Inheritance Pedigree Charts Unit 4 A2

People with night blindness have difficulty seeing in dim light People with night blindness have difficulty seeing in dim light. The allele for night blindness, N, is dominant to the allele for normal vision, n. These alleles are not carried on the sex chromosomes. Individual 12 is a boy. What is his phenotype? What is the genotype of individual 1? Explain the evidence for your answer. What is the probability that the next child born to individuals 10 and 11 will be a girl with night blindness? Show your working.

1. (a) Normal sight; 1 (b) Nn; Must have at least one N allele as she has the condition and must pass on an n allele to her normal sighted children; 2 (c) Two marks for correct answer of ¼ / 0.25 / 25%; One mark for incorrect answer that determines probability of next child having night blindness as ½ / 0.5 / 50%; 2 max  

The allele for Rhesus positive, R, is dominant to that for Rhesus negative, r. Haemophilia is a sex-linked condition. The allele for haemophilia, h, is recessive to the allele for normal blood clotting, H, and is carried on the X–chromosome. The diagram shows the Rhesus blood group phenotypes in a family tree where some individuals have haemophilia. Use the information in the diagram to give one piece of evidence that the allele for the Rhesus negative condition is recessive. Explain the evidence from the cross between individuals 3 and 4 that the gene controlling Rhesus blood group is not sex-linked

a) (i) specific cross identified - 3 ,4 & 8 or 10, 11 & 12 /Rh negative phenotype produced from parents which are both Rh positive; 1 (ii) with sex linkage daughter cannot have (recessive) condition unless male parent has the condition; as male passes X chromosome to his daughter; 2

Becker muscular dystrophy is a sex linked inherited condition caused by an allele of a gene on the X chromosome. Sufferers experience some loss of muscle strength. The diagram shows how members of one family were affected by the condition. Explain one piece of evidence from the diagram which shows that the allele for Becker muscular dystrophy is recessive. The allele for Becker muscular dystrophy is sex-linked. Explain how individual 9 inherited the condition from his grandfather.

(b) (i) 3 and 4 do not show the condition but 9/one male does; 4 must be carrier; OR 1 affected but not daughter/4; who gets X from father; 2 (ii) grandfather/1 passed on his (affected) X chromosome to his daughter/4; who was unaffected, because of the ‘normal’ X inherited from her mother/2; 9 inherited his X chromosome from his mother/4; 2 max

The diagram shows the inheritance of coat colour in pigs through three generations. Explain one piece of evidence from the diagram which shows that coat colour is not controlled by one gene with two codominant alleles.

sandy stated as heterozygous/suitable allusion to alleles; suitable cross chosen;(as in table) N.B. second two points linked, not stand-alone explained why could not be codominance; N.B. Second two points linked, not stand alone Suitable cross Reason why not codominance 3 and 4 Offspring should all be sandy 10 and 11 Offspring should all be sandy 7 and 8 Offspring should all be red BUT if candidate assumes sandy is homozygous, mark accordingly e.g. “look at cross 1 and 2; all their offspring would be sandy;” and not that, if red or white then identified as heterozygote, then full 3 marks are still possible.

Nail-patella syndrome is an inherited condition caused by a single gene. Sufferers have abnormal nail growth and underdeveloped kneecaps. The pedigree shows how members of one family were affected by the syndrome. Explain one piece of evidence from the pedigree which indicates that (i) the allele for the nail-patella syndrome is dominant; (ii) the gene is not sex-linked.

(b) (i) 3 and 4 produce unaffected male/8 / female/10, so must 3 must be a carrier recessive; if the condition is expressed in the heterozygous form, then the condition must be dominant 2 4 would inherit an X chromosome form her father which would carry the dominant allele and thus she would express the condition, as she does not it can not be on the X

The Rhesus blood group is genetically controlled The Rhesus blood group is genetically controlled. The gene for the Rhesus blood group has two alleles. The allele for Rhesus positive, R, is dominant to that for Rhesus negative, r. The diagram shows the inheritance of the Rhesus blood group in one family. Explain one piece of evidence from the diagram which shows that the allele for Rhesus positive is dominant. Explain one piece of evidence from the diagram which shows that the gene is not on the X chromosome. Sixteen percent of the population of Europe is Rhesus negative. Use the Hardy-Weinberg equation to calculate the percentage of this population that you would expect to be heterozygous for the Rhesus gene. Show your working. (3)

3 and 4 / two Rhesus positives produce Rhesus negative child/children / 7 / 9; Both Rhesus positives/3 and 4 carry recessive (allele)/ are heterozygous / if Rhesus positive was recessive, all children (of 3 and 4) would be Rhesus positive/recessive; 3 would not be/is Rhesus positive / would be Rhesus negative; 3 would receive Rhesus negative (allele) on X (chromosome) from mother / 3 could not receive Rhesus positive (allele) from mother / 3 would not receive Rhesus positive (allele)/X (chromosome) from father/1 / 3 will receive Y (chromosome) from father/1; OR 9 would be Rhesus positive / would not be/is Rhesus negative / 8 and 9/all daughters of 3 and 4 would be Rhesus positive; As 9 would receive X chromosome/dominant allele from father/3; Correct answer of 48(%) = 3 marks;;; q2/ p2= 16%/0.16 / p/q = 0.4; Shows that 2pq = heterozygotes / carriers;

A single gene controls the presence of hair on the skin of cattle A single gene controls the presence of hair on the skin of cattle. The gene is carried on the X chromosome. Its dominant allele causes hair to be present on the skin and its recessive allele causes hairlessness. The diagram shows the pattern of inheritance of these alleles in a group of cattle. Use evidence from the diagram to explain that hairlessness is caused by a recessive allele that hairlessness is caused by a gene on the X chromosome.

1. Animal 2 / 5 has hair but offspring do not; 2. So 2 / 5 parents must be heterozygous/carriers; OR 3. 4/7/8 are hairless but parents have hair; 4. So 2 / 5 must be heterozygous/carriers; Accept parents as alternative to animals 2 and 5, 1 + 3: Allow reference to children/offspring for animals 7 + 8 Hairless males have fathers with hair / 4 is hairless but 1 is hairy / 7 and/or 8 are hairless but 6 is hairy / only males are hairless

Tay-Sachs disease is a human inherited disorder Tay-Sachs disease is a human inherited disorder. Sufferers of this disease often die during childhood. The allele for Tay-Sachs disease t, is recessive to allele T, present in unaffected individuals. The diagram shows the inheritance of Tay-Sachs in one family. Explain one piece of evidence from the diagram which proves that the allele for Tay-Sachs disease is recessive. Explain one piece of evidence from the diagram which proves that the allele for Tay-Sachs disease is not on the X chromosome. In a human population, one in every 1000 children born had Tay-Sachs disease. Use the Hardy-Weinberg equation to calculate the percentage of this population you would expect to be heterozygous for this gene. Show your working.(3)

1. 3 and 4 and 9/11/affected offspring; 2. Both 3 and 4 are carriers/heterozygous; OR If dominant at least one of 3 and 4 would be affected; 1 Accept: 9/11 and their parents 1 Accept: unaffected parents have affected children 2 Accept: if 3 and 4 are unaffected all their children will be unaffected

1. 11 is affected, 3 is not; 2. 3/father of 11 does not have a recessive allele on his X chromosome/ Xt; OR (If on X) 11/affected female would not receive the recessive allele on X chromosome/Xt from 3/father; (If on X) 3/father (of 11) would pass on the dominant allele on his X chromosome/XT; Answer in range of 5.8 - 6.2% = 3 marks;;; If incorrect answer, then 2 max of following points 1. q2/p2/tt = 0.001 or 1 divided by 1000; 2. p/q/T = 0.968 – 0.97; 3. Understanding that heterozygous = 2pq;