theme of the 29th Lopez Grand Family Reunion? "Worthy Legacy" "My Family, My Home" "Reconnecting Roots" "Faith and Family" What traditional game is always played at Lopez reunions? [Insert correct game] [Option 2] [Option 3] [Option 4] Family Geography In which city was the 28th Lopez Family Reunion held? Tacurong City Davao City Cotabato City Pigcawayan Where was the 29th Lopez Grand Family Reunion held? Belle's Farm & Resort, Midpapan Bonboc Garden [Option 3] [Option 4] In which Philippine region did Agapito and Cemona Lopez originally settle? [Insert correct region] [Option 2] [Option 3] [Option 4] Which family branch has members living in the most countries? [Insert correct branch] [Option 2] [Option 3] [Option 4] Family Members Who is the oldest living Lopez family member? [Insert name] [Option 2] [Option 3] [Option 4] Which family member served as the speaker at the 29th reunion? Ptr. Christie Joy L. Manzinares Rev. Ronie Balboa Laud Ptr. Alma Lopez Rev. David Lopez Who gave the welcome song at the 2020 reunion? Dorce S. Divinagracia Tenizo Family Grande Siblings Dumaan Family How many Lopez family members are named after Biblical figures? [Insert correct number] [Option 2] [Option 3] [Option 4] Recent Family Events Which new family tradition was introduced at the last reunion? [Insert correct tradition] [Option 2] [Option 3] [Option 4] How many family members attended the last reunion (in person and virtually)? [Insert correct number] [Option 2] [Option 3] [Option 4] Bible Knowledge (Filipino Family Edition) Which Bible verse is the theme for the 2025 reunion? Colossians 2:5 Psalm 133:1 Proverbs 22:6 Philippians 2:1-2 In the Bible, who said "As for me and my house, we will serve the Lord"? Joshua Moses Abraham David Which biblical character is known for his wisdom and is often quoted in Filipino family gatherings? Solomon Paul Peter John Which psalm begins with "Blessed is everyone who fears the Lord, who walks in his ways"? Psalm 128 Psalm 23 Psalm 91 Psalm 119 What does Proverbs say is "the beginning of wisdom"? Fear of the Lord Knowledge Understanding Prudence
Introduction to Free Fall A free-falling object is an object that is falling under the sole influence of gravity. Any object that is being acted upon only by the force of gravity is said to be in a state of free fall. There are two important motion characteristics that are true of free-falling objects: • Free-falling objects do not encounter air resistance. • All free-falling objects (on Earth) accelerate downwards at a rate of 9.8 m/s/s (often approximated as 10 m/s/s for back-of-the-envelope calculations) Because free-falling objects are accelerating downwards at a rate of 9.8 m/s/s, a ticker tape trace or dot diagram of its motion would depict an acceleration. The dot diagram at the right depicts the acceleration of a free-falling object. The position of the object at regular time intervals - say, every 0.1 second - is shown. The fact that the distance that the object travels every interval of time is increasing is a sure sign that the ball is speeding up as it falls downward. Recall from an earlier lesson, that if an object travels downward and speeds up, then its acceleration is downward. Free-fall acceleration is often witnessed in a physics classroom by means of an ever-popular strobe light demonstration. The room is darkened and a jug full of water is connected by a tube to a medicine dropper. The dropper drips water and the strobe illuminate the falling droplets at a regular rate - say once every 0.2 seconds. Instead of seeing a stream of water free-falling from the medicine dropper, several consecutive drops with increasing separation distance are seen. The pattern of drops resembles the dot diagram shown in the graphic at the right. The Acceleration of Gravity It was learned in the previous part of this lesson that a free-falling object is an object that is falling under the sole influence of gravity. A free-falling object has an acceleration of 9.8 m/s/s, downward (on Earth). This numerical value for the acceleration of a free-falling object is such an important value that it is given a special name. It is known as the acceleration of gravity - the acceleration for any object moving under the sole influence of gravity. A matter of fact, this quantity known as the acceleration of gravity is such an important quantity that physicists have a special symbol to denote it - the symbol g. The numerical value for the acceleration of gravity is most accurately known as 9.8 m/s2. There are slight variations in this numerical value (to the second decimal place) that are dependent primarily upon on altitude. We will occasionally use the approximated value of 10 m/s2 in order to reduce the complexity of the many mathematical tasks that we will perform with this number. By so doing, we will be able to better focus on the conceptual nature of physics without too much of a sacrifice in numerical accuracy. g = 9.8 m/s2, downward Look It Up! Even on the surface of the Earth, there are local variations in the value of the acceleration of gravity (g). These variations are due to latitude, altitude and the local geological structure of the region. Recall from an earlier lesson that acceleration is the rate at which an object changes its velocity. It is the ratio of velocity change to time between any two points in an object's path. To accelerate at 9.8 m/s2 means to change the velocity by 9.8 m/s each second. If the velocity and time for a free-falling object being dropped from a position of rest were tabulated, then one would note the following pattern. Time (s) Velocity (m/s) 0 0 1 - 9.8 2 - 19.6 3 - 29.4 4 - 39.2 5 - 49.0 . Observe that the velocity-time data above reveal that the object's velocity is changing by 9.8 m/s each consecutive second. That is, the free-falling object has an acceleration of approximately 9.8 m/s2. Another way to represent this acceleration of 9.8 m/s2 is to add numbers to our dot diagram that we saw earlier in this lesson. The velocity of the ball is seen to increase as depicted in the diagram at the right. (NOTE: The diagram is not drawn to scale - in two seconds, the object would drop considerably further than the distance from shoulder to toes.) Representing Free Fall by Graphs • Early in Lesson 1 it was mentioned that there are a variety of means of describing the motion of objects. One such means of describing the motion of objects is through the use of graphs - position versus time and velocity vs. time graphs. In this part of Lesson 5, the motion of a free-falling motion will be represented using these two basic types of graphs. Representing Free Fall by Position-Time Graphs A position versus time graph for a free-falling object is shown below. Observe that the line on the graph curves. As learned earlier, a curved line on a position versus time graph signifies an accelerated motion. Since a free-falling object is undergoing an acceleration (g = 9.8 m/s/s), it would be expected that its position-time graph would be curved. A further look at the position-time graph reveals that the object starts with a small velocity (slow) and finishes with a large velocity (fast). Since the slope of any position vs. time graph is the velocity of the object (as learned in Lesson 3), the small initial slope indicates a small initial velocity and the large final slope indicates a large final velocity. Finally, the negative slope of the line indicates a negative (i.e., downward) velocity. Representing Free Fall by Velocity-Time Graphs A velocity versus time graph for a free-falling object is shown below. Observe that the line on the graph is a straight, diagonal line. As learned earlier, a diagonal line on a velocity versus time graph signifies an accelerated motion. Since a free-falling object is undergoing an acceleration (g = 9,8 m/s/s, downward), it would be expected that its velocity-time graph would be diagonal. A further look at the velocity-time graph reveals that the object starts with a zero velocity (as read from the graph) and finishes with a large, negative velocity; that is, the object is moving in the negative direction and speeding up. An object that is moving in the negative direction and speeding up is said to have a negative acceleration (if necessary, review the vector nature of acceleration). Since the slope of any velocity versus time graph is the acceleration of the object (as learned in Lesson 4), the constant, negative slope indicates a constant, negative acceleration. This analysis of the slope on the graph is consistent with the motion of a free-falling object - an object moving with a constant acceleration of 9.8 m/s/s in the downward direction. The Kinematic Equations The goal of this first unit has been to investigate the variety of means by which the motion of objects can be described. The variety of representations that we have investigated includes verbal representations, pictorial representations, numerical representations, and graphical representations (position-time graphs and velocity-time graphs). In Lesson 6, we will investigate the use of equations to describe and represent the motion of objects. These equations are known as kinematic equations. There are a variety of quantities associated with the motion of objects - displacement (and distance), velocity (and speed), acceleration, and time. Knowledge of each of these quantities provides descriptive information about an object's motion. For example, if a car is known to move with a constant velocity of 22.0 m/s, North for 12.0 seconds for a northward displacement of 264 meters, then the motion of the car is fully described. And if a second car is known to accelerate from a rest position with an eastward acceleration of 3.0 m/s2 for a time of 8.0 seconds, providing a final velocity of 24 m/s, East and an eastward displacement of 96 meters, then the motion of this car is fully described. These two statements provide a complete description of the motion of an object. However, such completeness is not always known. It is often the case that only a few parameters of an object's motion are known, while the rest are unknown. For example as you approach the stoplight, you might know that your car has a velocity of 22 m/s, East and is capable of a skidding acceleration of 8.0 m/s2, West. However you do not know the displacement that your car would experience if you were to slam on your brakes and skid to a stop; and you do not know the time required to skid to a stop. In such an instance as this, the unknown parameters can be determined using physics principles and mathematical equations (the kinematic equations). The BIG 4 The kinematic equations are a set of four equations that can be utilized to predict unknown information about an object's motion if other information is known. The equations can be utilized for any motion that can be described as being either a constant velocity motion (an acceleration of 0 m/s/s) or a constant acceleration motion. They can never be used over any time period during which the acceleration is changing. Each of the kinematic equations include four variables. If the values of three of the four variables are known, then the value of the fourth variable can be calculated. In this manner, the kinematic equations provide a useful means of predicting information about an object's motion if other information is known. For example, if the acceleration value and the initial and final velocity values of a skidding car is known, then the displacement of the car and the time can be predicted using the kinematic equations. Lesson 6 of this unit will focus upon the use of the kinematic equations to predict the numerical values of unknown quantities for an object's motion. The four kinematic equations that describe an object's motion are: There are a variety of symbols used in the above equations. Each symbol has its own specific meaning. The symbol d stands for the displacement of the object. The symbol t stands for the time for which the object moved. The symbol a stands for the acceleration of the object. And the symbol v stands for the velocity of the object; a subscript of i after the v (as in vi) indicates that the velocity value is the initial velocity value and a subscript of f (as in vf) indicates that the velocity value is the final velocity value. Each of these four equations appropriately describes the mathematical relationship between the parameters of an object's motion. As such, they can be used to predict unknown information about an object's motion if other information is known. In the next part of Lesson 6 we will investigate the process of doing this. Kinematic Equations and Problem-Solving The four kinematic equations that describe the mathematical relationship between the parameters that describe an object's motion were introduced in the previous part of Lesson 6. The four kinematic equations are: In the above equations, the symbol d stands for the displacement of the object. The symbol t stands for the time for which the object moved. The symbol a stand for the acceleration of the object. And the symbol v stands for the instantaneous velocity of the object; a subscript of i after the v (as in vi) indicates that the velocity value is the initial velocity value and a subscript of f (as in vf) indicates that the velocity value is the final velocity value. Problem-Solving Strategy In this part of Lesson 6 we will investigate the process of using the equations to determine unknown information about an object's motion. The process involves the use of a problem-solving strategy that will be used throughout the course. The strategy involves the following steps: 1. Construct an informative diagram of the physical situation. 2. Identify and list the given information in variable form. 3. Identify and list the unknown information in variable form. 4. Identify and list the equation that will be used to determine unknown information from known information. 5. Substitute known values into the equation and use appropriate algebraic steps to solve for the unknown information. 6. Check your answer to ensure that it is reasonable and mathematically correct. The use of this problem-solving strategy in the solution of the following problem is modeled in Examples A and B below. Example Problem A . Ima Hurryin is approaching a stoplight moving with a velocity of +30.0 m/s. The light turns yellow, and Ima applies the brakes and skids to a stop. If Ima's acceleration is -8.00 m/s2, then determine the displacement of the car during the skidding process. (Note that the direction of the velocity and the acceleration vectors are denoted by a + and a - sign.) The solution to this problem begins by the construction of an informative diagram of the physical situation. This is shown below. The second step involves the identification and listing of known information in variable form. Note that the vf value can be inferred to be 0 m/s since Ima's car comes to a stop. The initial velocity (vi) of the car is +30.0 m/s since this is the velocity at the beginning of the motion (the skidding motion). And the acceleration (a) of the car is given as - 8.00 m/s2. (Always pay careful attention to the + and - signs for the given quantities.) The next step of the strategy involves the listing of the unknown (or desired) information in variable form. In this case, the problem requests information about the displacement of the car. So d is the unknown quantity. The results of the first three steps are shown in the table below. Diagram: Given: Find: vi = +30.0 m/s vf = 0 m/s a = - 8.00 m/s2 d = ?? The next step of the strategy involves identifying a kinematic equation that would allow you to determine the unknown quantity. There are four kinematic equations to choose from. In general, you will always choose the equation that contains the three known and the one unknown variable. In this specific case, the three known variables and the one unknown variable are vf, vi, a, and d. Thus, you will look for an equation that has these four variables listed in it. An inspection of the four equations above reveals that the equation on the top right contains all four variables. vf2 = vi2 + 2 • a • d Once the equation is identified and written down, the next step of the strategy involves substituting known values into the equation and using proper algebraic steps to solve for the unknown information. This step is shown below. (0 m/s)2 = (30.0 m/s)2 + 2 • (-8.00 m/s2) • d 0 m2/s2 = 900 m2/s2 + (-16.0 m/s2) • d (16.0 m/s2) • d = 900 m2/s2 - 0 m2/s2 (16.0 m/s2)*d = 900 m2/s2 d = (900 m2/s2)/ (16.0 m/s2) d = (900 m2/s2)/ (16.0 m/s2) d = 56.3 m The solution above reveals that the car will skid a distance of 56.3 meters. (Note that this value is rounded to the third digit.) The last step of the problem-solving strategy involves checking the answer to assure that it is both reasonable and accurate. The value seems reasonable enough. It takes a car a considerable distance to skid from 30.0 m/s (approximately 65 mi/hr) to a stop. The calculated distance is approximately one-half a football field, making this a very reasonable skidding distance. Checking for accuracy involves substituting the calculated value back into the equation for displacement and insuring that the left side of the equation is equal to the right side of the equation. Indeed it is! Example Problem B Ben Rushin is waiting at a stoplight. When it finally turns green, Ben accelerated from rest at a rate of a 6.00 m/s2 for a time of 4.10 seconds. Determine the displacement of Ben's car during this time period. Once more, the solution to this problem begins by the construction of an informative diagram of the physical situation. This is shown below. The second step of the strategy involves the identification and listing of known information in variable form. Note that the vi value can be inferred to be 0 m/s since Ben's car is initially at rest. The acceleration (a) of the car is 6.00 m/s2. And the time (t) is given as 4.10 s. The next step of the strategy involves the listing of the unknown (or desired) information in variable form. In this case, the problem requests information about the displacement of the car. So d is the unknown information. The results of the first three steps are shown in the table below. Diagram: Given: Find: vi = 0 m/s t = 4.10 s a = 6.00 m/s2 d = ?? The next step of the strategy involves identifying a kinematic equation that would allow you to determine the unknown quantity. There are four kinematic equations to choose from. Again, you will always search for an equation that contains the three known variables and the one unknown variable. In this specific case, the three known variables and the one unknown variable are t, vi, a, and d. An inspection of the four equations above reveals that the equation on the top left contains all four variables. d = vi • t + ½ • a • t2 Once the equation is identified and written down, the next step of the strategy involves substituting known values into the equation and using proper algebraic steps to solve for the unknown information. This step is shown below. d = (0 m/s) • (4.1 s) + ½ • (6.00 m/s2) • (4.10 s)2 d = (0 m) + ½ • (6.00 m/s2) • (16.81 s2) d = 0 m + 50.43 m d = 50.4 m The solution above reveals that the car will travel a distance of 50.4 meters. (Note that this value is rounded to the third digit.) The last step of the problem-solving strategy involves checking the answer to assure that it is both reasonable and accurate. The value seems reasonable enough. A car with an acceleration of 6.00 m/s/s will reach a speed of approximately 24 m/s (approximately 50 mi/hr) in 4.10 s. The distance over which such a car would be displaced during this time period would be approximately one-half a football field, making this a very reasonable distance. Checking for accuracy involves substituting the calculated value back into the equation for displacement and insuring that the left side of the equation is equal to the right side of the equation. Indeed, it is! The two example problems above illustrate how the kinematic equations can be combined with a simple problem-solving strategy to predict unknown motion parameters for a moving object. Provided that three motion parameters are known, any of the remaining values can be determined. In the next part of Lesson 6, we will see how this strategy can be applied to free fall situations. Or if interested, you can try some practice problems and check your answer against the given solutions. Kinematic Equations and Free Fall As mentioned in Lesson 5, a free-falling object is an object that is falling under the sole influence of gravity. That is to say that any object that is moving and being acted upon only be the force of gravity is said to be "in a state of free fall." Such an object will experience a downward acceleration of 9.8 m/s/s. Whether the object is falling downward or rising upward towards its peak, if it is under the sole influence of gravity, then its acceleration value is 9.8 m/s/s. Like any moving object, the motion of an object in free fall can be described by four kinematic equations. The kinematic equations that describe any object's motion are: The symbols in the above equation have a specific meaning: the symbol d stands for the displacement; the symbol t stands for the time; the symbol a stands for the acceleration of the object; the symbol vi stands for the initial velocity value; and the symbol vf stands for the final velocity. Applying Free Fall Concepts to Problem-Solving There are a few conceptual characteristics of free fall motion that will be of value when using the equations to analyze free fall motion. These concepts are described as follows: • An object in free fall experiences an acceleration of -9.8 m/s/s. (The - sign indicates a downward acceleration.) Whether explicitly stated or not, the value of the acceleration in the kinematic equations is -9.8 m/s/s for any freely falling object. • If an object is merely dropped (as opposed to being thrown) from an elevated height, then the initial velocity of the object is 0 m/s. • If an object is projected upwards in a perfectly vertical direction, then it will slow down as it rises upward. The instant at which it reaches the peak of its trajectory, its velocity is 0 m/s. This value can be used as one of the motion parameters in the kinematic equations; for example, the final velocity (vf) after traveling to the peak would be assigned a value of 0 m/s. • If an object is projected upwards in a perfectly vertical direction, then the velocity at which it is projected is equal in magnitude and opposite in sign to the velocity that it has when it returns to the same height. That is, a ball projected vertically with an upward velocity of +30 m/s will have a downward velocity of -30 m/s when it returns to the same height. These four principles and the four kinematic equations can be combined to solve problems involving the motion of free-falling objects. The two examples below illustrate application of free fall principles to kinematic problem-solving. In each example, the problem solving strategy that was introduced earlier in this lesson will be utilized. Example Problem A Luke Autbeloe drops a pile of roof shingles from the top of a roof located 8.52 meters above the ground. Determine the time required for the shingles to reach the ground. The solution to this problem begins by the construction of an informative diagram of the physical situation. This is shown below. The second step involves the identification and listing of known information in variable form. You might note that in the statement of the problem, there is only one piece of numerical information explicitly stated: 8.52 meters. The displacement (d) of the shingles is -8.52 m. (The - sign indicates that the displacement is downward). The remaining information must be extracted from the problem statement based upon your understanding of the above principles. For example, the vi value can be inferred to be 0 m/s since the shingles are dropped (released from rest; see note above). And the acceleration (a) of the shingles can be inferred to be -9.8 m/s2 since the shingles are free-falling (see note above). (Always pay careful attention to the + and - signs for the given quantities.) The next step of the solution involves the listing of the unknown (or desired) information in variable form. In this case, the problem requests information about the time of fall. So t is the unknown quantity. The results of the first three steps are shown in the table below. Diagram: Given: Find: vi = 0.0 m/s d = -8.52 m a = - 9.8 m/s2 t = ?? The next step involves identifying a kinematic equation that allows you to determine the unknown quantity. There are four kinematic equations to choose from. In general, you will always choose the equation that contains the three known and the one unknown variable. In this specific case, the three known variables and the one unknown variable are d, vi, a, and t. Thus, you will look for an equation that has these four variables listed in it. An inspection of the four equations above reveals that the equation on the top left contains all four variables. d = vi • t + ½ • a • t2 Once the equation is identified and written down, the next step involves substituting known values into the equation and using proper algebraic steps to solve for the unknown information. This step is shown below. -8.52 m = (0 m/s) • (t) + ½ • (-9.8 m/s2) • (t)2 -8.52 m = (0 m) *(t) + (-4.9 m/s2) • (t)2 -8.52 m = (-4.9 m/s2) • (t)2 (-8.52 m)/(-4.9 m/s2) = t2 1.739 s2 = t2 t = 1.32 s The solution above reveals that the shingles will fall for a time of 1.32 seconds before hitting the ground. (Note that this value is rounded to the third digit.) The last step of the problem-solving strategy involves checking the answer to assure that it is both reasonable and accurate. The value seems reasonable enough. The shingles are falling a distance of approximately 10 yards (1 meter is pretty close to 1 yard); it seems that an answer between 1 and 2 seconds would be highly reasonable. The calculated time easily falls within this range of reasonability. Checking for accuracy involves substituting the calculated value back into the equation for time and insuring that the left side of the equation is equal to the right side of the equation. Indeed it is! Example Problem B Rex Things throws his mother's crystal vase vertically upwards with an initial velocity of 26.2 m/s. Determine the height to which the vase will rise above its initial height. Once more, the solution to this problem begins by the construction of an informative diagram of the physical situation. This is shown below. The second step involves the identification and listing of known information in variable form. You might note that in the statement of the problem, there is only one piece of numerical information explicitly stated: 26.2 m/s. The initial velocity (vi) of the vase is +26.2 m/s. (The + sign indicates that the initial velocity is an upwards velocity). The remaining information must be extracted from the problem statement based upon your understanding of the above principles. Note that the vf value can be inferred to be 0 m/s since the final state of the vase is the peak of its trajectory (see note above). The acceleration (a) of the vase is -9.8 m/s2 (see note above). The next step involves the listing of the unknown (or desired) information in variable form. In this case, the problem requests information about the displacement of the vase (the height to which it rises above its starting height). So d is the unknown information. The results of the first three steps are shown in the table below. Diagram: Given: Find: vi = 26.2 m/s vf = 0 m/s a = -9.8 m/s2 d = ?? The next step involves identifying a kinematic equation that would allow you to determine the unknown quantity. There are four kinematic equations to choose from. Again, you will always search for an equation that contains the three known variables and the one unknown variable. In this specific case, the three known variables and the one unknown variable are vi, vf, a, and d. An inspection of the four equations above reveals that the equation on the top right contains all four variables. vf2 = vi2 + 2 • a • d Once the equation is identified and written down, the next step involves substituting known values into the equation and using proper algebraic steps to solve for the unknown information. This step is shown below. (0 m/s)2 = (26.2 m/s)2 + 2 •(-9.8m/s2) •d 0 m2/s2 = 686.44 m2/s2 + (-19.6 m/s2) •d (-19.6 m/s2) • d = 0 m2/s2 -686.44 m2/s2 (-19.6 m/s2) • d = -686.44 m2/s2 d = (-686.44 m2/s2)/ (-19.6 m/s2) d = 35.0 m The solution above reveals that the vase will travel upwards for a displacement of 35.0 meters before reaching its peak. (Note that this value is rounded to the third digit.) The last step of the problem-solving strategy involves checking the answer to assure that it is both reasonable and accurate. The value seems reasonable enough. The vase is thrown with a speed of approximately 50 mi/hr (merely approximate 1 m/s to be equivalent to 2 mi/hr). Such a throw will never make it further than one football field in height (approximately 100 m), yet will surely make it past the 10-yard line (approximately 10 meters). The calculated answer certainly falls within this range of reasonability. Checking for accuracy involves substituting the calculated value back into the equation for displacement and insuring that the left side of the equation is equal to the right side of the equation. Indeed, it is! Kinematic equations provide a useful means of determining the value of an unknown motion parameter if three motion parameters are known. In the case of a free-fall motion, the acceleration is often known. And in many cases, another motion parameter can be inferred through a solid knowledge of some basic kinematic principles.
• There are two groups of animals which are important in agriculture. • The groups are domestic and wild animals. 1. Domestic animals • These are animals which are trained to live with people in their homes. • Some examples includes : Uses of domestic animals • Source of food, for example milk, eggs and meat. • For cultural purposes, for example paying lobola. • Some are used for transport and labour. • Domestic animals can also be a source of income. 2 . Wild animals • These are animals which are found in game reserves and in the forests • They are also called game animals. • Examples of wild animals are: Uses of wild animals • Some of the wild animals give us meat, hides and ivory. • Wild animals attract visitors from other countries, so the country gets money. Wednesday 06 September 2023 Exercise: Domestic animals 1. What is a domestic animal? [2] 2. What is a wild animal? [2] 3. Name any 4 domestic animals that you know. [4] 4. State any 2 uses of domestic animals. [2] 5. Name any 3 wild animals that you know. [3] 6. Give 2 uses of wild ani mals. [2] 7. Wild animals can also be called animals. [1] • Animals, like human beings need good food to help them to grow and reproduce. • The main sources of food for animals include stock feeds, pastures, veld grass, crop remains and cereal grains. • These foods contain the most needed essential nutrients. Nutrient Function carbohydrates Gives energy Fats Give energy and warmth Proteins Helps in growth and repair of body parts Minerals Help in the formation of bones and teeth Vitamins Help develop good sight, improve fertility and help animals fight diseases Water Transports food in the body, cools the body and remove waste from the body. • Livestock are domesticated animals that are kept for food, use or for sale. • Small livestock has many uses. • They are a good source of food. • They also give us manure to use in gardens. • We get clothes and medicine as well from small livestock • Examples of small livestock are rabbits, sheep, goats and all form of poultry Poultry • All animals that are kept by farmers which have wings and feathers and lay eggs are called poultry. • They are a good source of white meat, eggs and manure. • They are also a source of income when we well them and their products. • Poultry includes chicken, guinea fowl, ducks, turkey, pigeons and quail birds. Rabbits • Rabbits have 3 known uses which are: They are kept for meat Kept for pelts. (pelt is animal skin used to make blankets, hats and jackets) They are used at agricultural shows • Rabbits are cheap to buy, easy to keep and feed. • They take about 4 months to mature. • Rabbits are fed using green vegetables and rabbit pellets. Wednesday 20 September 2023 Small livestock 1. Give 3 examples of small livestock. [3] 2. State 3 things that are provided by small livestock. [3] 3. What are the 2 uses of rabbits? [2] 4. Pelts are used to make _________________ [1] 5. State any 4 examples poultry. [4] 6. Rabbits take ___________ months to mature. [1] 7. What is poultry? • Apiculture is the keeping of bees in order for them to produce honey for sale. • Apiculture is very important because: (i) Provides honey - a valuable nutritional food (ii) Provides bees wax - which has many uses in industry Uses of wax For making candles Polish furniture Make crayons Prevent tools from rusting (iii) Honey bees are excellent pollinating agents, thus increasing agricultural yields. BEE COLONY Inhabitants of the bee colony and their roles • A honey bee colony typically consists of three kinds of adult bees: workers, drones and the queen 1. Workers • Workers are the smallest and constitute the majority of bees occupying the colony. • They do not lay eggs. • Workers have specialized structures, such as brood food glands, scent glands, wax glands, and pollen baskets. • these allow them to perform all the labors of the hive. Roles of the worker bees they forage for nectar, pollen, water, and plant sap. They clean and polish the cells. feed the brood. care for the queen. remove debris. handle incoming nectar. build beeswax combs. guard the entrance. 2. Drones • Drones (male bees) are the largest bees in the colony. • They are generally present only during late spring and summer. • The drone’s head is much larger than that of either the queen or worker. • Drones have no stinger, pollen baskets, or wax glands. • Their main function is to mate with the queen. 3 . Queen • Each colony has only one queen. • The queen is the largest of the bees in a bee colony. • The Queen Bee plays a vital role in the hive because she is the only female with fully developed ovaries. • She produces both fertilized and unfertilized eggs. • Queens lay the greatest number of eggs in the spring and early summer. • The queen also produce chemical scents that help regulate the unity of the colony. 1. What is apiculture? [2] 2. Give 3 reasons why apiculture is important in Zimbabwe. [3] 3. Name the 3 inhabitants of the bee colony. [3] 4. Briefly explain the roles of each inhabitant named in number 3. • Apart from using hand tools, farmers also use some farm implements and machinery to carry out their field work. • Machines help farmers do their work more easily and quickly. • The most common implements used by farmers to grow, harvest and transport their produces are: mouldboard plough Cultivator Scotch cart Harrow Planter Maize sheller combine harvesters Boom sprayers Disc harrow Spike toothed harrow KNAPSACK SPRAYER • Farming is a business. • Communal farmers grow crops and keep animal mainly for their own use. • If there is any extra they sell to get money. • Commercial farmers grow crops and keep animals for sale. • Crops and animals produced are called farm produce. • There are places were farmers have to sell their produce. • Farmers can take their produce to local markets. • A market is a place where buying and selling occurs. • Some of the local markets includes: A shopping centre A school A nearest bus stop A local village A school Local Grain Marketing Board depot (GMB) Types of farm produce Beef Fruits Mutton Eggs Vegetables Pork Milk Chicken Cereals/grains beans
All living things are made up of one or more cells. A cell is the smallest unit that can carry on all of the processes of life. Beginning in the 17th century, curious naturalists were able to use microscopes to study objects too small to be seen with the unaided eye. Their studies led them to propose the cellular basis of life. Hooke In 1665, English scientist Robert Hooke studied nature by using an early light microscope, such as the one in Figure 4-1a. A light micro- scope is an instrument that uses optical lenses to magnify objects by bending light rays. Hooke looked at a thin slice of cork from the bark of a cork oak tree. “I could exceedingly plainly perceive it to be all perforated and porous,” Hooke wrote. He described “a great many little boxes” that reminded him of the cubicles or “cells” where monks live. When Hooke focused his microscope on the cells of tree stems, roots, and ferns, he found that each had similar little boxes. The drawings that Hooke made of the cells he saw are shown in Figure 4-1b. The “little boxes” that Hooke observed were the remains of dead plant cells, such as the cork cells shown in Figure 4-1c. SECTION 1 OBJECTIVES ● Name the scientists who first observed living and nonliving cells. ● Summarize the research that led to the development of the cell theory. ● State the three principles of the cell theory. ● Explain why the cell is considered to be the basic unit of life. VOCABULARY cell cell theory Robert Hooke used an early microscope (a) to see cells in thin slices of cork. His drawings of what he saw (b) indicate that he had clearly observed the remains of cork cells (300) (c). FIGURE 4-1 (a) (b) (c) Copyright © by Holt, Rinehart and Winston. All rights reserved. 70 CHAPTER 4 Leeuwenhoek The first person to observe living cells was a Dutch trader named Anton van Leeuwenhoek. Leeuwenhoek made microscopes that were simple and tiny, but he ground lenses so precisely that the magnification was 10 times that of Hooke’s instruments. In 1673, Leeuwenhoek, shown in Figure 4-2a, was able to observe a previ- ously unseen world of microorganisms. He observed cells with green stripes from an alga of the genus Spirogyra, as shown in Figure 4-2b, and bell-shaped cells on stalks of a protist of the genus Vorticella, as shown in Figure 4-2c. Leeuwenhoek called these organisms animalcules. We now call them protists. THE CELL THEORY Although Hooke and Leeuwenhoek were the first to report observ- ing cells, the importance of this observation was not realized until about 150 years later. At this time, biologists began to organize information about cells into a unified understanding. In 1838, the German botanist Matthias Schleiden concluded that all plants were composed of cells. The next year, the German zoologist Theodor Schwann concluded the same thing for animals. And finally, in his study of human diseases, the German physician Rudolf Virchow (1821–1902) noted that all cells come from other cells. These three observations were combined to form a basic theory about the cel- lular nature of life. The cell theory has three essential parts, which are summarized in Table 4-1. Anton van Leeuwenhoek (1632–1723) is shown here with one of his hand-held lenses (a). Leeuwenhoek observed an alga of the genus Spirogyra (b) and a protist of the genus Vorticella (c). FIGURE 4-2 TABLE 4-1 The Cell Theory All living organisms are composed of one or more cells. Cells are the basic units of structure and function in an organism. Cells come only from the reproduction of existing cells. (a) (b) (c) www.scilinks.org Topic: Cell Theory Keyword: HM60241 mb06se_csfs01.qxd 5/18/07 10:54 AM Page 70
What is a Plant Cell? Plant cells are eukaryotic cells that vary in several fundamental factors from other eukaryotic organisms. Both plant and animal cells contain a nucleus along with similar organelles. One of the distinctive aspects of a plant cell is the presence of a cell wall outside the cell membrane. Plant Cell Structure Just like different organs within the body, plant cell structure includes various components known as cell organelles that perform different functions to sustain itself. These organelles include: Cell Wall It is a rigid layer which is composed of polysaccharides cellulose, pectin and hemicellulose. It is located outside the cell membrane. It also comprises glycoproteins and polymers such as lignin, cutin, or suberin. The primary function of the cell wall is to protect and provide structural support to the cell. The plant cell wall is also involved in protecting the cell against mechanical stress and providing form and structure to the cell. It also filters the molecules passing in and out of it. The formation of the cell wall is guided by microtubules. It consists of three layers, namely, primary, secondary and the middle lamella. The primary cell wall is formed by cellulose laid down by enzymes. Cell membrane It is the semi-permeable membrane that is present within the cell wall. It is composed of a thin layer of protein and fat. The cell membrane plays an important role in regulating the entry and exit of specific substances within the cell. For instance, cell membrane keeps toxins from entering inside, while nutrients and essential minerals are transported across. Nucleus The nucleus is a membrane-bound structure that is present only in eukaryotic cells. The vital function of a nucleus is to store DNA or hereditary information required for cell division, metabolism and growth. 1. Nucleolus: It manufactures cells’ protein-producing structures and ribosomes. 2. Nucleopore: Nuclear membrane is perforated with holes called nucleopore that allow proteins and nucleic acids to pass through. Plastids They are membrane-bound organelles that have their own DNA. They are necessary to store starch and to carry out the process of photosynthesis. It is also used in the synthesis of many molecules, which form the building blocks of the cell. Some of the vital types of plastids and their functions are stated below: Leucoplasts They are found in the non-photosynthetic tissue of plants. They are used for the storage of protein, lipid and starch. Chromoplasts They are heterogeneous, colored plastid which is responsible for pigment synthesis and for storage in photosynthetic eukaryotic organisms. Chromoplasts have red-, orange- and yellow-colored pigments which provide color to all ripe fruits and flowers. Central Vacuole It occupies around 30% of the cell’s volume in a mature plant cell. Tonoplast is a membrane that surrounds the central vacuole. The vital function of the central vacuole apart from storage is to sustain turgor pressure against the cell wall. The central vacuole consists of cell sap. It is a mixture of salts, enzymes and other substances. Golgi Apparatus They are found in all eukaryotic cells, which are involved in distributing synthesized macromolecules to various parts of the cell. Ribosomes They are the smallest membrane-bound organelles which comprise RNA and protein. They are the sites for protein synthesis, hence, also referred to as the protein factories of the cell. Mitochondria They are the double-membraned organelles found in the cytoplasm of all eukaryotic cells. They provide energy by breaking down carbohydrate and sugar molecules, hence they are also referred to as the “Powerhouse of the cell.” Lysosome Lysosomes are called suicidal bags as they hold digestive enzymes in an enclosed membrane. They perform the function of cellular waste disposal by digesting worn-out organelles, food particles and foreign bodies in the cell. In plants, the role of lysosomes is undertaken by the vacuoles. Chloroplasts It is an elongated organelle enclosed by phospholipid membrane. The chloroplast is shaped like a disc and the stroma is the fluid within the chloroplast that comprises a circular DNA. Each chloroplast contains a green colored pigment called chlorophyll required for the process of photosynthesis. The chlorophyll absorbs light energy from the sun and uses it to transform carbon dioxide and water into glucose. Structure of Chloroplast Chloroplasts are found in all higher plants. It is oval or biconvex, found within the mesophyll of the plant cell. The size of the chloroplast usually varies between 4-6 µm in diameter and 1-3 µm in thickness. They are double-membrane organelle with the presence of outer, inner and intermembrane space. There are two distinct regions present inside a chloroplast known as the grana and stroma. • Grana are made up of stacks of disc-shaped structures known as thylakoids or lamellae. The granum of the chloroplast consists of chlorophyll pigments and are the functional units of chloroplasts. • Stroma is the homogenous matrix which contains grana and is similar to the cytoplasm in cells in which all the organelles are embedded. Stroma also contains various enzymes, DNA, ribosomes, and other substances. Stroma lamellae function by connecting the stacks of thylakoid sacs or grana. The chloroplast structure consists of the following parts: Membrane Envelope It comprises inner and outer lipid bilayer membranes. The inner membrane separates the stroma from the intermembrane space. Intermembrane Space The space between inner and outer membranes. Thylakoid System (Lamellae) The system is suspended in the stroma. It is a collection of membranous sacs called thylakoids or lamellae. The green colored pigments called chlorophyll are found in the thylakoid membranes. It is the sight for the process of light-dependent reactions of the photosynthesis process. The thylakoids are arranged in stacks known as grana and each granum contains around 10-20 thylakoids. Stroma It is a colorless, alkaline, aqueous, protein-rich fluid present within the inner membrane of the chloroplast present surrounding the grana. Grana Stack of lamellae in plastids is known as grana. These are the sites of conversion of light energy into chemical energy. Chlorophyll It is a green photosynthetic pigment that helps in the process of photosynthesis. Functions of Chloroplast Following are the important chloroplast functions: • The most important function of the chloroplast is to synthesize food by the process of photosynthesis. • Absorbs light energy and converts it into chemical energy. • Chloroplast has a structure called chlorophyll which functions by trapping the solar energy and is used for the synthesis of food in all green plants. • Produces NADPH and molecular oxygen (O 2 ) by photolysis of water. • Produces ATP – Adenosine triphosphate by the process of photosynthesis. • The carbon dioxide (CO2) obtained from the air is used to generate carbon and sugar during the Calvin Cycle or dark reaction of photosynthesis. Mitochondria “Mitochondria are membrane-bound organelles present in the cytoplasm of all eukaryotic cells, that produce adenosine triphosphate (ATP), the main energy molecule used by the cell.” What are Mitochondria? Popularly known as the “Powerhouse of the cell,” mitochondria (singular: mitochondrion) are a double membrane-bound organelle found in most eukaryotic organisms. They are found inside the cytoplasm and essentially function as the cell’s “digestive system.” They play a major role in breaking down nutrients and generating energy-rich molecules for the cell. Many of the biochemical reactions involved in cellular respiration take place within the mitochondria. The term ‘mitochondrion’ is derived from the Greek words “mitos” and “chondrion” which means “thread” and “granules-like”, respectively. It was first described by a German pathologist named Richard Altmann in the year 1890. Structure of Mitochondria • The mitochondrion is a double-membraned, rod-shaped structure found in both plant and animal cell. • Its size ranges from 0.5 to 1.0 micrometers in diameter. • The structure comprises an outer membrane, an inner membrane, and a gel-like material called the matrix. • The outer membrane and the inner membrane are made of proteins and phospholipid layers separated by the intermembrane space. • The outer membrane covers the surface of the mitochondrion and has a large number of special proteins known as porins. Cristae The inner membrane of mitochondria is rather complex in structure. It has many folds that form a layered structure called cristae, and this helps in increasing the surface area inside the organelle. The cristae and the proteins of the inner membrane aid in the production of ATP molecules. The inner mitochondrial membrane is strictly permeable only to oxygen and ATP molecules. A number of chemical reactions take place within the inner membrane of mitochondria. Mitochondrial Matrix The mitochondrial matrix is a viscous fluid that contains a mixture of enzymes and proteins. It also comprises ribosomes, inorganic ions, mitochondrial DNA, nucleotide cofactors, and organic molecules. The enzymes present in the matrix play an important role in the synthesis of ATP molecules. Functions of Mitochondria The most important function of mitochondria is to produce energy through the process of oxidative phosphorylation. It is also involved in the following process: 1. Regulates the metabolic activity of the cell 2. Promotes the growth of new cells and cell multiplication 3. Helps in detoxifying ammonia in the liver cells 4. Plays an important role in apoptosis or programmed cell death 5. Responsible for building certain parts of the blood and various hormones like testosterone and estrogen 6. Helps in maintaining an adequate concentration of calcium ions within the compartments of the cell 7. It is also involved in various cellular activities like cellular differentiation, cell signaling, cell senescence, controlling the cell cycle and in cell growth. Disorders Associated with Mitochondria Any irregularity in the way mitochondria function can directly affect human health, but often, it is difficult to identify because symptoms differ from person to person. Disorders of the mitochondria can be quite severe; in some cases, they can even cause an organ to fail.
CARBOHYDRATES Carbohydrates are organic compounds composed of carbon, hydrogen, and oxygen in a ratio of about one carbon atom to two hydrogen atoms to one oxygen atom. The number of carbon atoms in a carbohydrate varies. Some carbohydrates serve as a source of energy. Other carbohydrates are used as structural materials. Carbohydrates can exist as monosaccharides, disaccharides, or polysaccharides. Monosaccharides A monomer of a carbohydrate is called a monosaccharide (MAHN-oh-SAK-uh-RIED). A monosaccharide—or simple sugar— contains carbon, hydrogen, and oxygen in a ratio of 1:2:1. The gen- eral formula for a monosaccharide is written as (CH2O)n, where n is any whole number from 3 to 8. For example, a six-carbon mono- saccharide, (CH2O)6, would have the formula C6H12O6. The most common monosaccharides are glucose, fructose, and galactose, as shown in Figure 3-6. Glucose is a main source of energy for cells. Fructose is found in fruits and is the sweetest of the monosaccharides. Galactose is found in milk. Notice in Figure 3-6 that glucose, fructose, and galactose have the same molecular formula, C6H12O6, but differing structures. The different structures determine the slightly different properties of the three compounds. Compounds like these sugars, with a single chemical formula but different structural forms, are called isomers (IE-soh-muhrz). SECTION 2 OBJECTIVES ● Distinguish between monosaccharides, disaccharides, and polysaccharides. ● Explain the relationship between amino acids and protein structure. ● Describe the induced fit model of enzyme action. ● Compare the structure and function of each of the different types of lipids. ● Compare the nucleic acids DNA and RNA. VOCABULARY carbohydrate monosaccharide disaccharide polysaccharide protein amino acid peptide bond polypeptide enzyme substrate active site lipid fatty acid phospholipid wax steroid nucleic acid deoxyribonucleic acid (DNA) ribonucleic acid (RNA) nucleotide C HO H C H OH C OH H C CH2OH H C H OH O Glucose C OH C O H OH C OH H CH2OH C H CH2OH Fructose C H HO C OH H C OH H C CH2OH H C H OH O Galactose Glucose, fructose, and galactose have the same chemical formula, but their structural differences result in different properties among the three compounds. FIGURE 3-6 Copyright © by Holt, Rinehart and Winston. All rights reserved. 56 CHAPTER 3 Disaccharides and Polysaccharides In living things, two monosaccharides can combine in a condensa- tion reaction to form a double sugar, or disaccharide (die-SAK-e-RIED). For example in Figure 3-4, the monosaccharides fructose and glu- cose can combine to form the disaccharide sucrose. A polysaccharide is a complex molecule composed of three or more monosaccharides. Animals store glucose in the form of the polysaccharide glycogen. Glycogen consists of hundreds of glucose molecules strung together in a highly branched chain. Much of the glucose that comes from food is ultimately stored in your liver and muscles as glycogen and is ready to be used for quick energy. Plants store glucose molecules in the form of the polysaccha- ride starch. Starch molecules have two basic forms—highly branched chains that are similar to glycogen and long, coiled, unbranched chains. Plants also make a large polysaccharide called cellulose. Cellulose, which gives strength and rigidity to plant cells, makes up about 50 percent of wood. In a single cellu- lose molecule, thousands of glucose monomers are linked in long, straight chains. These chains tend to form hydrogen bonds with each other. The resulting structure is strong and can be broken down by hydrolysis only under certain conditions. PROTEINS Proteins are organic compounds composed mainly of carbon, hydrogen, oxygen, and nitrogen. Like most of the other biological macromolecules, proteins are formed from the linkage of monomers called amino acids. Hair and horns, as shown in Figure 3-7a, are made mostly of proteins, as are skin, muscles and many biological catalysts (enzymes). Amino Acids There are 20 different amino acids, and all share a basic structure. As Figure 3-7b shows, each amino acid contains a central carbon atom covalently bonded to four other atoms or functional groups. A single hydrogen atom, highlighted in blue in the illustration, bonds at one site. A carboxyl group, —COOH, highlighted in green, bonds at a second site. An amino group, —NH2, highlighted in yel- low, bonds at a third site. A side chain called the R group, high- lighted in red, bonds at the fourth site. The main difference among the different amino acids is in their R groups. The R group can be complex or it can be simple, such as the CH3 group shown in the amino acid alanine in Figure 3-7b. The differences among the amino acid R groups gives different proteins very different shapes. The different shapes allow pro- teins to carry out many different activities in living things. Amino acids are commonly shown in a simplified way such as balls, as shown in Figure 3-7c. (a) Many structures, such as hair and horns are made of proteins. (b) Proteins are made up of amino acids. Amino acids differ only in the type of R group (shown in red) they carry. Polar R groups can dissolve in water, but nonpolar R groups cannot. (c) Amino acids have complex structures, so, in this and other textbooks, they are often simplified into balls. FIGURE 3-7 (b) Alanine (an amino acid) (c) Simplified version of amino acid CH3 H N OH C C H O H (a) Copyright © by Holt, Rinehart and Winston. All rights reserved. BIOCHEMISTRY 57 H H N C C OH H O H CH3 H2O Glycine Alanine H N OH C C H O H H H N C C H O H CH3 N OH C C H O H (a) (b) (a) The peptide bond (shaded blue) that binds amino acids together to form a polypeptide results from a condensation reaction that produces water. (b) Poly- peptides are commonly shown as a string of balls in this textbook and elsewhere. Each ball represents an amino acid. FIGURE 3-8 Substrate Products Enzyme 1 2 3 In the induced fit model of enzyme action, the enzyme can attach only to a substrate (reactant) with a specific shape. The enzyme then changes and reduces the activation energy of the reaction so reactants can become products. The enzyme is unchanged and is available to be used again. 3 2 1 FIGURE 3-9 Dipeptides and Polypeptides Figure 3-8a shows how two amino acids bond to form a dipeptide (die-PEP-TIED). In this condensation reaction, the two amino acids form a covalent bond, called a peptide bond (shaded in blue in Figure 3-8a) and release a water molecule. Amino acids often form very long chains called polypeptides (PAHL-i-PEP-TIEDZ). Proteins are composed of one or more polypep- tides. Some proteins are very large molecules, containing hun- dreds of amino acids. Often, these long proteins are bent and folded upon themselves as a result of interactions—such as hydrogen bonding—between individual amino acids. Protein shape can also be influenced by conditions such as temperature and the type of solvent in which a protein is dissolved. For exam- ple, cooking an egg changes the shape of proteins in the egg white. The firm, opaque result is very different from the initial clear, runny material. Enzymes Enzymes—RNA or protein molecules that act as biological catalysts—are essential for the functioning of any cell. Many enzymes are proteins. Figure 3-9 shows an induced fit model of enzyme action. Enzyme reactions depend on a physical fit between the enzyme molecule and its specific substrate, the reactant being catalyzed. Notice that the enzyme has folds, or an active site, with a shape that allows the substrate to fit into the active site. An enzyme acts only on a specific substrate because only that substrate fits into its active site. The linkage of the enzyme and substrate causes a slight change in the enzyme’s shape. The change in the enzyme’s shape weakens some chemical bonds in the substrate, which is one way that enzymes reduce activation energy, the energy needed to start the reaction. After the reaction, the enzyme releases the products. Like any catalyst, the enzyme itself is unchanged, so it can be used many times. An enzyme may not work if its environment is changed. For example, change in temperature or pH can cause a change in the shape of the enzyme or the substrate. If such a change happens, the reaction that the enzyme would have catalyzed cannot occur.