inheritance

Question 1

State three causes of genetic variation

Answer 1

Mutation
Crossing over
Independent segregation / assortment (of homologous chromosomes)
Random fusion of gametes / fertilisation / mating

Question 2

What is meant by a genome?

Answer 2

(All) the DNA in a cell/organism;
‘(all) the ‘genes’/alleles’ ‘genetic material/code’ in a cell/organism/ person’
‘the total number of DNA bases in a cell/organism

Question 3

In genetic crosses, the observed phenotypic ratios obtained in the offspring are often not the same as the expected ratios.

Suggest two reasons why.

Answer 3

Small sample size;
Fusion/fertilisation of gametes is random;
Linked Genes; Sex-linkage / crossing over;
Epistasis;
Lethal genotypes;

Question 4

How do multiple alleles of a gene arise?

Answer 4

mutations;
which are different / at different positions in the gene;

Question 5

What is a gene pool?

Answer 5

All the alleles in a population;

Question 6

What does Hardy Weinberg’s equation predict

Answer 6

The frequency/proportion of alleles (of a particular gene);
Will stay constant from one generation to the next/over generations / no genetic change over time;
Providing no mutation/no selection/population large/population genetically isolated/mating at random/no migration;

Question 7

What is meant by a recessive allele?

Answer 7

Only expressed in the homozygote / not expressed in the heterozygote / not expressed if dominant present;1

Question 8

Define gene linkage

Answer 8

(Genes/loci) on same chromosome;

Question 9

Define codominance

Answer 9

Both alleles expressed in the phenotype;

Question 10

Describe why observed phenotypes don’t match expected values

Answer 10

Fertilisation is random
OR
Fusion of gametes is random;
Small/not-large population/sample;
Selection advantage/disadvantage/lethal alleles;

Question 11

Define epistasis

Answer 11

The allele of one gene affects or masks the expression of another in the phenotype;

Question 12

Expected offspring phenotype ratios from heterozygous parents:
Monohybrid
Dihybrid
Epistasis
Autosomal linkage

Answer 12

Dominant : recessive

Monohybrid 3:1
Dihybrid 9:3:3:1
Epistasis 9:4:3 or 15:1 or 9:7
Autosomal linkage 3:1 (no x over) (no other pattern other than 4 phenotypes with recombination of alleles)

Question 13

Male offspring are more likely than females to show recessive sex-linked characteristics. Explain why.

Answer 13

(Recessive) allele is always expressed in males / males have one (recessive) allele;
Females need two recessive alleles / females need to be homozygous recessive / females could have dominant and recessive alleles / be heterozygous;

Question 14

Rules for recessive alleles

Answer 14

Unaffected parents can have an affected offspring (if they are Heterozygous)

Question 15

Rules for Dominant alleles

Answer 15

Affected offspring MUST have at least one affected parent.
Unaffected parents ONLY have unaffected offspring.
If both parents are affected and have an unaffected offspring, both parents must be Heterozygous

Question 16

Genotype:

Answer 16

The genetic constitution of an organism (All the alleles that an organism

Question 17

Explain how a single base substitution causes a change in the structure of a polypeptide

Answer 17

Change in (sequence of) amino acid(s)/primary structure;
Change in hydrogen/ionic/disulfide bonds;
Alters tertiary/30 structure;

Question 18

What is meant by the term phenotype (

Answer 18

Expression / appearance / characteristic due to) genetic constitution / genotype / allele(s);

(Expression / appearance / characteristic due to) interaction with environment;

Question 19

2 equations in hardy weinburg

Answer 19

(alleles);P2 (homozygous dominant ) + 2Pq + q2 (homozygous recessive)= 1.0
P + q = 1.0 (genes)

Question 20

In human populations, cystic fibrosis is caused by a recessive allele.

People who are unaffected maybe homozygous dominant (FF) or heterozygous (Ff) but anyone with the disease must be homozygous recessive (ff)

In a population of 10 000 people, 5 people had the disease and the rest did not. How many of these people were heterozygotes?

Answer 20

The frequency of the disease is 5 in 10 000 = = 0.0005

i.e. the frequency of homozygous recessive individuals = q2 = 0.0005

Therefore the frequency  of the recessive allele 	=   q	=  

Therefore the frequency of the dominant allele 	= P	=  1 – q
							=  1 – 0.22	= 0.978 Therefore the frequency of heterozygous individuals  (2Pq)	
							= 2 x 0.978 x 0.022
								
						2Pq= 0.044

Therefore the number of heterozygotes			 = 10 000 x 0.044   = 440

Question 21

assumptions made in hardy weinburg

Answer 21

⦁ Migration, i.e. the gene pool is isolated and there is no flow of other alleles in or out
⦁ Gene mutations
⦁ Selection for or against a particular allele

IN ADDITION:

⦁ There should be a large population
⦁ Mating within the population should be random (all individuals have an equal chance of mating)

Question 22

A student investigated the monohybrid inheritance of eye shape in fruit flies. Two fruit flies with bar (narrow) eyes were crossed. Of the offspring, 1538 had bar eyes and 462 had round (normal) eyes.

(a) Using suitable symbols, give the genotypes of the parents. (2)

b)The ratio of bar-eyed flies and round-eyed flies in the student’s results were not the same as the ratio she had expected.(1)

c)What ratio of bar-eyed to round-eyed flies was the student expecting? (2)
d)The student wished to test her results with the ones she had expected.

Which statistical test should she use

Answer 22

a)Bb / suitable equivalent;
Both parents have bar eyes, but have some offspring with round eyes, so parents must be carriers of recessive allele for round eyes;

b )3:1;

(c) Fertilisation is random
OR
Fusion of gametes is random;
Small / not large population / sample;
Selection advantage / disadvantage / lethal alleles;
d)χ2 / chi squared;

Question 23

This fruit fly has another characteristic controlled by a pair of codominant alleles, WN and WV.

What is meant by codominant alleles?

Answer 23

Both alleles expressed in the phenotype (if both are present);

Question 24

In another population of 950 fruit flies, the frequency of the WV allele was 0.2.

Use the Hardy-Weinberg equation to calculate the number of insects that would be expected to have the genotype WNWV.

Answer 24

304

Question 25

There were 850 fruit flies in one population. In this population, 510 fruit flies had the genotype WNWN, 255 had the genotype WNWV and 85 had the genotype WVWV.

Calculate the actual frequency of the allele WV. Do not use the Hardy-Weinberg equation in your calculation

Answer 25

0.25;

Question 26

There were 500 insects in the total population. In this population, 300 insects had the genotype CM CM, 150 had the genotype CM CN and 50 had the genotype CN CN. Calculate the actual frequency of the allele CN by using these figures. Show your working.
b)Use your answer to (ii) and the Hardy-Weinberg equation to calculate the number of insects that would be expected to have the genotype CN CN.

Answer 26

a)0.25 / 25%; = 2 marks CN = 250 / 1000; = 1 mark
P2 = (0.25)2 / 0.0625 / square of calculated figure for CN; = 2 marks

p2 +2pq + q2 = 1.0; = 1 mark

= 31.25 / 31;

[Accept: Derived from either p2 or q2]

Question 27

The inheritance of body colour in fruit flies was investigated. Two fruit flies with grey bodies were crossed. Of the offspring, 152 had grey bodies and 48 had black bodies.

(a) Using suitable symbols, give the genotypes of the parents. Explain your answer.

Answer 27

Gg / suitable equivalent;

Grey : black about 3: 1;

[Note: Can be in table / diagram]

Question 28

The shells of this snail may be unbanded or banded. The absence or presence of bands is controlled by a single gene with two alleles. The allele for unbanded, B, is dominant to the allele for banded, b.

A population of snails contained 51% unbanded snails. Use the Hardy-Weinberg equation to calculate the percentage of this population that you would expect to be heterozygous for this gene. Show your working.

Answer 28

Correct answer of 42% = 3 marks

Answer of 0.42 = 2 marks

Award one mark maximum for answer of

49.9 / 49.98 / 50% or 0.49 / 0.5

2. q2 = 0.49 / 49% OR q = 0.7 / 70%

Award one mark maximum for answer of 40.8 / 41% or 0.41

3. Shows understanding that 2pq = heterozygotes / carriers / shows answer is derived from 2pq;

Accept: b2 = 0.49 / 49% or b = 0.7 / 70% for mark point 2

Question 29

A breeder crossed a black male cat with a black female cat on a number of occasions. The female cat produced 8 black kittens and 4 white kittens.

(a) (i) Explain the evidence that the allele for white fur is recessive.
b)Predict the likely ratio of colours of kittens born to a cross between this black male and a white female.

Answer 29

Parents are heterozygous;

Accept carriers / carries white allele
Kittens receive white allele from parents / black cat;
b)1:1;

Answer must be expressed as a ratio that could be reduced to 1 : 1

Question 30

The fruit fly is a useful organism for studying genetic crosses. Female fruit flies are approximately 2.5 mm long. Males are smaller and possess a distinct black patch on their bodies. Females lay up to 400 eggs which develop into adults in 7 to 14 days. Fruit flies will survive and breed in small flasks containing a simple nutrient medium consisting mainly of sugars.

(a) Use this information to explain two reasons why the fruit fly is a useful organism for studying genetic crosses

Answer 30

Large number of eggs / offspring / flies (therefore) improves reliability / can use statistical tests / are representative / large sample (size) / reduces sampling error;

Each mark point requires a feature linked in mark scheme (by therefore) to an explanation

Do not accept a large number of eggs produces a large number of flies unless the term sample is used

Ignore references to accuracy or precision

2. Small size / (breed) in small flasks / simple nutrient medium (therefore) reduces costs / easily kept / stored;

Accept small size so can be kept in small flasks

3. Size / markings / phenotypes (therefore) males / females easy to identify;

Answers must relate to size, markings or use the term phenotype

4. Short generation time / 7 - 14 days / develop quickly / reproduce quickly (therefore) results obtained quickly / saves times / many generations;

Question 31

Polydactyly in cats is an inherited condition in which cats have extra toes. The allele for polydactyly is dominant.

(i) In a population, 19% of cats had extra toes. Use the Hardy-Weinberg equation to calculate the frequency of the recessive allele for this gene in this population. Show your working.
b)Some cat breeders select for polydactyly. Describe how this would affect the frequencies of the homozygous genotypes for this gene in their breeding populations over time.

Answer 31

Correct answer of 0.9 = 2 marks;

Incorrect answer but shows q2 = 0.81 = one mark.

Note: 0.9% = one mark
b)Homozygous dominant increases and homozygous recessive decreases.

Question 32

Male fruit flies have the sex chromosomes XY and the females have XX. In the fruit fly, a gene for eye colour is carried on the X chromosome. The allele for red eyes, R, is dominant to the allele for white eyes, r. The genetic diagram shows a cross between two fruit flies.
Male fruit flies are more likely than female fruit flies to show a phenotype produced by a recessive allele carried on the X chromosome. Explain why.

Answer 32

Males have one allele;

Answers should be in context of alleles rather than chromosomes

2. Females need two recessive alleles / must be homozygous recessive / could have dominant and recessive alleles / could be heterozygous / carriers

Question 33

Mutation is one cause of genetic variation in organisms.

Give two other causes of genetic variation.

Answer 33

Crossing over;

2. Independent segregation/assortment (of homologous chromosomes);

Accept independent assortment of alleles.

Accept meiosis as an alternative for 1 or 2 if neither of these marks is awarded.

3. Random fusion of gametes

OR

Random fertilisation;

Accept random mating.

Question 34

In birds, males are XX and females are XY.

(a) Use this information to explain why recessive, sex-linked characteristics are more common in female birds than in male birds.

Answer 34

(Recessive) allele is always expressed in females / females have one (recessive) allele / males need two recessive alleles / males need to be homozygous recessive / males could have dominant and recessive alleles / be heterozygous / carriers;

Accept: Y chromosome does not carry a dominant allele. Other answers must be in context of allele not chromosome or gene.

Question 35

In genetic crosses, the observed phenotypic ratios obtained in the offspring are often not the same as the expected ratios.

Suggest two reasons wh

Answer 35

Small sample size;

2. Fusion/fertilisation of gametes is random;

Ignore breeding is rando
Genes;
Accept crossing over / sex linkage

4. Epistasis;
5. Lethal genotypes

Question 36

Feather colour in one species of chicken is controlled by a pair of codominant alleles which are not sex-linked. The allele CB codes for black feathers and the allele CW codes for white feathers. Heterozygous chickens are blue-feathered.

On a farm, 4% of the chickens were black-feathered. Use the Hardy-Weinberg equation to calculate the percentage of this population that you would expect to be blue-feathered. Show your working

Answer 36

Correct answer of 32 (%) = 3 marks;;;

Accept: 0.32 = 2 marks

If incorrect answer, allow following points

1. p2 / q2 = 4% / 0.04 / or p / q = 0.2;
2. Shows understanding that 2pq = heterozygotes / carriers;

Accept: answer provided attempts to calculate 2pq. This can be shown mathematically i.e. 2 x two different numbers.

Question 37

Meiosis results in cells that have the haploid number of chromosomes and show genetic variation. Explain how.

Answer 37

Homologous chromosomes pair up / bivalents form;

2. Crossing over / chiasmata form;
3. Produces new combination of alleles;
4. Chromosomes separate;
5. At random;
6. Produces varying combinations of chromosomes / genes / alleles (not twice) ;
7. Chromatids separated at meiosis II / later;

Independent assortment / random segregation = marking points 4 and 5