In birds, males are XX and females are xy



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Q1.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.

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(1)


(b)     In chickens, a gene on the X chromosome controls the rate of feather production.
The allele for slow feather production, F, is dominant to the allele for rapid feather production, f. The following figure shows the results produced from crosses carried out by a farmer.

 

(i)      Explain one piece of evidence from the figure which shows that the allele for rapid feather production is recessive.

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(2)


(ii)     Give all the possible genotypes of the following chickens from the figure.

Chicken 5 ..............................................................................................

Chicken 7 ..............................................................................................

(2)

(iii)     A cross between two chickens produced four offspring. Two of these were males with rapid feather production and two were females with slow feather production. Give the genotypes of the parents.

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(1)

(c)     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 ...................................... %



(3)

(Total 9 marks)
Q2.Researchers investigated some characteristics of people from different parts of England. In the north of England they selected 200 people and recorded their phenotypes for three different characteristics.

Their results are shown in the figure below.

 

 

Phenotype produced
by dominant allele


Number of
people


Phenotype produced
by recessive allele


Number of
people


 

Tongue roller

131

Non-tongue roller

58

 

Right-handed

182

Left-handed

14

 

Straight thumb

142

Hitch-hiker thumb

50

(a)     Calculate the ratio of straight thumb to hitch-hiker thumb in this study.

 

 

 



 

Ratio = .......................................................................



(1)



(b)     The numbers for the tongue rolling and thumb characteristics do not add up to 200.
For each characteristic suggest one reason why the numbers do not add up to 200.

Tongue rolling ................................................................................................

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Thumb ............................................................................................................

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(2)



(c)     One student looked at the researchers’ results and concluded that 91% of people in the UK are right-handed.


Do you agree with this conclusion? Give reasons for your answer.

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(2)

(Total 5 marks)

Q3.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.

1......................................................................................................................

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2......................................................................................................................

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(2)




(b)     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.

(i)      Complete the genetic diagram for this cross.

 


 

Phenotypes of parents

red-eyed female

 

white-eyed male

 

Genotype of parents

..................................

×

..................................

 

Gametes

..............and................

 

..............and................

 

Phenotypes of offspring

red-eyed females

and

red-eyed males

 

Genotype of offspring

..................................

 

..................................


(3)




(ii)     The number of red-eyed females and red-eyed males in the offspring was counted. The observed ratio of red-eyed females to red-eyed males was similar to, but not the same as, the expected ratio. Suggest one reason why observed ratios are often not the same as expected ratios.

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(1)




(c)     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.

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(2)

(Total 8 marks)






 

M1.(a)     (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.

1




(b)     (i)      1.      1, (2) and 5;



Accept: for 1 mark that 1 and 2 have slow (feather production) but produce one offspring with rapid (feather production).

Neutral: any reference to 3 being offspring of 1.

2.      1 must possess / pass on the recessive allele / 1 must be a carrier / heterozygous / if slow (feather production) is recessive all offspring of (1 and 2) would be slow (feather production) / if rapid (feather production) was dominant 1 would have rapid (feather production);

Reject: both parents must be carriers / possess the recessive allele.

Reject: one of the parents (i.e. not specified) must be a carrier / heterozygous.

2




(ii)      5 = XfY / XfY- / f / f- / fY ;

7 = XFXf and XFXF (either way round) /

or XfXF and XFXF (either way round) /

or XFXf, XfXF and XFXF(in any order);

Note: allow 5 = XfY, XfY.

Accept: for both 5 and 7 a different letter than F. However, lower case and capital letter must correspond to that shown in the answer. For example accept 7 = XRXr and XRXR.

2




(iii)     XFXf and XfY or XfXF and XfY



or XFXf and XfY- or XfXF and XfY- /

or Ff and fY /

or Ff and fY- /

or Ff and f- /

or Ff and f;

Accept: a different letter than F. However, lower case and capital letter must correspond to that shown in the answer.

Accept: each alternative either way round.

1




(c)     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.

3

[9]










M2.(a)     2.84:1;

Accept ‘2.84 to 1’ or (just) 2.84

Do not accept 1:2.84 or 142:50

1




(b)     1.      Some embarrassed / some not willing to show tongue / cannot tell;

2.      Could not decide whether thumb was straight or not / thumb bending is judgemental / subjective;

2




(c)     1.      (No) - should be 92.9% / should be calculated from 182 out of 196 / should not be calculated from 182 out of 200;



Allow either no or yes approach but no mark awarded for no or yes on its own

2.      (Yes) – assumes 4 out of 200 use either hand;



Accept ambidextrous

3.      (But) sample may not be representative;



This could be expressed in other ways e.g. only based on one part of the country / might not be the same in different parts of the UK / might not be representative of UK

4.      Small sample size / only sampled 200;



2 max

[5]










M3.(a)     1.      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;



2 max




(b)     (i)      1.      XRXR and XrY;



All marking points are completely independent. Allow crosses from the following parents for a possible three marks:

XRXR and Xr-

XRXR and XrY;

RR and rY / rY

RR and r− or RR and r

2.      XR and XR plus X r and Y;

3.      XRXr and XRY;

OR

1.      XRXr and XrY;



OR

XRXr and Xr

XRXr and XrY;

2.      XR and Xr plus Xr and Y;



Rr and rY / rY

Rr and r or Rr and r

Accept different symbols e.g. W and w

2. Accept gametes in a punnet square

3.      XRXr and XRY;



3




(ii)     Fertilisation is random / fusion of gametes is random / small / not large population / sample / selection advantage / disadvantage / lethal alleles;



Mutation = neutral

Random mating = neutral

Accept fertilisation / fusion of gametes is due to chance

1




(c)     1.      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;



2

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