Paper 2 Flashcards
Wong length in fruit flies is controlled by two alleles
Vestigial( short) wings are caused by the allele n
Normal length wings are caused by the allele N
Vestigial wing allele is recessive to the normal wing allele
What does it mean it allele is recessive?
Two copies of the allele to be present for the characteristic to be displayed.
Two fruit flies with Normal length wings are crossed. Complete the Pinnett in figure 1 to show this cross.
Two fruit flies with Normal length wings are crossed. Complete the Pinnett in figure 1 to show this cross.
The two fruit flies crossed in figure 1 have 200 offspring’s. Compare how many of the offspring would be expected to have vestigial wings?
Proportion of offspring expected to have vestigial wings =
25%/0.25/1 in 4
So out of 200 fruit flies, the expected number of offspring with vestigial wings = (200 ÷ 100) × 25 / 200 × 0.25 / 200 ÷ 4 = 50 [2 marks for the correct answer, otherwise 1 mark for the correct proportion of offspring expected to have vestigial wings)
Remember, the allele for vestigial wings is recessive, so only offspring with the genotype ‘nn’ will have vestigial wings.
Fruit flies are eukaryotic organisms. Describe how the genetic material of fruit fly is stored in its cells.
Genetic material is stored as DNA molecules which are arranged into chromosomes. The chromosomes are found in the nucleus of each cell.
Quadrats can be used to investigate the distribution of organisms in a habitat.
A group of students used 1 m? quadrats to investigate the plant species that live in a small field next to their school.
They placed three quadrats at random in each section of the field and counted the number of buttercups, clover and dandelions in each quadrat.
Their results are shown in Table 1.
Explain why the quadrats were placed randomly?
To make sure the result were representative of the whole population in each section.
To prevent bias
What is the median number of buttercups per m2 in section A
52
What is the mean number of of clover per m2 in section B. Give you answer in 2 significant figures.
90.333
90
Section C measures 5m by 3m
Use the data in table 1 to estimate the total population of dandelions in section C
Area of section C=5x3=15m2
12x15=180
All you have to do is to multiply the number of dandelions found in 1m2 by the total area of section C
Use the data in table 1 to give a conclusion about the distribution of the three plant species across the field.
The number of plants per m2/ density of plants/ frequency of plants of all three species increases with the increasing distance from the school.
Give three examples of abiotic factors that may affect the number of plants of each species growing at different locations in the field.
Any three from: e.g. differences in temperature / differences in the mineral content of the soil / differences in the soil pH / differences in the light intensity received by the plants / differences in the moisture level of the soil / differences in the wind intensity/direction 3 marks — 1 mark for each correct answer|
Remember, abiotic factors are non-living parts of an ecosystem.
Students carried out an experiment to investigate the effect of age on reaction time.
This is the method they used.
1. They recruited 35 volunteers with ages between 20 and 60.
2. The volunteers were asked to complete a task on a computer. In the task they had to press the enter key every time a certain shape appeared on the screen.
Each volunteer completed the task 10 times and the mean reaction rate for each volunteer was calculated.
4. The students then divided the volunteers’ results into four age groups and calculated a mean reaction time for each group.
The results of the experiment are shown in Table 2.
The student’s experiment tested a nervous response in the volunteers.
The effector in the response is muscle in the hand. Describe how muscles respond when they are stimulated by a nervous impulse.
They contract.
A motor neurone is involved in this nervous response. Describe the role of the motor neurones in the human body.
Motor neurones carry Electrical/ nervous impulses from the central nervous system to effectors.
Calculate the percentage increase in mean reaction time between the 20-29 age group and the 50-60 age group.
Percentage increase= final value-original value/original value X100
0.39-0.30/0.30X100=30%
Use data from Table 2 to draw a bar chart on Figure 3, showing the mean reaction time for each age group.
Complete the y-axis. Include a label and use a suitable scale.
Plot the mean reaction time data.