Questions on monohybrid inheritance and genetic diagrams

Question 1

In tomatoes, red fruit is dominant to yellow fruit. Complete a full genetic diagram for a cross between a heterozygous red-fruited plant and a yellow-fruited plant, showing the parental genotypes, the gametes, the offspring genotypes and the expected phenotype ratio.

Accepted Answer

B1 the symbols are defined, for example R for the allele for red fruit and r for the allele for yellow fruit, using one letter for both alleles B1 parental genotypes given as Rr for the red-fruited plant and rr for the yellow-fruited plant B1 gametes shown as circled single alleles rather than pairs: R and r from the red-fruited parent, r and r from the yellow-fruited parent B1 offspring genotypes obtained from the gametes as Rr and rr, in equal numbers A1 offspring phenotypes and ratio given as 1 red-fruited : 1 yellow-fruited

Question 2

A breeder has a tall pea plant and needs to know whether it is homozygous or heterozygous. Explain how a test cross would answer the question, and explain why a result suggesting the plant is homozygous can never be certain.

Accepted Answer

B1 cross the tall plant with a homozygous recessive dwarf plant, which can only contribute a recessive allele to every offspring B1 if any offspring is dwarf, the tall parent must have supplied a recessive allele, so it is heterozygous B1 if a large number of offspring are all tall, the tall parent is very probably homozygous, because a heterozygote would be expected to give a 1 : 1 ratio A1 each fertilisation is independent, so a heterozygote can give a run of dominant offspring by chance: eight in a row happens about once in every 256 attempts

Question 3

Cystic fibrosis is caused by a recessive allele. Two parents are both carriers and neither has the condition. Calculate the probability that their first child has cystic fibrosis, and calculate the probability that their first two children both have it.

Accepted Answer

M1 both parents are heterozygous, so the cross is Ff × Ff and the offspring genotypes are 1 FF : 2 Ff : 1 ff A1 one of the four equally likely outcomes is homozygous recessive, so the probability is 1 in 4, or 0.25 M1 the two fertilisations are independent events, so multiply the separate probabilities: 0.25 × 0.25 A1 1 in 16, or 0.0625

Question 4

Haemophilia is caused by a recessive allele carried on the X chromosome. Explain why the condition is far more common in males than in females, and explain why a man with haemophilia can never pass it to his sons.

Accepted Answer

B1 a male is XY, and the Y chromosome is much smaller and carries no equivalent of the gene B1 a male therefore carries only one allele of the gene and expresses it whichever it is, because there is no second copy to mask it, so one recessive allele is enough B1 a female is XX and needs two copies of the recessive allele to show the condition; with one she is an unaffected carrier, and inheriting two is much less likely A1 a father passes his X to every daughter and his Y to every son, so his sons receive no allele of the gene from him at all and every daughter is at least a carrier

Question 5

A woman of blood group AB has children with a man of blood group O. Explain why none of their children can be blood group AB, and explain why none of them can be blood group O.

Accepted Answer

B1 the woman's genotype is I^AI^B, so every egg carries either I^A or I^B and never both B1 the man's genotype is I^OI^O, so every sperm carries I^O B1 every child is therefore I^AI^O or I^BI^O, giving blood group A or blood group B, because I^A and I^B are each dominant to I^O A1 group AB would need I^A and I^B together, which the father cannot supply, and group O would need two copies of I^O, which the mother cannot supply

Question 6

Explain why the albinism allele of TYR is recessive while the Huntington's disease allele of HTT is dominant, referring to the proteins the two genes code for.

Accepted Answer

B1 TYR codes for the enzyme tyrosinase, and an enzyme is a catalyst that is not used up, so a small quantity converts a great deal of tyrosine into melanin B1 a heterozygote transcribes and translates its one functional allele and makes enough tyrosinase for normal pigmentation, and the faulty allele contributes nothing but does no harm, so it is recessive B1 the expanded HTT allele produces an altered huntingtin that actively damages neurones rather than simply failing to work A1 the normal huntingtin made from the other allele cannot undo that damage, so one copy of the allele is enough to give the phenotype and it is dominant

Question 7

A woman of blood group A whose mother was group O has children with a man of blood group B whose father was group O. Complete the two empty cells of the Punnett square, and give the blood groups of the children with the ratio in which they are expected.

Accepted Answer

M1 the two empty cells are AO and OO, taking one allele from the gamete above the column and one from the gamete beside the row B1 AB gives blood group AB, because A and B are codominant and both are expressed B1 BO gives group B and AO gives group A, because O is recessive to both A1 OO gives group O, so the four groups AB, B, A and O are expected in a 1 : 1 : 1 : 1 ratio

Question 8

Red-green colour blindness is caused by a recessive allele on the X chromosome. A colour-blind man has children with a woman who is homozygous for normal colour vision. Calculate the probability that a daughter of this couple is colour-blind, and calculate the probability that a son is colour-blind.

Accepted Answer

M1 the man is X^bY and the woman is X^BX^B, so daughters receive X^b from their father and X^B from their mother, and sons receive X^B from their mother and Y from their father A1 every daughter is X^BX^b, a carrier, so the probability that a daughter is colour-blind is 0 M1 every son is X^BY, because a son never receives his father's X chromosome A1 the probability that a son is colour-blind is also 0, since no son can inherit the father's X^b

Question 9

The pedigree shows the inheritance of a condition in two families. Explain, using evidence from the pedigree, whether the allele causing the condition is dominant or recessive, and whether the gene is on an autosome or on the X chromosome.

Accepted Answer

B1 the allele is recessive B1 I-1 and I-2 are unaffected but their daughter II-1 is affected (or I-3 and I-4 with II-4), so both parents carried the allele without showing it B1 the gene is on an autosome B1 II-1 is an affected daughter of an unaffected father, I-1; a father gives his X to every daughter, so if the gene were on the X he would have given her the allele and would be affected himself

Question 10

The condition in the pedigree is caused by a recessive allele of an autosomal gene. II-2 and II-3 plan to have a child. Calculate the probability that the child will have the condition.

Accepted Answer

M1 I-1 and I-2 are both heterozygous, because their daughter II-1 is affected, so the unaffected II-2 is a carrier with probability 2/3 (ff is ruled out, leaving FF, Ff and fF) M1 I-3 and I-4 are both heterozygous, because their son II-4 is affected, so II-3 is also a carrier with probability 2/3 M1 the child is affected only if both parents are carriers and it receives the recessive allele from each: 2/3 × 2/3 × 1/4 A1 1/9, or 0.11

Question 11

The pedigree shows the inheritance of a condition through three generations. Explain, using evidence from the pedigree, why the condition is caused by a dominant allele of a gene on an autosome, and explain what the genotypes of II-3 and II-4 must be.

Accepted Answer

B1 II-3 and II-4 are both affected but their daughter III-1 is unaffected, so each parent carried an allele that does not cause the condition and was masked: the allele causing it is dominant B1 I-1 is an affected father with an unaffected daughter, II-2, or I-1 and the unaffected I-2 have an affected son, II-1 B1 a father gives his X to every daughter and to no son, so a dominant allele on his X would have made II-2 affected, or could not have reached II-1 from him; the gene is autosomal B1 II-3 and II-4 are both heterozygous, because each must have given III-1 the allele that does not cause the condition

Question 12

The condition in the pedigree is caused by a recessive allele on the X chromosome. Identify the two females in the pedigree who must be carriers, giving your reasons, and calculate the probability that III-2 is a carrier.

Accepted Answer

B1 I-2 is a carrier: her son II-1 is affected, a son's single X comes from his mother, and she is unaffected so she is heterozygous B1 II-3 is a carrier: her son III-1 is affected and she is unaffected, for the same reason M1 with X^A as the normal allele, II-3 is X^AX^a and II-4 is X^AY; every daughter receives X^A from her father and either X from her mother A1 the probability that III-2 is a carrier is 1/2

Question 13

Name the protein coded for by each of the genes TYR, F8 and HTT, and state the phenotype associated with a faulty allele of each.

Accepted Answer

B1 TYR codes for tyrosinase, and an allele giving no working enzyme means no melanin is made, which is albinism B1 F8 codes for factor VIII, and a faulty allele stalls the clotting cascade, which is haemophilia A B1 HTT codes for huntingtin, and an allele with an expanded CAG repeat gives a protein that damages neurones, which is Huntington's disease

Question 14

In peas, the allele Le codes for a working enzyme that makes an active gibberellin and the allele le codes for a version that does not work. Suggest why plants of genotype le le are dwarf, and suggest what would happen if active gibberellin were applied to such a plant.

Accepted Answer

B1 a le le plant has no working copy of the enzyme, so very little active gibberellin is made B1 gibberellin lengthens the internodes of a stem by promoting cell division and cell elongation between the nodes, so without it the internodes stay short and the plant is dwarf A1 applying active gibberellin supplies the product the plant could not make, so the internodes lengthen and the plant grows tall; a Le plant treated the same way changes little, because it already has enough

Question 15

In codominance, both alleles are fully expressed in the heterozygote, so the products of both appear side by side. In incomplete dominance, the heterozygote has a phenotype intermediate between those of the two homozygotes. Roan cattle, from a cross between red and white parents, have a coat of red hairs mixed with white hairs. Pink snapdragons come from a cross between red-flowered and white-flowered plants. Explain which relationship each example shows, and explain why a cross between two heterozygotes gives a 1 : 2 : 1 phenotype ratio in both cases.

Accepted Answer

B1 roan cattle show codominance, since each hair is fully red or fully white, while pink snapdragons show incomplete dominance, since the colour is intermediate; both are needed for the mark B1 each heterozygous parent passes either allele to half its gametes and gametes fuse at random, so the offspring genotypes are 1 homozygous : 2 heterozygous : 1 homozygous for the other allele B1 in both relationships the heterozygote looks different from either homozygote, so each genotype has its own phenotype and the genotype ratio appears unchanged as the phenotype ratio

Question 16

A person with genotype Hb^AHb^S makes both normal haemoglobin and sickle haemoglobin. Explain, in terms of transcription and translation, why both alleles are expressed in this heterozygote, and explain why the alleles are described as codominant rather than one being dominant and the other recessive.

Accepted Answer

B1 both alleles of the HBB gene are present in the nucleus of every red-cell precursor and both are transcribed into mRNA and translated into their respective beta-globin polypeptides, since neither allele is switched off by the presence of the other B1 the result is that the heterozygote's red cells contain both normal haemoglobin and sickle haemoglobin, made from the two different alleles at once B1 this is codominance rather than complete dominance because both phenotypes, both proteins, are simultaneously present and detectable in the heterozygote, rather than one allele's effect being completely masked by the other

Question 17

A student writes a genetic diagram for a monohybrid cross using the symbols S for the dominant allele and s for the recessive allele, without stating what either allele stands for. Suggest two reasons why this is poor practice, even though S and s are legitimate genetic symbols.

Accepted Answer

B1 without a stated definition, a reader cannot know what characteristic S and s refer to, so the diagram cannot be understood or marked on its own B1 S and s are hard to tell apart in handwriting, unlike letters such as T and t whose capital and lower-case forms look clearly different, so the diagram risks being misread even once the meaning is known B1 defining the symbols in a stated line, such as 'let S represent...', is treated as part of a complete genetic diagram, so omitting it loses a mark

Question 18

State what each of the following shows in a pedigree diagram: an open circle, a shaded square, and the label III-2.

Accepted Answer

B1 an open circle is a female who does not show the characteristic, an unaffected female B1 a shaded square is a male who shows the characteristic, an affected male B1 III-2 is the second person from the left in the third generation

Question 19

A student says that the pedigree proves the allele is on the X chromosome, because only males are affected. Suggest why the pedigree on its own does not prove this.

Accepted Answer

B1 the pedigree is also consistent with an autosomal recessive allele: I-1 and I-2 could both be heterozygous, and so could II-3 and II-4 B1 the autosomal explanation needs II-4, who married into the family, to be a carrier as well; that is less likely for a rare allele but not impossible, so the pedigree makes X-linkage more probable without proving it B1 only two people are affected, so both being male could be chance; with an autosomal allele each affected child is male with probability 1/2, so both are male with probability 1/4

Question 20

State what is meant by the term allele, and state what is meant by describing someone as a carrier of a recessive condition.

Accepted Answer

B1 an allele is one of the alternative versions of a gene, found at the same locus on a chromosome B1 a carrier is an individual heterozygous for a recessive allele, so they do not show the condition but can pass the allele on

Question 21

State the difference between a gene and an allele, and state what is meant by the locus of a gene.

Accepted Answer

B1 a gene is a length of DNA coding for a polypeptide or a functional RNA, whereas an allele is one of the alternative versions of that gene B1 the locus is the fixed position on a chromosome at which a particular gene is found, the same position on both members of a homologous pair