| A | B |
|---|
| aerobic respiration is the | complete breakdown of glucose in ATP |
| anaerobic alcoholic respiration | incomplete breakdown of glucose- fermentation |
| anaerobic respiration uses | no oxygen |
| respiration rate is measured by | CO2 consumption and O2 consumption |
| pyruvate is a | product of glycolysis |
| magnesium sulfate activates | enzymes in glycolysis |
| sodium fluoride is an | enzyme inhibitor |
| sodium fluoride | stops glycolysis |
| glucose is | an energy source for respiration (food) |
| during anaerobic fermentation experiment the dependant variable was | CO2 production |
| During the anaerobic fermentation experiment what was the independant variable | presence of inhibitors or activators |
| yeast rises because of | CO2 |
| yeast is an organism that eats | glucose |
| During the exp. using peas the dependant variable was | O2 consumption |
| During the peas exp. what was the independant variable | temperature |
| during the peas exp. what was the negative control | plastic beads |
| what was the postive control in the peas exp. | peas |
| in the respiration lab why was sodalime used | to absorb CO2 |
| what happens when temp. slows down to enzyme activity | temporarily denatures enzyme |
| what happens when temp. speeds up to enzyme activity | more effective collisions happen increasing enzyme activity |
| Peas use O2 during aerobic respiration and | release CO2 |
| smaller the animal | larger the surface area to volume ratio |
| smaller animals | lose more heat |
| smaller animals have to respire more to | replace lost heat |
| cold blooded | poikilotherm |
| in cold blooded animals body temp is determined by | the environment |
| cold blooded animals become active when | external environment is cold |
| the most important byproduct of cellular respiration | heat |
| a byproduct of eating is | breaking food down into ATP |
| in cold blooded animals less | heat is needed to maintain bodytemp |
| cold blooded animals eat less because | they don't need to produce heat |
| cold blooded animals have | have a slower respiration rate |
| hemeotherms have stable | internal temp. |
| body temp is determined by | heat produced during cellular respiration |
| more heat is needed to | maintain body temp. |
| warm blooded organisms have a stable internal temp reguardless of | external influence |
| homeotherms retard heat loss by | insulating layers |
| > the internal heat needed | > the respiration and O2 used |
| if you did not restrain an active rodent what would you expect to happen to his oxygen | animals need more ATP and heat is a byproduct |
| in a leaf structure what is visible | cell wall |
| in a leaf structure what stuctures are not visible | everything but the cell wall |
| what distinguishes chloroplasts from other structures | they are green |
| the shape of chloroplasts | circle |
| heat causes plant cells to heat up causing the chloroplast to | gather energy from heat to make food |
| where photosynthesis takes place | guard cell, palisades, and spongy layer |
| the cuticle or upper epidermis is also known as | the cuticle |
| what does the cuticle do | waxy protective layer decreases water loss & injury |
| the veins of a leaf acts as | muscle and skeleton |
| the spongy layer stores glucose and does what else | gas exchange, & some photosynthesis |
| the guard cell does what | opens and closes stomata |
| the stomata is where | gas exchange takes place |
| as absorbance increases | transmittance decreases |
| the greatest amount of light absorbtion | 380-440 |
| the colors of the wavelegnths that are the greatest amount of absorbtion | violet and blue |
| the spectroscope uses | numbers |
| the spectrophotometer uses | color |
| fluorescence are electrons that move out to higher energy levels | by light absorbtion that falls back and gives off light absorbed |
| the spectrum | R O Y G B I V |
| what colors are removed by the pigments | blue and violet are removed |
| blue and violet were removed because | the colors were absorbed by the chlorophyll |
| in our exp. why did we use a fishbowl | to elminate the effect of temperature |
| why did we use spring water for the elodea | because tap water could have inhibitors |
| what roles does sodium bicabonate have in our exp. | makes CO2 |
| greater the distance, lower the light intensity | less photosynthesis, less O2 bubbles |
| less distance, higher the light intensity | more photsynthesis, more O2 bubbles |
| why do plants grow more during the summer months | plants are closer to the sun & light intensity is increased |
| o2 is a byproduct of | photosynthesis |
| energy of photons is inversely | proportional to their wavelegnths |
| light that is absorbed in not | transmitted |
| light that is transmitted is not | absorbed |
| absorbance and transmittance are | complemnetary phenomena |
| light that is absorbed is converted into | chemical bond energy: glucose |
| in our exp. why was alcohol used | it was the solvent |
| why did we use alcohol as the solvent | because alcohol is nonpolar & the pigment would dissolve |
| rezeroing the spectrophotometer is done because why | to be able to subtract the reading from the solvent allowing for pigment reading only |
| chlorophyll b | light yellow/ green |
| chlorophyll a | darker blue/ green |
| xanthophyll | yellow |
| carotenes | orange |
| mitosis | one cell divides into another |
| somatic cells | all cells except egg & sperm |
| a plant cell is | anastral, no centrioles, but has a cell plate |
| plant cells divide from | inside out |
| animal cell has a | cleavage furrow |
| animal cells divide from | the outside in |
| dyad (x) | double chromosome; made of 2 sister chromatids |
| monad (-) | single chrosome: made of 1 chromotid |
| diploid (2n) | 2 of every type of chromosome |
| haploid | 1 of every type of chromosome |
| In mitosis in G1 what doubles | organelles |
| In mitosis in G2 what doubles | proteins |
| In mitosis DNA does what | doubles |
| In mitosis monads become | dyads |
| In mitosis in prophase centrioles go | to opposite poles and form spindles |
| In mitosis what happens to the nuclear membrane | it disappears |
| In mitosis in prophase chromatin | coils up & thickens & dyads appear |
| In mitosis in metaphase spindles | go across |
| In mitosis in metaphase dyads attach to | center of spindle at kinetochore |
| In mitosis in metaphase dyads | line up single file |
| In mitosis in metaphase centromeres | double to help sister chromatids easily separate |
| In mitosis during anaphase what begins | cytokenesis |
| In anaphase in mitosis dyads become | monads |
| In mitosis in anaphase sister chromatids | separate |
| In mitosis in telophase there are | 2 separate identical cells |
| meiosis is | reduction division |
| homologue | 2 pairs of the same type |
| bivalent | 4 chromatids |
| gametogenesis | spermatogenesis & oogenesis |
| spermatogenesis occurs in the | testes |
| oogenesis occurs in the | ovaries |
| In meosis 1 in interphase1 what happens in G1 | organelles double |
| In meosis interphase1 what doubles in G2 | proteins |
| in meosis in interphase1 DNA | doubles |
| in meosis 1 in interphase1 monads become | dyads |
| In meosis 1 in prophase there are how many spindles | 2 |
| in meosis 1 prophase1 what takes place | synapse |
| in meosis 1 what important step happens | crossing over |
| in meosis in metaphase what important event happens | random/ independant alignment |
| in meosis 1 in prophase1 bivalents are in the | middle |
| In meosis 1 in anaphase 1 the homolgus pairs | separate |
| In meosis 1 in anaphase 1 the diploid becomes a | haploid & are still dyads |
| In meosis 1 teleophase 1 the cells are not | identical |
| what phase is skipped in meosis II | s phase- DNA is not doubled |
| what doubles in interphase of meosis II | organelles and proteins double |
| in metaphase II of Meosis II what doubles | centromeres |
| in anaphase II of meosis II dyads | become monads and sister chromosomes separate |
| In telophase II there are | 4 monads and no identical cells |
| number of divisions in mitosis | 1 |
| number of divisions in meiosis | 2 |
| number of daughter cells in mitosis | two |
| number of daughter cells in meosis | four |
| important events that occure during meiosis I | paiting up and crossing over |
| in telophase there are | monads |
| in meosis I there are | dyads |
| in telophase in mitosis and meosis there are | monads in both phases |
| allele | alternate form of a gene |
| degrees of freedom equation | n-1 |
| monohybrid | one trait |
| dihybrid | 2 different traits |
| homozygous | having 2 identical alleles LL ll |
| heterozygous | unlike allele for trait Ll |
| phenotype | physical appearance- tall short |
| genotype | genes of organism for trait |
| In chi square if O & E were equal X2 is | 0 |
| In chi squared a small value for X2 indicates | that observed & expected ratios are close |
| In chi square if there is a large value in X2 that means | there is a marked deviation from ratio |
| the probabilty that 2 events occur simultaneously | is the product of each occuring by itself |
| if the calculated value of X@ in chi square is < the X2 table value | accept hypothesis |
| If the X2 value is > the X2 table value | reject hypothesis |
| the reason white albino plants die | they cannot do photosynthesis |
| the chance of getting an albino plant | 25% |
| the reason albino plants continue and do not die off | they are protected in the heterozygous allele |
| ratio for dihybrid cross | 9:3:3:1 |
| how many chromosome pairs in humand | 23 |
| how many chromosome pairs are autosomes in human | 22 |
| How many pairs of sex chromosomes | 1 pair |
| female | XX |
| male | XY |
| turner syndrone | monosomy XO |
| how many chromosomes in Turners syndrome | 45 |
| poly -X/ triplo X | trisomny- XXX- 47 chromosomes |
| kleinfelters | trisomy- XXY- 47 chromosomes |
| klinefelters & Jacob syndrome are found in | males |
| Turner and Poly X are found in | females |
| syndrome that the problem is with the father in meosis II | Jacobs |
| a clue for sex linked patterns | primarily found in males |
| a clue for sex linked patterns | passed from granfather to grandson |
| 1 pair of chromosomes that is not homolgus in a normal karyotype | 23 sex chromosome in males |
| all members of a species living in one local make up | a population |
| genetic drift refers to changes in | gene pool frequencies due to change alone |
| hardy weinberg law calculates | gene pool frequencies |
| conditions for hardy weinberg law to be valid | no mutations, no gene flow, random mating, no genetic drift, no natural selection |
| genetic drift is totally due to | chance |
| how does the end result of genetic drift differ from natural selection | populations become different |
| an organisms fitness is based on | reproductive success |
