IB Biology D3.2: Inheritance worked solutions

Alleles are alternative forms of one gene, so they occupy the same locus on a pair of homologous chromosomes. They differ in base sequence, not necessarily in amino acid sequence, they are not on different chromosomes, and a pair need not be one dominant and one recessive (codominant alleles exist).

Each fertilisation is independent and the sperm carries X or Y with equal probability, so the chance of a girl is \tfrac12 whatever the sex of earlier children.

Bb \times Bb gives BB, Bb, Bb and bb in equal proportions, so \tfrac14=25\% are bb and brown.

The heterozygote is intermediate between the two homozygotes, which is incomplete dominance. Codominance would show red and white patches together, and complete dominance would give all red or all white.

Group O needs ii, so both parents carry i. Group AB needs I^{\mathrm A} from one parent and I^{\mathrm B} from the other. Only I^{\mathrm A}i\times I^{\mathrm B}i gives A, B, AB and O.

(a) Recessive means that the disease appears only when both alleles are the faulty one: aa (1). (b) A diet low in phenylalanine, started after newborn screening, stops phenylalanine accumulating in the blood and damaging the brain (1).

Adult height depends on many genes and on nutrition, and it takes any value in a range. Blood group, sex and the presence of haemophilia fall into distinct categories.

A test cross uses a homozygous recessive partner, tt, so every offspring phenotype reveals the gamete that came from the unknown plant.

The mother gives her son either X^{\mathrm B} or X^{\mathrm b} with equal probability and the father gives Y. The son is colour blind with probability \tfrac12. The probability for any child would be \tfrac14, but the sex of this one is already given.

(a) II-1 is aa, so each parent carries a; both are unaffected so both are Aa (1). (b) II-3 is the unaffected child of two carriers: Aa, Aa gives AA, Aa, Aa, aa, and aa is excluded, so Aa has probability \tfrac23 (1). (c) II-2 is a carrier with probability \tfrac23. His partner is Aa. The child is aa with probability \tfrac14, so the chance is \tfrac23\times\tfrac14=\tfrac16 (2).

(a) Her father is X^{\mathrm b}Y, so she received X^{\mathrm b} from him; she has normal vision, so her other X carries B: X^{\mathrm B}X^{\mathrm b} (1). Gametes X^{\mathrm B} and X^{\mathrm b} against X^{\mathrm B} and Y give X^{\mathrm B}X^{\mathrm B}, X^{\mathrm B}Y, X^{\mathrm B}X^{\mathrm b}, X^{\mathrm b}Y, one mark for correct gametes and one for the grid (2). (b) Half of the sons are X^{\mathrm b}Y, so \tfrac12 (1); one child in four is a colour-blind son.

(a) Her mother is ii, so the woman received i; she is group A, so her other allele is I^{\mathrm A} (1). (b) Gametes I^{\mathrm A} and i against I^{\mathrm A} and I^{\mathrm B} (1) give I^{\mathrm A}I^{\mathrm A}, I^{\mathrm A}I^{\mathrm B}, I^{\mathrm A}i, I^{\mathrm B}i (1). (c) Group B is I^{\mathrm B}i, one box in four: \tfrac14. I^{\mathrm A} and I^{\mathrm B} are codominant, so a person with both has both antigens on the red blood cells and is group AB (1).

(a) The median is the mean of the fifth and sixth values, \frac{156+158}{2}=157 cm. The lower half is 148 to 156, so Q_1=153 cm; the upper half is 158 to 175, so Q_3=161 cm (2). IQR =161-153=8 cm. The upper fence is 161+1.5\times8=173 cm, and 175 cm lies beyond it, so it is plotted as an outlier (1). (b) Height is polygenic: several genes each add a small effect, and the environment, such as diet, also acts, so the values form a continuous range (2).

(a) Two faulty alleles mean no working enzyme (1). Phenylalanine is not converted to tyrosine and accumulates in the blood, damaging the developing brain (1). (b) The baby is still aa; the diet lowers the phenylalanine supplied, so the phenotype is normal. Genotype alone does not fix the phenotype (1).

(a) Tall smooth is T_hh: \tfrac34\times\tfrac14=\tfrac3{16} (1), and \tfrac3{16}\times320=60 (1). (b) The second parent is tthh and makes only th gametes. A 1 : 1 : 1 : 1 ratio shows that the tall hairy parent makes four kinds of gamete in equal numbers, so it is the double heterozygote TtHh (2). (c) The T/t pair and the H/h pair lie on different chromosomes. Each bivalent orients at random at the metaphase I equator, so which allele of one gene goes to a pole does not affect the other gene (2).

(a) The parental gametes are AB and ab. Aabb and aaBb carry new combinations, so they are the recombinants (1); \frac{9+9}{100}=18\% (1). (b) Non-sister chromatids of the homologous pair crossed over in prophase I and exchanged segments between the loci, so one chromatid carries A with b and the other a with B (2). The two large classes and two small ones show that the genes are linked on one chromosome.

(a) Null hypothesis: the genes assort independently and the ratio is 1 : 1 : 1 : 1 (1). Expected number E=\frac{400}{4}=100 per class (1). \chi^2=\frac{12^2}{100}+\frac{8^2}{100}+\frac{2^2}{100}+\frac{2^2}{100}=\frac{144+64+4+4}{100}=2.16 (1). (b) Degrees of freedom =4-1=3, and 2.16<7.81 (1). The null hypothesis is not rejected: the difference between observed and expected is not significant at 5%, so independent assortment is probable. The test does not prove it (1).

Each of the following earns 1, up to 6, with at least one example given. (1) Parents pass alleles to offspring in haploid gametes; a diploid zygote has two copies of each autosomal gene. (2) Dominant alleles show in the heterozygote; recessive alleles only when homozygous, as in PKU, so two carriers have a \tfrac14 chance of an affected child. (3) Codominant alleles are both expressed: I^{\mathrm A}I^{\mathrm B} gives group AB. (4) Incomplete dominance gives an intermediate phenotype. (5) A gene can have more than two alleles in the population, such as I^{\mathrm A}, I^{\mathrm B} and i, though one person carries two. (6) Genes on the X chromosome give sex linkage: a male has one X, so he is affected by one recessive allele, as in haemophilia. (7) Traits such as height are continuous because many genes each add an effect and the environment also acts. (8) Mutation, including SNPs, makes new alleles. Marks are for distinct ideas, not a list of definitions.

(a) The condition appears in every generation and each affected person has an affected parent: dominant (1). The affected father I-1 has an unaffected daughter II-1; under X-linked dominant inheritance all his daughters would be affected, so that is ruled out (1). X-linked recessive would need both unaffected mothers I-2 and II-4 to be carriers, which is unlikely. Most likely autosomal dominant (1). (b) I-2 is unaffected, so aa (1). II-3 is affected and has an unaffected parent, so Aa (1). (c) II-2 is Aa and her partner is aa, so \tfrac12 (1). (d) II-1 is aa and her unaffected partner is aa, so no child can be affected: 0(1).

(a) Nine of the sixteen boxes of an AaBb \times AaBb grid have at least one A and one B; the other seven lack one of them. The ratio is 9 purple : 7 white (2). (b) E_{\text{purple}}=\tfrac9{16}\times640=360 and E_{\text{white}}=\tfrac7{16}\times640=280 (2). (c) \chi^2=\frac{(376-360)^2}{360}+\frac{(264-280)^2}{280}=0.711+0.914=1.63 (3). (d) With two classes there is one degree of freedom and 1.63<3.84, so the null hypothesis is not rejected; the data fit a 9 : 7 ratio (1).

(a) If the genes were unlinked the ratio would be 1 : 1 : 1 : 1, so E=\frac{280}{4}=70 (1). \chi^2=\frac{48^2+48^2+50^2+50^2}{70}=\frac{9608}{70}=137 (2). (b) With 3 degrees of freedom the critical value is 7.81. Since 137>7.81 the null hypothesis is rejected: the genes do not assort independently and are probably linked (2). (c) The recombinants are the small classes: \frac{22+20}{280}=15\% (1). (d) A plant Ab/aB makes the parental gametes Ab and aB most often, so a test cross gives Aabb and aaBb most often (1).

(a) Her father is ii, so she is I^{\mathrm A}i (1). Her father is X^{\mathrm h}Y and her mother X^{\mathrm H}X^{\mathrm H}, so she is X^{\mathrm H}X^{\mathrm h}. The man is I^{\mathrm B}i and X^{\mathrm H}Y (1). (b) Group O is ii: \tfrac14. An affected son is X^{\mathrm h}Y: \tfrac12 of children are sons and \tfrac12 of those are affected, so \tfrac14. Joint probability \tfrac14\times\tfrac14=\tfrac1{16} (3). (c) AB needs I^{\mathrm A} from the mother and I^{\mathrm B} from the father: \tfrac12\times\tfrac12=\tfrac14 (1). (d) The genes are on different chromosomes, so the two events are independent and probabilities multiply(1).

(a) He has no B allele, so X^{\mathrm b}X^{\mathrm b}Y (1). (b) The father is X^{\mathrm B}Y and cannot give X^{\mathrm b}, so both X^{\mathrm b} came from the mother, who must be X^{\mathrm B}X^{\mathrm b} (1). Non-disjunction in meiosis I would give an egg with both different X chromosomes, X^{\mathrm B}X^{\mathrm b}. Two identical X^{\mathrm b} in the egg means the sister chromatids did not separate in anaphase II (2). (c) A son of a carrier mother and unaffected father is colour blind with probability \tfrac12 (1). (d) The SRY gene on the Y chromosome makes the gonads develop as testes (1).

(a) The base substitution changes one codon in the mRNA (1). A different amino acid is placed in the polypeptide (1). This changes the folding and the active site, so phenylalanine does not bind and the enzyme does not work (1). (b) Aa carriers make enough working enzyme to be healthy, so the allele a is not removed by selection and is passed to half of their gametes (2). (c) A substitution that gives another codon for the same amino acid is a silent SNP: a new allele with no change in the polypeptide or the phenotype (1).

(a) q^2=\frac{4}{100\,000}=4.0\times10^{-5}, so q=\sqrt{4.0\times10^{-5}}=0.0063 (2). (b) p=1-0.0063=0.9937, so 2pq=2\times0.9937\times0.0063=0.0126, and 0.0126\times100\,000\approx1257 carriers (2). (c) Carriers are healthy, so most copies of the recessive allele are in heterozygotes where selection does not act against them (1). (d) Conditions for the equilibrium: a large population, random mating, no mutation, no migration, no selection (1).

(a) Methyl groups added to cytosine reduce transcription of the gene (1), for example by changing how the DNA is packaged or the binding of transcription factors (1). (b) The larvae have the same alleles. Diet changes the pattern of methylation and so of gene expression, producing different phenotypes: the queen and the worker (2). (c) The base sequence is not altered, so this is phenotypic plasticity and not a change of genotype (1).