Non-Mendelian Genetics: Study Guide (Biology 5.4)

When the ratios do not fit

In Topic 5.3, Mendel’s laws predicted neat ratios: 3 : 1, 9 : 3 : 3 : 1, 1 : 1 : 1 : 1. Many real traits give different ratios. How do you know a difference is real and not just chance? You use the chi-square test. If χ² is greater than the critical value, the observed ratio differs significantly from the Mendelian prediction, and you look for a biological reason.

This topic covers the main reasons:

Cause	What changes
Linked genes	genes on one chromosome do not assort independently
Incomplete dominance or codominance	heterozygotes look different from both homozygotes
Sex-linked genes	inheritance depends on the sex of parent and offspring
Pleiotropy	one gene affects several traits, so those traits are inherited together
Non-nuclear genes	genes in mitochondria or chloroplasts come from one parent only

In every case the alleles of nuclear genes still segregate in meiosis. What changes is where the gene sits or how its alleles act.

Incomplete dominance and codominance

With complete dominance, Aa looks like AA. Two other kinds of allele interaction give the heterozygote its own phenotype.

• Incomplete dominance. Neither allele masks the other, so the heterozygote is a blend of the two homozygous phenotypes. A classic example is snapdragon flower colour: red × white gives pink.
• Codominance. Both alleles are fully expressed in the heterozygote. In the human ABO blood groups, alleles Iᴬ and Iᴮ are codominant: an IᴬIᴮ person has both A and B antigens on their red blood cells (blood group AB). A third allele, i, is recessive to both.

Because the heterozygote has its own phenotype, a cross between two heterozygotes gives a 1 : 2 : 1 phenotypic ratio, the same as the genotypic ratio. With complete dominance it would be 3 : 1.

Codominance example. A parent with genotype IᴬIᴮ has a child with a partner who is Iᴬi. The possible children are IᴬIᴬ, Iᴬi, IᴬIᴮ and Iᴮi, each with probability 1/4. So P(group A) = 1/4 + 1/4 = 1/2 (sum rule), P(group AB) = 1/4, P(group B) = 1/4, and P(group O) = 0.

Linked genes and gene mapping

Each chromosome carries many genes. Genes on the same chromosome are genetically linked: they tend to go into gametes together, so they do not assort independently.

Linkage is not complete. In prophase I, non-sister chromatids exchange segments by crossing over. If a crossover happens between two linked genes, it produces chromosomes with new combinations of their alleles. Offspring carrying these are recombinants. Offspring with the original combinations are parental types.

The farther apart two genes are, the more likely a crossover falls between them. So the percentage of recombinants measures distance:

Recombination frequency (%) = number of recombinant offspring ÷ total offspring × 100

1% recombination = 1 map unit. Using these distances to place genes in order along a chromosome is gene mapping.

How to spot linkage in a dihybrid test cross (AaBb × aabb):

• Unlinked genes give four phenotypes in a 1 : 1 : 1 : 1 ratio.
• Linked genes give two large classes (the parental types) and two small classes (the recombinants).
• A chi-square test against 1 : 1 : 1 : 1 gives a large χ², so you reject the hypothesis of independent assortment.

Recombination frequency cannot exceed about 50%. At that point, recombinants and parental types are equally common, which looks the same as independent assortment.

Sex-linked traits

In humans and other mammals, females are XX and males are XY. The X chromosome carries many genes; the small Y carries few. A gene on a sex chromosome is sex-linked.

• X-linked genes. A male has one X, so he has only one allele for each X-linked gene. A single recessive allele is therefore expressed in a male. A female needs two copies. This is why X-linked recessive traits, such as red-green colour blindness and haemophilia, are much more common in males.
• A father passes his X to all his daughters and his Y to all his sons. So an affected father never passes an X-linked trait to his sons.
• Y-linked genes pass from a father to all his sons, and never to daughters.

Pedigree clues for X-linked recessive traits: affected individuals are mostly male; affected sons usually have unaffected (carrier) mothers; there is no father-to-son transmission; and every daughter of an affected father is at least a carrier.

Write X-linked genotypes with the allele as a superscript: Xᴮ for normal colour vision, Xᵇ for colour blindness, and Y with no allele.

Other sex-determination systems. Not every species uses X and Y.

• Birds use the ZW system: males are ZZ and females are ZW. Here the female has only one Z, so Z-linked recessive traits are more common in females.
• Bees use haplodiploidy: females develop from fertilized eggs and are diploid; males develop from unfertilized eggs and are haploid. A male has no father and passes on his whole genome to every daughter.

Pleiotropy

Pleiotropy is when one gene affects several traits. The best-known human example is the sickle-cell allele of the gene for a haemoglobin subunit. People with two copies make haemoglobin that makes red blood cells sickle-shaped. This one change causes anaemia, blocked small blood vessels, pain and organ damage. Heterozygotes are more resistant to malaria.

Because all these effects come from one gene, they are inherited together. They cannot segregate independently, and crossing over can never separate them. This is how you tell pleiotropy from linkage: two linked genes are occasionally separated by crossing over, giving recombinants.

Non-nuclear inheritance

Mitochondria and chloroplasts contain their own small circular DNA. Their genes follow different rules.

• When a cell divides, its mitochondria and chloroplasts are shared out at random between daughter cells and gametes. So their traits do not follow Mendel’s ratios. In some plants with variegated (patchy green and white) leaves, cells with mixed chloroplasts divide, and some cell lines end up with mostly faulty chloroplasts, giving white patches.
• In animals, the egg supplies the zygote’s mitochondria; sperm usually contribute none. Traits coded by mitochondrial DNA are therefore typically maternally inherited: an affected mother can pass the trait to any of her children, while an affected father usually passes it to none. (How severely each child is affected can vary, because mitochondria are shared out at random.)
• In plants, mitochondria and chloroplasts are usually passed on in the ovule, not the pollen, so their traits are also typically maternally inherited.

Reciprocal crosses reveal this. Cross a female with trait X and a male without it, then the reverse. Nuclear genes (not sex-linked) give the same result both ways. Organelle genes follow the mother.

Worked example 1: incomplete dominance and chi-square

Question. In a hypothetical ornamental flower, true-breeding red plants are crossed with true-breeding white plants. All F₁ plants are pink. Two F₁ plants are crossed, giving 240 F₂ plants: 52 red, 131 pink and 57 white. Do the data support incomplete dominance?

1. Model. Red = CᴿCᴿ, white = CᵂCᵂ, pink = CᴿCᵂ. F₁ × F₁ = CᴿCᵂ × CᴿCᵂ, which predicts 1 red : 2 pink : 1 white.
2. Null hypothesis. “Flower colour is controlled by one gene with incompletely dominant alleles, so the F₂ occurs in a 1 : 2 : 1 ratio of red : pink : white. Differences are due to chance.”
3. Expected counts. 240 × 1/4 = 60 red; 240 × 1/2 = 120 pink; 60 white.
4. χ².

Phenotype	o	e	(o − e)²	(o − e)² ÷ e
red	52	60	64	1.067
pink	131	120	121	1.008
white	57	60	9	0.150
Total	240	240		χ² = 2.225

5. Degrees of freedom = 3 − 1 = 2. Critical value at p = 0.05 = 5.99.
6. Conclusion. 2.225 < 5.99, so fail to reject the null hypothesis. The data are consistent with incomplete dominance.

Interpretation. The pink F₁ shows that neither allele masks the other. If red were completely dominant, the F₁ would be red and the F₂ would show 3 red : 1 white. Notice that the pink heterozygote is a blend: this is incomplete dominance, not codominance. A codominant heterozygote would show red and white, for example in separate patches.

Worked example 2: mapping linked genes

Question. In a hypothetical beetle, black body (B) is dominant to brown (b), and long wings (L) are dominant to short wings (l). A BbLl beetle, whose parents were BBLL and bbll, is test crossed with a bbll beetle. The 1000 offspring are: 412 black long, 418 brown short, 87 black short and 83 brown long. (a) Show that the genes are linked. (b) Calculate the map distance. (c) A third gene, E, is 9 map units from B and 8 map units from L. Give the gene order.

(a) Linked? Independent assortment would give 250 in each class. The observed classes are two large (412, 418) and two small (87, 83). A chi-square test against 1 : 1 : 1 : 1 gives χ² = (162² + 168² + 163² + 167²) ÷ 250 = 108 926 ÷ 250 ≈ 436, far above 7.81 (3 df). Reject independent assortment. The large classes match the parents’ combinations (B with L, b with l), so these alleles sit together on the same chromosomes.

(b) Map distance.

1. Parental types: black long and brown short (412 + 418 = 830).
2. Recombinants: black short and brown long (87 + 83 = 170).
3. Recombination frequency = 170 ÷ 1000 × 100 = 17%.
4. Map distance = 17 map units.

(c) Gene order. B–E = 9 and E–L = 8. Since 9 + 8 = 17 = B–L, E must lie between B and L: the order is B–E–L (Figure 1). In general, the gene in the middle is the one whose two distances add up to the largest distance.

Check. The recombinant classes are similar in size (87 and 83), as expected, because each crossover makes one of each.

Worked example 3: an X-linked cross

Question. A woman with normal colour vision, whose father was colour-blind, has children with a man with normal vision. (a) Give both genotypes. (b) Find the probability that a son is colour-blind, and that a child is a colour-blind son. (c) Could a daughter be colour-blind?

1. Her genotype. Her father gave her his only X, which carried Xᵇ. She has normal vision, so her other X is Xᴮ. She is XᴮXᵇ (a carrier). The man is XᴮY.
2. Sons get Y from the father and an X from the mother: Xᴮ or Xᵇ with equal chance. P(colour-blind, given a son) = 1/2.
3. Any child. P(son) = 1/2, so P(colour-blind son) = 1/2 × 1/2 = 1/4 (Figure 2).
4. Daughters always receive the father’s Xᴮ, so none can be colour-blind. Each has a 1/2 chance of being a carrier.

Interpretation. The trait passed from the woman’s father, through her, to her sons: it “skipped” a generation through a carrier daughter. This is the typical pattern of X-linked recessive inheritance, and it explains why more males than females are affected.

Common misconceptions

• “Incomplete dominance and codominance are the same.” Incomplete: a blend. Codominance: both phenotypes appear fully.
• “Linked genes never recombine.” Crossing over separates them at a rate that depends on their distance apart.
• “Recombination frequency is recombinants ÷ parental types.” Divide by the total number of offspring.
• “Sons inherit X-linked traits from their fathers.” Sons get their X from their mothers; fathers give sons a Y.
• “Females are always XX.” In birds, females are ZW; in bees, sex depends on whether the egg was fertilized.
• “Pleiotropy is the same as linkage.” Pleiotropy is one gene with several effects; linkage is two genes close together. Only linked genes can be separated by crossing over.
• “Mitochondrial traits come from both parents.” They are typically inherited from the mother only.
• “A significant chi-square tells you why the ratio is different.” It only shows that the difference is unlikely to be chance. You must then propose a biological explanation.

Where this leads

Topic 5.5, Environmental Effects on Phenotype, shows that the same genotype can produce different phenotypes in different environments. Test yourself first with the practice questions, then use the revision notes and the checklist. The previous topic, Mendelian Genetics, has the basic cross and chi-square methods.