essay :o Flashcards
Bonding and bonds in organisms
- Carbohydrates- cellulose, starch and glycogen
- DNA
- Muscle contraction
- Proteins
- Lipids
The importance of the control of movement in cells and organisms.
- Mitosis
- Photosynthesis
- Muscle contraction
- Regulation of transcription and translation
- Control of blood glucose
The importance of diffusion in organisms
- transport across membranes
- DNA and protein synthesis
- photosynthesis
- muscle contraction
- control of blood glucose conc
The importance of proteins in the control of processes and responses in organisms.
- Enzymes as catalysts
- Proteins/enzymes in photosynthesis
- Control of movement across Membranes
- Haemoglobin
- Enzymes in gene transcription and Translation
The membranes of different types of cells are involved in many different functions
- Membrane function as selectively permeable barrier
- Photosynthesis, chloroplast, thylakoids
- Pacinian corpuscle
- Hormones - eg Blood glucose regulation – insulin and glucagon
- Absorption and co-transport of sodium ions and glucose
The importance of shapes fitting together in cells and organisms
- Enzyme properties and digestion
- Structure of DNA
- Transcription & translation
- Muscle contraction
- Haemoglobin
The movement of substances within living organisms
- DNA Transcription and translation
- Mitosis and Meiosis
- Muscle contraction
- Photosynthesis
- Blood glucose conc
chi squared
test to compare the pattern in data collected or observed with the pattern that would be expected by chance
commonly used to check results of genetic crosses
-calculate chi squared with equation
-the probability value can be found using the table of chi squared distribution
-work out the degrees of freedom, which is the number of categories minus 1
-the critical value found at p=5% for that degree of freedom
-if calculated value is less than critical value it gives a p value of >5% of the data being just chance. so the null hypothesis is accepted as there is no evidence of a significant difference between the observed and expected numbers
student t test
test used to judge the significance of any difference between the means of sets of data that are collected from two groups
- calculate value of t using formula
- calculate degrees of freedom, for an unpaired test this is (n1 + n2)-2 where n is numbers samples at each site
- t value looked up on table
- if calculate t value is less than critical value, there is a more than 5% probability that chance caused the difference. the null hypothesis is accepted and there is no significant difference between the sites
correlation coefficient
determines whether two variables correlate in any way
e.g. does rock pool algal diversity increase as pool surface area increases?
the variables compared need to be plotted on scatter graphs which will indicate possible relationships that can be tested
-degrees of freedom= the number of values for the two variables n-2
-if calculated value greater than critical value, gives p value of <0.001 or <0.1% of observed data being due only to chance. thus the correlation is 99.9% certain
Explain why the scientists sterilised the surfaces of the seeds and grew them in soil that had been heated to 85 oC for 2 days.
- To kill any fungus/bacteria on surface of seeds
or in soil; - So only the added fungus has any effect;
Explain why it was important that the soil contained no mineral ions useful to the plant
So that only nitrate or ammonia/type of fertiliser
affects growth;
The pea plants were divided into four groups, A, B, C and D.
Group A – heat-treated mycorrhizal fungus added, nitrate fertiliser
Group B – mycorrhizal fungus added, nitrate fertiliser
Group C – heat-treated mycorrhizal fungus added, ammonium fertiliser
Group D – mycorrhizal fungus added, ammonium fertiliser
The heat-treated fungus had been heated to 120 oC for 1 hour.
Explain how groups A and C act as controls
- So that effects of nitrate or ammonium alone
could be seen; - So that effects of fungus can be seen;
Suggest what the scientists should have done during the drying process to be sure that all of the water had been removed from the plant samples
- Weigh samples at intervals during drying;
- To see if weighings became constant (by 3
days) ;
What conclusions can be drawn from the data in Table 3 about the effects of the fungus on growth of the pea plants.
With live fungus – showing effects of the fungus:
1. Fungus increases growth of roots and shoots in
both;
2. Produces greater growth with nitrate;
What conclusions can be drawn from the data in Table 3 about the effects of nitrate fertiliser and ammonium fertiliser on growth of the pea plants
With heat-treated fungus – showing effects of
fertiliser:
1. Similar dry masses for roots and shoots;
2. (Probably) no significant difference because
SDs overlap;
Explain why determination of dry mass was an appropriate method to use in this investigation
- Dry mass measures/determines increase in
biological/organic material; - Water content varies;
Which treatment gave the best result in commercial terms? Justify your answer.
- Fungus with nitrate-containing fertiliser gave
largest shoot: root ratio; - And largest dry mass of shoot;
The student wanted to determine the rate of water loss per mm2 of surface area of the leaves of the shoot in Figure 5.
Outline a method she could have used to find this rate. You should assume that all water loss from the shoot is from the leaves.
- Method for measuring area;
eg draw round (each) leaf on graph paper and count squares; - Of both sides of (each) leaf;
- Divide rate (of water loss/uptake from
potometer) by (total) surface area (of leaves);
The rate of water movement through a shoot in a potometer may not be the same as the rate of water movement through the shoot of a whole plant.
Suggest one reason why
Plant has roots
OR
xylem cells very narrow;
Suggest two reasons why water molecules and carbon dioxide molecules can both pass through PIP1.
- Both small/similar size (so fit channel);
2. Have a similar shape (so bind to/fit channel);
The scientists first produced transgenic poplar trees. These trees all had a length of foreign DNA inserted into them. This DNA led to the production of singlestranded RNA that specifically inhibited expression of the gene for PIP1.
The scientists then measured the difference in the amount of PIP1 in leaves of transgenic poplars and in leaves of wild type poplars without the foreign DNA. The amount of PIP1 in the transgenic poplars was approximately 15% of that in the wild type poplars.
Using this information, what can you conclude about the effect of the foreign DNA in the transgenic poplar trees?
- Single-stranded RNA (has base sequence)
complementary to PIP1 mRNA; - Binds to mRNA (of PIP1)/leads to destruction
of mRNA; - Prevents/reduces translation (of PIP1);
- Reduces photosynthesis/named process that
uses water;
The transgenic poplars still produced some PIP1.
Suggest why
Not all of mRNA bound to single-stranded
RNA/there is more mRNA than interfering RNA
OR
Not all mRNA destroyed/disabled;
Using only Figure 7, evaluate the importance of PIP1 in the movement of carbon dioxide and water through leaves of poplar trees.
- Loss of PIP reduces water and carbon dioxide
movement; - Differences significant because SDs don’t
overlap
OR
Need stats test to see whether significant differences (or not); - Greater (proportional) effect on carbon dioxide
transport; - Not all movement through PIP;
The scientists used units of µg g−1 for the concentration of ammonia in soil.
Suggest why, in this investigation, the scientists used these units.
- (µg because) very little ammonia (in soil);
- (µg because) avoids use of (lots of) decimal
places (in their results) / avoids the use of powers of 10 / avoids the use of standard form; - (g-1) to allow comparisons (between samples);
The scientists concluded that the soil mixture experiment showed there were different communities of bacteria in soils A and B.
What evidence from Figure 6 supports their conclusions? Give reasons for your answer
- pH 4.3 / B has fastest rate of breakdown (of
ammonia); - A + B / mixture at pH 6.9 slowest / slower (than
A or B); - Suggests (community / bacteria at) pH 4.3 / B
doesn’t work (well) at pH 6.9 / pH of mixture;
This method allowed the scientists to estimate the expression of the amoA gene in each culture but not the growth of the bacterial population in each culture.
Explain why.
- They didn’t count bacteria / cells / population(s);
- Copies / number of mRNA related to amount of
enzyme / amoA produced / translated; - Don’t know how much mRNA / amoA produced
by each cell; - Don’t know if amoA (mRNA / enzyme) is linked
to cell division / growth (of population);
The scientists set up their cultures in sterile glass bottles.
Suggest one suitable method for sterilising the bottles and explain why it was necessary to sterilise them
- Suitable method;
eg in boiling water / steam / autoclave / wash in disinfectant / wash in alcohol - (Reason) to remove / kill other bacteria /
organisms that might break down ammonia
The importance of responses to changes in the internal and external environment of an organism.
Haemoglobin Blood glucose conc Pacinian corpuscle- AP caused by pressure Co-transport of glucose in ileum muscle contraction
Ions and organisms
Haemoglobin dissociation- Fe2+ photosynthesis- H+ Muscle contraction- Ca2+ Pacinian corpuscle- Na+ Co-transport of glucose
DNA and the transfer of information
DNA structure Cell division - Mitosis and meiosis Transcription and translation Mutation Genetic engineering
