Question 1

Which process releases heat to maintain a constant body temperature?

Accepted Answer

Thermoregulation is a mechanism by which mammals maintain body temperature with tightly controlled self-regulation independent of external temperatures. Temperature regulation is a type of homeostasis and a means of preserving a stable internal temperature in order to survive. In mammals, one way the body dissipates excess heat to maintain homeostasis is through exhalation. Air that enters the lungs is warmed by body heat and then exhaled. So, the correct answer is respiration.

Question 2

Three parts of a chromosome and their functions are listed.Which part is matched with its correct function?

Accepted Answer

Each chromosome has two short arms called p arms and two long arms called q arms. These arms are held together at the center by the centromere. The tips of the chromosome are capped by sections of DNA called telomeres. Telomeres prevents loss of genes and protect chromosomes during DNA replication. Histones bind to DNA, help give chromosomes their shape, they function as spools for the thread-like DNA to wrap around.

Question 3

Rifampicin is an antibiotic used to treat tuberculosis (TB). It works by inhibiting RNA polymerase in bacteria.Which processes are directly inhibited by this antibiotic?1 DNA replication2 transcription3 ATP synthesis

Accepted Answer

Rifampin specifically inhibits bacterial RNA polymerase, the enzyme responsible for DNA transcription, by forming a stable drug-enzyme complex.

Question 4

The hormone glucagon is an example of a cell signaling molecule. The table below lists the main events that occur when the blood glucose concentration decreases below the set point.The events are not listed in the correct order.Complete Table to show the correct order in which these events occur.Three of the events have already been placed in their correct order.

Accepted Answer

J and F in correct position ; A and G in correct position ; B and H in correct position ; C and I in correct position ;

Question 5

During an immune response, only B-lymphocytes with receptors that are specific to the antigens present are activated. Activation occurs when an antigen binds to a receptor of a B-lymphocyte. Activated B-lymphocytes grow in size and then divide by mitosis. Many further mitotic cell divisions occur, increasing the number of B-lymphocytes with receptors specific to the antigen. Eventually, cells produced in this process will develop into either plasma cells that secrete antibodies or memory B-cells.Fig. given below is a summary of B-lymphocyte activation and the events that follow:The development of plasma cells and memory B-cells in this process depends on transcription factors.Explain the role of transcription factors in gene expression in eukaryotic cells.

Accepted Answer

The role of transcription factors in gene expression in eukaryotic cells:1. binds to DNA 2. allows RNA polymerase / other transcription factors, to bind (to, DNA / gene / promoter) .3. regulates / initiates / inhibits, gene expression / transcription.

Question 6

During an immune response, only B-lymphocytes with receptors that are specific to the antigens present are activated. Activation occurs when an antigen binds to a receptor of a B-lymphocyte. Activated B-lymphocytes grow in size and then divide by mitosis. Many further mitotic cell divisions occur, increasing the number of B-lymphocytes with receptors specific to the antigen. Eventually, cells produced in this process will develop into either plasma cells that secrete antibodies or memory B-cells.Fig. given below is a summary of B-lymphocyte activation and the events that follow:The development of plasma cells and memory B-cells in this process depends on transcription factors.Transcription factors are proteins. Genes that code for proteins can become mutated. Describe how different types of gene mutation can cause changes in the protein that is synthesised.

Accepted Answer

1 base substitution / mis-sense mutation 2 changes in triplet/codon. 3 base deletion / insertion.4 results in a frameshift mutation.5 change in protein primary structure and amino acid sequence.6 protein folds incorrectly and changes tertiary or 3-D structure.7 changes protein function, prevents protein function and makes protein unstable.

Question 7

During an immune response, only B-lymphocytes with receptors that are specific to the antigens present are activated. Activation occurs when an antigen binds to a receptor of a B-lymphocyte. Activated B-lymphocytes grow in size and then divide by mitosis. Many further mitotic cell divisions occur, increasing the number of B-lymphocytes with receptors specific to the antigen. Eventually, cells produced in this process will develop into either plasma cells that secrete antibodies or memory B-cells.Fig. given below is a summary of B-lymphocyte activation and the events that follow:BLIMP-1 is a transcription factor that is essential for the development of plasma cells and memory B-cells in the process shown in the above figure. BLIMP-1 reduces the synthesis of c-Myc in B-lymphocytes. c-Myc is a protein that is required for the mitotic cell cycle to continue.Suggest and explain how a mutation in the gene coding for BLIMP-1 can prevent the development of plasma cells and memory B-cells.

Accepted Answer

Mutation in the gene coding for BLIMP-1 can prevent the development of plasma cells and memory B-cells because this results in:1 BLIMP-1 is not synthesised and is non-functional.2 expression of the c-Myc gene is not reduced / continues.3 synthesis and concentration of c-Myc (protein) is maintained / increases.4 B-lymphocytes continue to divide by mitosis / proliferate.

Question 8

Microarrays can be used to analyse the effect of transcription factors, such as BLIMP-1, on gene expression.Describe how a microarray is used in the study of gene expression.

Accepted Answer

DNA microarrays can simultaneously measure the expression level of thousands of genes within a particular mRNA sample.∙ First, mRNA is extracted from cells of interest.∙ After the extraction, mRNA is used as a template strand to synthesise cDNA.∙ cDNA is tagged with a fluorescent dye.∙ fluorescent tagged cDNA binds to hybridize with the probe.∙ Each probe is unique to a different gene. They bind to detect the presence of complementary nucleic acid sequences (target sequences) by hybridization.∙ Fluorescence indicates gene is expressed.

Question 9

BCL6 is another transcription factor found in B-lymphocytes. The effect of BCL6 on gene expression was compared in two samples of B-lymphocytes.- Sample 1 consisted of B-lymphocytes that were producing BCL6.- Sample 2 consisted of B-lymphocytes that were not producing BCL6.Suggest why a microarray is suitable for identifying the function of the transcription factor BCL6 in these two samples.

Accepted Answer

A microarray is suitable for identifying the function of the transcription factor BCL6 in these two samples because:∙ BCL6 is a transcription factor that regulates the expression of genes.∙ Microarray can detect the expression of genes. DNA microarrays can simultaneously measure the expression level of thousands of genes within a particular mRNA sample.∙ Fluorescence-based gene expression, between the two samples, helps in the identification.∙ Difference in gene expression indicates regulation by BCL6.

Question 10

Mexican spadefoot toads, Spea multiplicata, live on land but return to ponds to breed. Eggs are laid in water and hatch into tadpoles, which feed in ponds before developing into adults.The tadpoles can be classified into two main types: omnivore-type tadpoles and carnivore-type tadpoles. Differences between the phenotypes of these two types of tadpole are related to their different feeding behaviours.∙ Omnivore-type tadpoles feed on tiny pieces of detritus (dead material from plants and animals) and algae (microscopic photosynthetic organisms) at the bottom of ponds. These tadpoles grow slowly.∙ Carnivore-type tadpoles feed on small animals in the water, such as fairy shrimp and small omnivore tadpoles. These tadpoles grow quickly.Fig. (a) shows two tadpoles of the same age, one of each type. A fairy shrimp is also shown. All three organisms are at the same distance from the camera.Between these two main types of tadpole there is a continuous range of tadpoles with intermediate body phenotypes and feeding behaviours.For any individual tadpole, regardless of age, it is possible to calculate a phenotype score depending on the features of the tadpole. A tadpole with a phenotype score close to 3 is a typical omnivore type and a tadpole with a phenotype score close to 7 is a typical carnivore type.The phenotype scores were determined for a large number of tadpoles sampled from two ponds. The availability of detritus and algae was high for one pond and low for the other pond. All other conditions in the two ponds were similar.For both ponds, phenotype scores were determined shortly after the tadpoles had hatched from eggs and ten days later.The results are shown in Fig. (b).Using the results shown in Fig. (b), describe and suggest an explanation for the change in frequency of tadpole phenotypes in the pond with low availability of detritus and algae.

Accepted Answer

∙ Frequency of omnivore types and carnivores types phenotype remains high even after 10 days in pond with high availability of detritus and algae.∙ Frequency of intermediate phenotypes decreases after 10 days  in pond with low availability of detritus and algae.∙ Disruptive selection, also called diversifying selection, describes changes in population genetics in which extreme values for a trait are favored over intermediate values. In this case, the variance of the trait increases and the population is divided into two distinct groups.∙ Selection pressure is food availability.∙ Omnivore-type and carnivore-type  extreme phenotypes are more likely to survive and have a selective advantage or intermediate phenotypes, are less likely to survive and selected against.∙ Omnivore-type and carnivore-type  extreme phenotypes are more likely to survive because they are good at accessing available food and intermediate phenotypes, are less likely to survive and selected against because they are poor at accessing available food.omnivore-type = phenotype score 3carnivore-type = phenotype score 7

Question 11

Eleven days after eggs had hatched, the mean body masses of omnivore-type tadpoles and carnivore-type tadpoles in the pond with low availability of detritus and algae were measured. The results are shown in the figure given below:Using the data in the Figure above, calculate the mean body mass of an omnivore-type tadpole as a percentage of the mean body mass of a carnivore-type tadpole.

Accepted Answer

Mean body mass of omnivore =0.58Mean body mass of carnivore =1.1The mean body mass of an omnivore-type tadpole as a percentage of the mean body mass of a carnivore-type tadpole =1.10.58​×100=52.7≈53%

Question 12

Mexican spadefoot toad tadpoles develop into adult toads that do not live in water. In some years, the ponds where Mexican spadefoot toad tadpoles live, dry out quickly. Suggest why the carnivore-type tadpoles have a selective advantage in the years when ponds dry out quickly.

Accepted Answer

Carnivore-type tadpoles have a selective advantage in the years when ponds dry out quickly because they grow quicker so develop into adults faster and in a shorter period of time.

Question 13

Genetic modifications in crops can provide resistance to insect pests. Bt maize is one type of GM crop that has been produced to be resistant to insect pests. State two benefits to farmers of insect resistance in crops.

Accepted Answer

Benefits to farmers of insect resistance in crops.∙ This will help maintain the soil fertility for a longer period of time, as harmful chemicals present in insecticides will not be added to the soil.∙ It will also reduce the cost of crop production by removing the expenditure of buying insecticides.∙ Helps in increasing the yield of the crop.

Question 14

Describe the difference between Bt maize and non-GM maize that explains why Bt maize is resistant to insects.

Accepted Answer

Bt maize is resistant to insects because:∙  Contains gene from Bacillus thuringiensis. Bacillus thuringiensis is a species of bacteria that produces proteins that are toxic to certain insects.∙ It produces Bt toxin that is harmful to insects. This leads to pore formation in the insect's intestine and death. A "lock and key" analogy is useful to explain this specificity. If the midgut receptor is considered the "lock" and the Cry protein is the "key" then insect death only occurs when the "lock and key" match.

Question 15

State two reasons why people may have objections to the growth of insect-resistant GM crops.

Accepted Answer

Some people may have objections to the growth of insect-resistant GM crops because:∙ Resistance to insects may be transferred to wild plants.∙ contamination of food marketed as organic.∙ Use of this may kill useful insects or pollinators.∙ decrease in biodiversity.∙ potential health risks of humans after eating GM crops.∙ insects may become resistant to toxin.

Question 16

The figure given below shows the area of land that was used to grow genetically modified (GM) crops in the USA, Brazil, India and China from 2004 to 2015.In 2015, the percentage of the USA population that objected to the use of GM crops was lower than in 2005. Suggest why a smaller percentage of the USA population objected to the use of GM crops in 2015.

Accepted Answer

A smaller percentage of the USA population objected to the use of GM crops in 2015 because of:- education and awareness increase with time.- consumption of GM foods shows no ill effects that's why people didn't objected to the use of GM crops in 2015.

Question 17

Some plants, such as rice, grow with their roots submerged in water.Figure given below shows a group of rice plants.Explain how rice is adapted to grow with its roots submerged in water.

Accepted Answer

∙ Rice stems contain a large number of air spaces (hollow aerenchyma) running the length of the stem and into the roots. This allows oxygen (some formed in the plant from photosynthesis) to penetrate through to the roots which are submerged in water, supplying oxygen for aerobic respiration. This decreases the chance that anaerobic respiration will occur.∙ Some leaves are trap air underwater due to ridges on leaves.∙ Ethanol produced from respiration under anaerobic conditions. Root cells of can tolerate ethanol and produces dehydrogenase to break down ethanol.∙ Also, there are some varieties of rice have high rate of respiration under anaerobic conditions to generate more ATP.

Question 18

Explain what is meant by the term biodiversity

Accepted Answer

The term “biodiversity” refers to the variety of living organisms. Biodiversity brings together the different species and forms of life (animal, plant, entomological and other) and their variability, that is to say, their dynamics of evolution in their ecosystems.Biodiversity refers to:'∙ variation in ecosystems or habitats.∙ number/variety of different species.∙ relative abundance of each species.∙ genetic variations within each species.

Question 19

Describe the sliding filament model of muscular contraction.

Accepted Answer

The sliding filament theory explains the process of muscle contraction during which the thin filaments slide over the thick filaments, which shortens the myofibril. Each muscle fibre has an alternate light and dark band, which contains a special contractile protein, called actin and myosin respectively. Actin is a thin contractile protein present in the light band and is known as the I-band, whereas myosin is a thick contractile protein present in the dark band and is known as the A-band. There is an elastic fibre called z line that bisects each I-band. The thin filament is firmly anchored to the z-line. The central part of the thick filament that is not overlapped by the thin filament is known as the H-zone. During muscle contraction, the myosin heads or cross bridges come in close contact with the thin filaments. As a result, the thin filaments are pulled towards the middle of the sarcomere. The Z line attached to the actin filaments is also pulled leading to the shortening of the sarcomere. Hence, the length of the band remains constant as its original length and the I-band shortens and the H-zone disappears.∙ The sliding filament contraction occurs in the sarcomere region.∙ The myosin filaments ratchet over actin filaments contracting the sarcomere.∙ The I and H bands within the sarcomere compress and expand to facilitate the movement.∙ The myofilaments do not expand and contract on their own.∙ ATP releases myosin from the actin filaments. During contraction, myosin attaches to the actin filaments. ATP attaches to the myosin head and releases it from the actin molecule, thereby, causing muscle relaxation.∙ Troponin attaches to the protein tropomyosin and lies between actin filaments. Tropomyosin blocks the attachment site for myosin head and prevents contraction in a relaxed muscle.∙ A cross-bridge Is the attachment of myosin to actin in a muscle cell. All types of muscles contract by cross-bridge cycling. This means repeated attachment of actin and myosin within the cell.

Question 20

Explain the roles of the hormones FSH, LH, oestrogen and progesterone in the human menstrual cycle.

Accepted Answer

Role of FSH:∙ It initiates follicular growth.∙ It controls the menstrual cycle.∙ FSH secreted by anterior pituitary.∙ It stimulates the ovary to produce an egg.∙ It regulates the development, growth, and development of secondary sexual characteristics and reproductive processes.∙ This hormone stimulates the growth of ovarian follicles in the ovary before the release of an egg from one follicle at ovulation.Role of LH:∙ LH stimulates ovulation and the release of oocyte.∙ LH stimulates the development of corpus luteum.∙ LH stimulates estrogen and progesterone production from the ovary.∙ It subsequently stimulates the corpus luteum to produce progesterone.Role of oestrogen:∙ oestrogen stimulates repair of endometrium.∙ oestrogen inhibits furthur release of FSH.∙ oestrogen maintains the lining of the uterus during the second half of the menstrual cycle.Role of progesterone:∙ corpus luteum secretes progesterone.∙ progesterone continues build-up of the endometrium or maintains endometrium.∙ progesterone inhibits secretion of LH / FSH.∙ corpus luteum degenerates so concentration of progesterone falls.

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