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EOC Final Review

EOC Final Review. SHORT-TERM (quick) ENERGY. 1. BENEDICTS SOLUTION. 1. SUGAR. 2. STARCH. 1. IODINE SOLUTION. 1 GLYCEROL & 3 FATTY ACIDS. LONG-TERM (quick) ENERGY. BROWN PAPER BAG TEST. MAINTAINING HOMEOSTASIS IN THE BODY. AMINO ACIDS. BIURET’S SOLUTION. NUCLEOTIDE

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EOC Final Review

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  1. EOC Final Review

  2. SHORT-TERM (quick) ENERGY 1. BENEDICTS SOLUTION 1. SUGAR 2. STARCH 1. IODINE SOLUTION 1 GLYCEROL & 3 FATTY ACIDS LONG-TERM (quick) ENERGY BROWN PAPER BAG TEST MAINTAINING HOMEOSTASIS IN THE BODY AMINO ACIDS BIURET’S SOLUTION NUCLEOTIDE (1 phosphate, 1 sugar, 1 nitrogen base) HOLD GENETIC INFORMATION HOLD RECIPE TO MAKE PROTEINS 1. ALL LIVING THINGS 2. DNA FINGERPRINTING

  3. PROTEINS AMINO ACIDS A I J D H E G C F B ALL PROTEINS!!!!!

  4. BOTH Stores ALL of the important information for the cell Decides what comes in and out of cell BOTH Extra support and protection for plant cells PLANTS ONLY Provides energy for the cell (ATP) BOTH Plants have 1 LARGE vacuole BOTH Stores material Animals have multiple small ones Photosynthesis to make food for plant (glucose) PLANTS ONLY Makes proteins (site of protein synthesis) BOTH

  5. PLANT ANIMAL CELL MEMBRANE NUCLEUS RIBOSOMES RIBOSOMES VACUOLE CELL WALL CELL MEMBRANE MITOCHONDRIA CHLOROPLAST

  6. CELL WALL CHLOROPLAST VACUOLE (large) PROKARYOTIC EUKARYOTIC NO NUCLEUS HAS A NUCLEUS NO MEMBRANE-BOUND ORGANELLES (ONLY RIBOSOMES) HAS MEMBRANE-BOUND ORGANELLES PLASMIDS (circular DNA) DNA IN NUCLEUS (in chromosomes) LARGER, COMPLEX, YOUNGER SMALL, SIMPLE, OLDER

  7. Unit 3 Reminders: • Cell specialization/differentiation: • Where do all cell originally come from (what type of cell)? • Do all cells have the same DNA? • Do all cells have the same function (job)? • What makes a muscle cell and a nerve cell different? • REMEMBER: all cells have the same DNA, but different jobs! The only difference are the GENES that are turned on or off in each cell – this determines their job!

  8. LOW HIGH HIGH LOW HIGH LOW PROTEIN LOW HIGH PROTEIN

  9. YES NO HIGH to LOW LOW to HIGH

  10. Transport Reminders: • ALWAYS draw you box-circle model • When the molecules CANNOT move it is OSMOSIS • WATER moves high to low • Use the salt concentration. Subtract from 100% (inside and outside). The left remaining amount is the water concentration. Move the water from the high concentration to the low concentration. 3. If water moves… • OUT = the cell will SHRINK or SHRIVEL • IN = the cell will SWELL or BURST

  11. 20% salt LOW (water) Water will move out of the cell and it will SHRINK 80% water 100% water 0% salt HIGH (water) 100% water HIGH (water) 0% salt Water will move into of the cell and it will SWELL 20% salt 80% water LOW (water)

  12. MAINTAINING BALANCE WITHIN AN ORGANISM

  13. 70% water L 90% water H 90% water 90% water SHRINK EQUILLIBRIUM 98% water H 90% water L SWELL

  14. Real life application of osmosis… • What happens when you don’t water your plants… The VACUOLE loses water (water leaves the cell), so the cell SHRINKS or SHRIVELS, causing the plant to wilt and die.

  15. ENZYME ENZYME PRODUCTS SUBSTRATE (reactant) ENZYME- SUBSTRATE COMPLEX ACTIVE SITE

  16. PROTEINS THE SHAPE OF THE ACTIVE SITE YES – ENZYMES ARE REUSED FOR THE SAME SPECIFIC TYPE OF REACTIONS, UNLESS THE ACTIVE SITE IS DENATURED (the shape is changed). pH AND TEMPERATURE CAN DENATURE THE ACTIVE SITE OF THE ENZYME. IF THE ACTIVE SITE CHANGES SHAPE, THE ENZYME CAN NO LONGER PERFORM AT AN OPTIMAL LEVEL AND MAY STOP WORKING.

  17. ACID 3 9 BASE

  18. SUN (energy) + CARBON DIOXIDE (CO2) + WATER (H2O)  OXYGEN (O2) + GLUCOSE (C6H12O6) CHLOROPLAST CO2 O2

  19. OXYGEN (O2) + GLUCOSE (C6H12O6)  36 ATP + CARBON DIOXIDE (CO2) + WATER (H2O) MITOCHONDRIA

  20. AEROBIC REPIRATION USES OXYGEN AND CREATES 36 ATP ANAEROBIC REPIRATION DOES NOT USE OXYGEN AND CREATES 2 ATP IN ANIMAL MUSCLE Remember: FERMENTATION is another name for ANAEROBIC RESPIRATION IN YEAST AND BACTERIA THE PRODUCTS ARE: 2 ATP, CARBON DIOXIDE, and ETHYL ALCOHOL

  21. SUN going IN WATER going IN OXYGEN being RELEASED

  22. C T G G C T SUGAR

  23. CGA GAC Arg- Asp- 3 mRNA letters (nitrogen bases)

  24. RIBOSE DEOXYRIBOSE A, G, C, T A, G, C, U 1 (single helix) 2 (double helix) NUCLEUS & CYTOPLASM NUCLEUS DELIVERS GENETIC MESSAGES TO MAKE PROTEINS HOLDS GENETIC INFORMATION TO CODE FOR PROTEINS PEPTIDE BONDS

  25. PROTEINS TRANSCRIPTON TRANSLATION TRANSPORT DNA MESSAGE FROM NUCLEUS TO RIBOSOME READ mRNA MESSAGE (anticodon) AND BRING CORRECT AMINO ACID TO THE RIBOSOME STORES GENETIC INFORMATION FOR LIFE MAINTAIN HOMEOSTASIS IN BODY (health, repair, communication, digestion, speed up reactions) mRNA NUCLEUS CYTOPLASM RIBOSOME tRNA PROTEIN ACID AMINO TRUE EVERY CELL HAS THE SAME DNA, BUT A DIFFERENT JOB. THE DIFFERENT JOBS ARE DETERMINED BY THE GENES THAT ARE TURNED ON OR OFF IN A CELL.

  26. SEXUAL ASEXUAL 2N = DIPLOID 2N = DIPLOID 2N = DIPLOID N = HAPLOID 2 1 2 4 50 chromosomes 25 chromosomes BEFORE CELL DIVISION BEFORE 1st CELL DIVISION NO YES YES NO YES YES YES YES NO YES

  27. MITOSIS MEIOSIS MEIOSIS BOTH! MITOSIS MEIOSIS GROWTH REPAIR REPLACING DEAD OR WORN OUT CELLS

  28. C Interphase= DNA Replication INTERPHASE Prophase = chromosomes form; nucleus breaks down; spindle fibers appear B ANAPHASE (away) PROPHASE (first) E Metaphase = chromosomes line up in the middle of the cell Anaphase = chromosomes pulled apart by spindle fibers A TELOPHASE (2 new cells) METAPHASE (middle) Telophase= nucleus reforms; cytoplasm divides; 2 new cells D MEIOSIS GAMETE (egg) FERTILIZATION ZYGOTE (1st diploid cell) MITOSIS EMBRYO ADULT GAMETE (sperm) MEIOSIS

  29. Phenotype = Tall Tt tt Genotype = _____ _____ ______ TT 1 (25%) 2 (50%) 1 (25%) SHORT Phenotype = _________ _________ TALL 1 (25%) 3 (75%) 1:2:1 3:1 1. Nutrition and health 2. Environment may favor tall trait = trees (food) may grow taller favoring tall organism. Tall would be able to eat, survive and reproduce. Short ones would die off.

  30. 100% PINK FLOWERS RR’ Genotype = _____ _____ ______ RR R’R’ 0 (0%) 0 (0%) 4 (100%) PINK WHITE 100% RR’ Phenotype = ______ _______ _______ RED 0 (0%) 4 (100%) 0 (0%) R R Phenotypes: RED, WHITE, PINK R’ R R’ R R’ Genotypes: RR= RED RR’= PINK R’R’= WHITE R’ R R’ R R’ Parents: R’R’ ______ x _______ RR

  31. 25% RED; 50% PINK; 25% WHITE 25% RR; 50% RR’; 25% R’R’ R’ R Parents: ______ x _______ RR’ RR’ R R R R’ R Genotype = _____ _____ ______ RR RR’ R’R’ R’ R’ R R’ R’ 1 (25%) 2 (50%) 1 (25%) Phenotype = ______ _______ _______ RED PINK WHITE 1 (25%) 2 (50%) 1 (25%)

  32. 100% Black-and-Tan offspring BT Genotype = _____ _____ ______ BB TT 100% BT 4 (100%) 0 (0%) 0 (0%) Tan Black+Tan Phenotype = _____ _____ ______ Black BB x TT 4 (100%) 0 (0%) 0 (0%) Phenotypes: Black Black-and-Tan Tan B B T B Genotypes: BB = Black T B T BT = Black + Tan TT = Tan T B T Parents: B T _______ x _______ BB TT

  33. XY XX Sex-linked traits travel on the X-chromosome 25% chance of child with hemophilia (1 son) 0% chance of daughter with hemophilia 25% chance of daughter being a carrier 1 0 1 Males only have 1 X – so they have it or they don’t 1 1 0 Phenotype: Healthy, carrier, Hemophilia (sick) XR Xr Xr XR XR XR XR Genotype: XRXR = healthy female XRY = healthy male XRXr = carrier female XrY = sick male Y XrXr = sick female XR Y Xr Y Parents: XRY ______ x ______ XRXr

  34. 25% change of child with Type O 0% change of child with homozygous Type A 25% change of child with Type AB Genotype: ____ ____ ____ ____ ____ ____ oo AA Ao BB Bo AB Phenotype: Type A, B, AB, or O 0 (0%) 1 (25%) 0 (0%) 1 (25%) 1 (25%) 1 (25%) Type B Phenotype: _______ _______ _______ _______ Type A Type AB Type O 1 (25%) 1 (25%) 1 (25%) 1 (25%) Genotype: A o Type A: AA or Ao Type B: BB or Bo B A B B o Type AB: AB Type O: oo A o Bo What we know: Parents: _____ x _____ Ao o o o Mom: Type A (AA or Ao) Dad: Type B (BB or Bo) Baby: Type O (oo)

  35. Narrowing it down: Phenotype: Type A, B, AB, or O What we know: If the child is blood Type O, then both parents have to give an ‘o’ allele. Mom: Type A (AA or Ao) Baby: Type O (oo) Genotype: Dad: 1. Type AB (AB) 2. Type A (AA or Ao) 3. Type O (oo) Type A: AA or Ao This leaves means Mom has to be heterozygous Type A (Ao). Type B: BB or Bo Dad 3 could be the father because he is Type O (oo) and can give an ‘o’ allele. Type AB: AB Type O: oo Dad 2 could be the father only is he is heterozygous Type A (Ao), because he must have an ‘o’ allele to give. Dad 1 could NOT be the father because he is blood Type AB (AB). He does not have an ‘o’ allele to give, so he cannot be the father of a child with Type O (oo) blood.

  36. SKIN COLOR, HAIR COLOR, EYE COLOR MULTIPLE ALLELES HAVE MORE THEN TWO ALLELES THAT CAN CODE FOR A DIFFERENT TRAITS, BUT ALL OF THE ALLELES ARE LOCATED ON THE SAME GENE. AN EXAMPLE OF MULTIPLE ALLELES ARE BLOOD TYPES (A, B, O). POLYGENIC TRAITS HAVE MORE THEN TWO ALLELES THAT CAN CODE FOR A DIFFERENT TRAITS, BUT THE ALLELES ARE LOCATED ON THE DIFFERNT GENES. THIS CREATES A VERY WIDE RANGE OF PHENOTYPES. EXAMPLES OF POLYGENIC TRAITS ARE HAIR COLOR, EYE COLOR, AND SKIN COLOR.

  37. AUTOSOMAL RECESSIVE Autosomal = because there are an equal number of males and females affected. Recessive = because it is not present it every generation AND affected children do not have affected parents. Aa Aa aa A_ A_ Aa aa Aa Aa A_ aa A_

  38. FEMALE - XX DOWN SYNDROME – 3 chromosomes on #21 NON-DISJUNCTION – chromosomes do not separate properly during meiosis

  39. AN INTERNATIONAL RESEARCH EFFORT TO DETERMINE THE SEQUENCE OF HUMAN GENOME (all DNA) AND IDENTIFY THE GENES IT CONTAINS. THE PRODUCTION OF MULTIPLE, IDENTICAL OFFSPRING USING BIOTECHNOLOGY.

  40. GEL ELECTROPHORESIS A & C – 4 out of 6 strands in common

  41. GENETIC ENGINEERING THE DESIRED GENE (insulin) IS CUT OUT USING RESTRICTION ENZYMES. THE DESIRED GENE (insulin) IS THEN GLUED INTO A BACTERIAL PLASMID (circuluar DNA) USING THE ENZYME LIGASE. THE PLASMID (now recombinant DNA with the bacterial host and desired gene) IS INSERTED BACK INTO THE BACTERIAL HOST. What is this technology typically used to produce? THE BACTERIAL WILL NOW PRODUCE COPIES OF THE DESIRED GENE EVERY TIME IT DIVIDES (using binary fission). So, the insulin gene is reproduced every time the bacteria divides.

  42. FOSSILS ARE USED TO COMPARE AGE AND FETURES TO HELP DETERMINE COMMON ANCESTRY COMPARING DNA, AMINO ACIDS, AND PROTEIN SEQUENCES TO DETERMINE COMMON ANCESTRY. FEWER DIFFERENCE MEANS A CLOSER COMMON ANCESTOR. VARIATION IS NECESSARY FOR EVOLUTION. THERE MUST BE DIFFERENCES AMONG ORGANISMS IN ORDER FOR THERE TO BE COMPETITION. COMPETITION CREATES NATURAL SELECTION – THOSE WITH THE MOST FAVORABLE TRAITS FOR THE ENVIRONMENT SURVIVE AND REPRODUCE.

  43. GEOGRAPHIC ISOLATION CREATES A PHYSICAL DIVIDE BETWEEN ORGANISMS. NATURAL PHYSICAL BARRIERS INCLUDE MOUNTAINS, LAKES< RIVERS, OCEANS AND ISLANDS. THE PHYSICAL SEPARATION MEANS THE ENVIRONMENTS ARE DIFFERENT. THE DIFFERENT ENVIRONMENTS WILL SELECT DIFFERENT TRAITS AS BEING BETTER FIT FOR THE SPECIFIC ENVIRONMENT. SO, OVER TIME THE MOST SUCCESSFUL ORGANISMS WILL DIFFER IN APPEARANCE BASED UPON THE ENVIRONMENT THEY ARE IN. THE SPECIES ARE ALSO NO LONGER MATING, DUE TO THE PHYSICAL SEPARATION. OVER A LONG PERIOD OF TIME THE SPECIES ARE NO LONGER ABLE TO REPRODUCE WITH ONE ANOTHER TO PRODUCE FERTILE OFFSPRING. THIS CREATE A NEW SPECIES – SPECIATION. THE ENVIRONMENT PLAYS A VERY IMPORTANT ROLE IN EVOLUTION. THE ENVIRONMENT SELECTS WHICH TRAITS AND ADAPTATION ARE BENEFITIAL TO AN ORGANISM. THE ORGANISM WITH THE DESIRABLE TRAITS SURVIVE AND PASS ON THEIR GENES.

  44. EVOLUTION OF CELLS… No OXYGEN, which means NO PHOTOSYNTHESIS, which means organisms found food and did NOT make it Anaerobic Heterotrophic Prokaryotic Cells SUN, WATER & CARBON DIOXIDE available on early earth – the organisms use these to begin doing PHOTOSYNTHESIS Photosynthetic Prokaryotic Cells PHOTOSYNTHESIS creates OXYGEN – this allows heterotrophic organisms to make MORE ATP than with no Oxygen… Aerobic Heterotrophic Prokaryotic Cells Aerobic Heterotrophic Prokaryote (mitochondria) & Photosynthetic Prokaryote (chloroplast) form a partnership… (Endosynbiotic Theory) Eukaryotic Cells

  45. ABIOGENESIS = LIFE COMES FROM NON-LIVING THINGS BIOGENESIS = LIFE COMES FROM OTHER LIVING THINGS REDI PASTEUR

  46. VARIATION (Differences among members of the species) Most with SHORTER necks Occasionally some with LONGER necks LONGER neck = easier time getting food COMPETITION (More organisms than resources. Must compete for food, shelter and mates) Food = survival LONG NECK = BETTER TRAIT NATURAL SELECTION (Those with the best traits and adaptations for the environment are able to survive and reproduce.) LONG Neck giraffes more successful in obtaining food and mates LONG neck gene is passed on because it is the more successful trait Future generations look more and more like the successful traits – LONG NECK giraffes

  47. KING KINGDOM PHILIP PHYLUM CAME CLASS OVER ORDER FOR FAMILY GOOD GENUS USED FOR SCIENTIFIC NAMING: Genus species SOUP SPECIES

  48. SCIENTIFIC NAMING USED UNDERSTOOD BY ALL SCIENTIST ACROSS THE WORLD. TO NAME: GENUS = FIRST NAME (capitalize first letter) SPECIES = LAST NAME (all lowercase) EXAMPLE: Homo sapien GENUS SPECIES COMMON NAME = HUMAN

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