2.10

Question 1

Cystic fibrosis is an example of

Answer 1

monohybrid inheritance

Question 2

When 2 heterozygous individuals interbreed, we expect a ratio in the offspring to be

Answer 2

3:1
75% of the offspring should show phenotype determined by dominant allele
25% phenotype of recessive

Question 3

Mendelian ratio is the

Answer 3

expected ratio of phenotypes resulting from a genetic cross

Question 4

We don’t always obtain results that fit the Mendelian ratio exactly, this is because the ratio only tells you the

Answer 4

chance of an individual with a certain phenotype being produced from a particular cross

Question 5

2 parents who are heterozygous for CF have a 25% chance of having a child with CF at any

Answer 5

pregnancy

Question 6

In human families, because the numbers of children are fairly small, we do not usually find an exact

Answer 6

3:1 ratio in the offspring

Question 7

(TESTING RESULTS) Sometimes large numbers of offspring are produced in genetic crosses. This is more likely to happen in genetic crosses using animals rather than humans and plants. They may find that the ratio of phenotypes in the offspring does not fit the predicted ratio exactly, however, this could simply be the result of

Answer 7

chance

Question 8

(TESTING RESULTS) A statistical test to use is called the

Answer 8

X2 test/chi-squared test

Question 9

(TESTING RESULTS) To check whether resistance, for example, is caused by a dominant allele, and that the offspring were produced in a 3:1 ratio, we can check this by using the

Answer 9

chi-squared test

Question 10

chi-squared step 1:

Answer 10

null hypothesis - is always that ‘there is no difference between observed and expected numbers’

Question 11

chi-squared step 2:

Answer 11

expected numbers must be worked out, e.g. 27+13 =40, 3:1 ratio = expect 30 resistant and 10 non-resistant
observed and expected results are put in a table

Question 12

chi-squared step 3:

Answer 12

work out value using chi-squared formula:

X2= sum of (O-E)2/E

Question 13

(chi-squared table) For each genotype, we find the difference between the observed and expected numbers (O-E) and write them in the

Answer 13

Correct column (3rd)

Question 14

(chi-squared table) In the next column (4th) we square the figure we obtained from the previous column giving

Answer 14

(O-E)2

Question 15

(chi-squared table) Next column (5th) we divide (O-E)2 by

Answer 15

E

Question 16

(chi-squared table) By adding all the numbers in the last column (O-E)2/E we get the

Answer 16

value for x2

Question 17

chi-squared step 4:

Answer 17

look up value of x2 on a table

Question 18

values table (step 4) has a column labelled

Answer 18

degrees of freedom

Question 19

degrees of freedom is always one less than the number of categories you have in your

Answer 19

data

Question 20

the rows of x2 values have numbers at the top have numbers at the top called

Answer 20

probability (p)

Question 21

probability row tells you the probability of getting this value of x2 by

Answer 21

chance alone

Question 22

accept null hypothesis if our x2 value gives a probability (p) of

Answer 22

0.05 or higher

Question 23

reject null hypothesis if p is

Answer 23

less than 0.05

Question 24

reject null hypothesis then the observed and expected results are

Answer 24

significantly different