| A | B |
|---|
| equation for aerobic respiration | C6+ H12+ O6--> 6H2O + 36ATP +heat |
| aerobic respiration is | complete breakdown of glucose |
| fermentaion is | anaerobic and incomplete breakdown of glucose |
| what gas was responsible for the holes in baked bread | CO2 |
| what was the independant variable in the yeast exp. | activators and inhibitors |
| what dependant variable in the yeast exp. was used as a measure of fermentaion in this experiment | CO2 |
| what is the function of glucose added to the yeast suspension | food |
| why did we add yeast in the experiment | ate glucose in which fermentation took place. released CO2 |
| how did the concentration of NaF affect fermentaion in the exp. | inhibited fermentation |
| when NaF was added to yeast glucose =little fermentation NaF & pyruvated added what happened | no pyruvate conteracted Na Inhibitor |
| How did the concentration of MgSO4 affect fermentaion in exp. | activated fermentation |
| In the pea exp. if the temp was cold what happened | slows down enzyme activity, temp. denatures enzyme |
| In pea exp. what happened in room temp | enzyme activity had effective collisions |
| In the pea exp. what happened in the warm water bath | enzyme activity even more, even more effective collisions |
| plants such as peas use O2 during aerobic respiration and | release CO2 |
| In the pea exp. why were plastic beads used | negative control |
| To measure resp. rates of the gerbil and snake why was soda lime used | to remove CO2 |
| what was measured in order to calculate resp. rate for gerbils | oxygen used |
| which has a higher resp. rate small or large animal | small |
| the small animal has a higher respt rate because | of surface area volume ratio |
| an active animal has a higher respt rate because | it needs to make ATP |
| warmblooded | endotherm |
| mammals heat source is | internal |
| coldblooded | ectotherm |
| in coldblooded animals their heat source is | external |
| 2 ways resp. rate was measured | CO2 production in yeast exp. O2 consumption in yeast exp |
| distinguishing feature of a chloroplast | green |
| upper epidermis of a leaf is also known as | cuticle |
| guard cells are located where | lower epidermis |
| 3 functions of the spongy layer in the mesophyll | gas exchange, photosynthesis, storgae of glucose |
| the importance of the palisade layer | where 80% of photosynthesis takes place |
| the function of veins in a plant | transportaion of water and nutrients |
| 3 places where photosynthesis can occur | guard cells, mesophyll, palisades |
| the role sodium bicarbonate dissolved in water serve | source of CO2 for photosynthesis |
| what gas came from the elodea formed the bubbles that were counted | O2 |
| O2 is a by product of | photosynthesis |
| relationship between number of bubbles & light intensity | less distance-greater light intensity-more O2 bubbles |
| distance between light source and plant source increases | greater the distance-less light intensity-less photosynthesis& O2 bubbles |
| dependant variable in the elodea exp. | O2 bubbles |
| independant variable of elodea exp. | light intensity |
| why is the peak of growing season in the summer | earth is closer to the sun- more light intensity more photosynthesis |
| what variable was eliminated in the elodea exp. by using a fishbowl of water | temperature/ heat |
| as absorbance increases | transmittence decreases |
| light that is absorbed is converted into | chemical bond energy called glucose |
| greatest absorption peak | 380-420 blue violet |
| smallest absorption peak | 660- red |
| what colors of light does a plant use most | blue violet red |
| name the wavelegnth range of red | 600-700nm |
| name the wavelength range of blue | 400-500 |
| name the wavelegnth range of green | 500-600 |
| if plants receive only green light why doesn't photosynthesis occur | they reflect green do not use it for absorbtion |
| fluorescence | when eletrons move to higher energy level by light absorption, then fall back giving off light |
| spectrophotometer gives what kinds of results | numerical |
| spectroscope gives what kinds of results | colors |
| why did we use a blank in the spectrophotometer experiment | so the machine knows what #'s to subtract in its readings when pigment is added |
| why was alcohol used as a blank in the color lab and not ether | alcohol was the actual solvent |
| alcohol is | non polar |
| In the color lab what 4 pigments were seen | Chloro. b, chloro. a, xanthophyll and carotene |
| abreviation for colors in the spectrum | ROYGBIV |
| if chlorophyll extract was placed in front of spectroscope what colors are seen | ROYG |
| why are blue and violet not seen in the spectroscope if you use chlorophyll extract | BV absorbed and used in photosynthesis |
| In Mitosis what happens in interphase G1 | organelles double |
| what happens in Mitosis in interphase during the S phase | DNA doubles |
| In mitosis during interphase what doubles in G2 | proteins |
| name the 4 stages of mitosis | prophase, metaphase, anaphase, telophase |
| when does cytokenesis begin | anaphase |
| which phase does a cell spend most of its time | interphase |
| if a nucleus has 10 chromosomes during interphase how many does it have during metaphase | 10 |
| how many chromosomes at the end of telophase | 10 |
| chromatin | the form DNA takes |
| In prophase dyand chromosomes are | visible |
| In prophase nuclear membrane and organelles start to | dissappear |
| in prophase centrioles go to poles and start to make the | spindles |
| In metaphase dyads | line up in the middle single file |
| in metaphase centromeres double so | sister chromatids can separate easily |
| In anapphase sister chromatids go apart to | opposite poles |
| In anaphase dyads become | monads |
| cytokenesis begins in | anaphase |
| In telophase ends up with | 2 genetically identical cells |
| In telophase the cells are still | diploid |
| in animal cells cytokenesis occurs in the | cleavage furrow |
| in plant cells cytokenesis occurs in the | cell plate |
| animal cells have | centromeres, spindle fibers, and kinetochores |
| plant cells do not have | centrioles and asters |
| why is crossing over a good thing | genetic variation |
| plant cells have a cell | wall |
| 2 examples of asexual reproduction | plant cuttings, potato tuber |
| crossing over occurs in | prophase I |
| random alighnment occurs in | metaphase I |
| somatic cell has 20 chromosomes how many in the gamete | 10 |
| somatic cell 20 total chromosomes how many haploids | 10 |
| somatic cell 20 chromosomes what is the diploid number | 20 |
| double chromosomes are | dyads |
| an ovary poduces cells that are | haploid |
| what is the degree of freedom | n-1 |
| heterozygous | 1 dominant allele 1 recessive allele Bb |
| homozygous | 2 recessive alleles or 2 dominat alleles BB bb |
| phenotype | physical appearance |
| genotype | alleles present represented with letters |
| if the X2 value in chi square is greater than the chart | reject hypothesis |
| if the X2 value in Chi square is less that the table value | accept hypothesis |
| in chi square if the observed no. were equal the X2 value would be | 0 |
| chance of getting a white albino plant | 25% |
| the albino allele is protected | in the heterozygous |
| what law of probabilty is used in the chi square | 3rd product rule |
| f not colorblind but your father is your geneotype would be | XBxb |
| you are F your mother is color blind what is your genotype | XBxb |
| you are female no one in your family is colorblind what is your genotype | XBXB |
| pair of chromosomes that is not homozygous in a normal karyotype | male 23 sex chromosome |
| name a clue for an X linked trait | primarily found in males |
| name a clue for an X linked trait | passes from grandfather to grandson |
| an organisms fitness is based on | reproductive success |
| if a recessive allele is lethal how long will it take to completely dissappear | never always protected in the heterozygous |
| how long would it take for a dominant allele to dissappear | immediately |
| a dominant allele is always | expressed |
| In hardy weinburg P2 = | % of homozygous dominat individuals |
| in hardy weinburg P = | frequency of dominant alleles |
| in hardy weinburg Q2 = | % of homozygous recessive individuals |
| in hardy weinburg Q= | frequency of recessive alleles |
| in hardy weinburg 2 pq= | % of heterozygous individuals |
| all members of a species living in one locale | population |
| conditions that cause change in a gene pool | mutations, gene flow, nonrandom mating, genetic drift, natural selection |
| genetic drift is due to | chance alone |
| hardy weinberg equation is a way to calculate | genepool frequencies |
| conditions for hardy weinberg law to be valid | no mutations, no gene flow, no random mating, no genetic drift, and no natural selection |
| difference between genetic drift and natural selection | genetic drift- chance, natural selection some are better adapted more than others |
| how the end result of genetic drift differ from natural selection | populations become different- natural selection becomes the same |
| XO | Turner Syndrome |
| XXX | Triplo X |
| XXY | Kleinfelters |
| XYY | Jacobs |
| syndrome where the father is the problem in meosis II | Jacobs |
| chromosomes in Turners | 45 |
| chromosomes in triplo x | 47 |
| chromosomes in kleinfekters | 47 |
| cromosomes in jacobs | 47 |
