Unit 5 · Topic 5.4 Beta

Non-Mendelian Genetics

Mendel's ratios assume genes on different chromosome pairs, one dominant and one recessive allele, one gene per trait and nuclear genes.

Practice 4: Representing and Describing DataPractice 5: Statistical Tests and Data Analysis

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Question set for this topic

Part 1 · Hook

Why this matters

In 1910 Thomas Hunt Morgan's lab was breeding fruit flies to test Mendel's rules. Some pairs of traits refused to give 9:3:3:1: offspring kept the same combinations as their grandparents far more often than chance allowed. Elsewhere, red and white snapdragons gave pink flowers, people turned out to have blood types A, B, AB and O, and some plant traits were passed on only by the mother. None of this overturns Mendel. It shows what his simple ratios assumed, and what happens when those assumptions fail.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. In a dihybrid cross AaBb × AaBb with genes on different chromosome pairs, the phenotype ratio is

  1. 9:3:3:1
  2. 3:1
  3. 1:2:1
Show the answer

Independent assortment gives four gamete types in equal numbers, and 16 equally likely combinations give 9:3:3:1.

  • Correct: 9:3:3:1:
  • 3:1:
  • 1:2:1:

2. Crossing over between homologs in prophase I produces

  1. recombinant chromosomes carrying new combinations of alleles
  2. extra copies of a chromosome
  3. alleles that neither parent had
Show the answer

Crossing over swaps matching pieces between homologs, mixing their alleles onto one chromosome.

  • Correct: recombinant chromosomes carrying new combinations of alleles:
  • extra copies of a chromosome:
  • alleles that neither parent had:

3. In a chi-square test with 3 degrees of freedom, χ² = 9.4 and the critical value at p = 0.05 is 7.82. The conclusion is

  1. reject the null hypothesis
  2. fail to reject the null hypothesis
  3. accept the null hypothesis as proven
Show the answer

χ² is larger than the critical value, so a difference this large would be unlikely by chance if the null hypothesis were true: reject it.

  • Correct: reject the null hypothesis:
  • fail to reject the null hypothesis:
  • accept the null hypothesis as proven:

Part 4 · See it

See it first

Left: an F1 female fruit fly's homologous pair; one homolog carries B (gray body) and P (red eyes) close together, the other b (black body) and p (purple eyes). Middle: most of her eggs carry B P or b p (parental), a few carry B p or b P (recombinant), made by crossing over between the two genes. Right: crossed with a black, purple-eyed male (b p / b p), the offspring are 470 gray red and 466 black purple (parental), 31 gray purple and 33 black red (recombinant), 1,000 in total. Bottom: recombination frequency = recombinant offspring ÷ total × 100 = 64 ÷ 1,000 × 100 = 6.4%; 1% recombination is 1 map unit, so the genes are about 6.4 map units apart. A scale runs from 0% (always inherited together) to 50% (behaving like unlinked genes).
Linked genes in a testcross. Parental combinations dominate; the few recombinants come from crossing over between the genes, and their percentage measures the distance between them. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

  1. Mendel's 9:3:3:1 assumes the two genes are on different chromosome pairs, so they assort independently.When two genes sit close together on one chromosome (linked genes), their alleles go into gametes together and that assumption fails.
  2. In a testcross of a fly with B P on one homolog and b p on the other, most eggs carry one of those two chromosomes unchanged.Parental-type offspring far outnumber the other classes, instead of the 1:1:1:1 that independent assortment predicts.
  3. In some meioses, crossing over happens in the stretch between the two genes.A few gametes carry recombinant combinations (B p, b P), and their share, the recombination frequency, rises with the distance between the genes: 1% = 1 map unit.
  4. Alleles do not always act as simply dominant and recessive, and many genes can act on one trait or one gene on many.Crosses give other ratios: 1:2:1 phenotypes with incomplete dominance or codominance, continuous variation for polygenic traits, 9:3:4 with epistasis.
  5. Mitochondria and chloroplasts carry their own genes and are passed on in the egg's cytoplasm.Their traits are inherited from the mother, so reciprocal crosses give different results.
  6. A chi-square test compares observed counts with the counts expected from a hypothesis.A large χ² (above the critical value) rejects the hypothesis, for example independent assortment for linked genes.

Part 6 · Key ideas

Key ideas

  • Linked genes are close together on one chromosome and tend to be inherited together. Recombination frequency = recombinants ÷ total × 100; 1% = 1 map unit. Values near 50% look like unlinked genes. Frequencies give gene order on a linkage map.
  • Incomplete dominance: the heterozygote is in between (pink snapdragons). Codominance: both alleles show fully (AB blood type, roan cattle). The ABO gene has multiple alleles.
  • Polygenic traits (height, skin color) depend on many genes and show continuous variation. Pleiotropy: one gene, many traits. Epistasis: one gene masks another.
  • Nonnuclear inheritance: mitochondrial and chloroplast genes come from one parent, usually the mother. Sex-linked and cytoplasmic traits give different results in a reciprocal cross.
  • Compare the observed ratio with the expected ratio using a chi-square test before claiming a cross does or does not fit Mendel.

Part 7 · Misconception

A common mistake

The wrong idea: Genes on the same chromosome are always inherited together, so they never form recombinant offspring.

What actually happens: Crossing over between them makes recombinants in some meioses. The closer the genes, the rarer the recombinants; genes far apart on one chromosome cross over so often that they can give nearly 50% recombinants, like genes on different chromosomes.

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Data table

Offspring counts from a fruit fly testcross

A true-breeding gray-bodied, red-eyed fly was crossed with a true-breeding black-bodied, purple-eyed fly. Gray (B) is dominant to black (b) and red (P) to purple (p). An F1 female (BbPp) was then crossed with a black, purple-eyed male (bbpp), and the offspring were counted.

Offspring of the testcross BbPp (female) × bbpp (male)
Offspring phenotypeNumber of flies
Gray body, red eyes470
Black body, purple eyes466
Gray body, purple eyes31
Black body, red eyes33
Total1,000

1. Which offspring classes are recombinant?

  1. Gray with purple eyes, and black with red eyes
  2. Gray with red eyes, and black with purple eyes
  3. Gray with red eyes, and gray with purple eyes
  4. Black with purple eyes, and black with red eyes
Show the answer

The F1 female got B P from one parent and b p from the other. Offspring showing B with p, or b with P, carry combinations neither of her parents had: recombinants.

  • Correct: Gray with purple eyes, and black with red eyes: Gray-purple and black-red mix the grandparents' combinations, and they are the two small classes.
  • Gray with red eyes, and black with purple eyes: These match the original true-breeding parents' combinations: parental types.
  • Gray with red eyes, and gray with purple eyes: These share a body color; one class is parental, the other recombinant.
  • Black with purple eyes, and black with red eyes: These share a body color; one class is parental, the other recombinant.

2. Calculate the recombination frequency between the body-color and eye-color genes, in percent. Give one decimal place.

Type a number in %.

Show the answer

Recombinants: 31 + 33 = 64. 64 ÷ 1,000 × 100 = 6.4%, so the genes are about 6.4 map units apart.

  • Answer: 6.4 %

3. A chi-square test of the counts against the 1:1:1:1 ratio gives χ² ≈ 760. Which conclusion is correct?

  1. With 3 degrees of freedom the critical value is 7.82, so reject independent assortment; the genes are likely linked.
  2. With 4 degrees of freedom the critical value is 9.49, so reject independent assortment; the genes are likely linked.
  3. With 3 degrees of freedom the critical value is 7.82, so fail to reject independent assortment.
  4. χ² is far above the critical value, which shows that the genes are on different chromosomes.
Show the answer

Four classes give df = 4 − 1 = 3, critical value 7.82 at p = 0.05. 760 is far larger, so the deviation from 1:1:1:1 is very unlikely to be chance. Too many parental types fits linkage.

  • Correct: With 3 degrees of freedom the critical value is 7.82, so reject independent assortment; the genes are likely linked.: df = 3, and 760 > 7.82: reject; excess parental types point to linkage.
  • With 4 degrees of freedom the critical value is 9.49, so reject independent assortment; the genes are likely linked.: Degrees of freedom are the number of classes minus one: 4 − 1 = 3, not 4.
  • With 3 degrees of freedom the critical value is 7.82, so fail to reject independent assortment.: A χ² larger than the critical value means reject, not fail to reject.
  • χ² is far above the critical value, which shows that the genes are on different chromosomes.: Rejecting independent assortment points the other way: the genes are close on the same chromosome.

Data table

Coat colors of Labrador retriever puppies

In Labrador retrievers, gene B controls the dark pigment made in hair: B (black) is dominant to b (brown, called chocolate). A second gene, E, on another chromosome, controls whether that pigment is put into the hair at all: dogs with genotype ee cannot deposit it and are yellow. A breeder pooled the litters of several matings between dogs that were all BbEe.

Puppies from BbEe × BbEe matings
Coat colorNumber of puppies
Black92
Chocolate29
Yellow39
Total160

4. Calculate χ² for the puppy counts against an expected ratio of 9 black : 3 chocolate : 4 yellow. Give two decimal places.

Type a number.

Show the answer

Expected: 160 × 9/16 = 90 black, 160 × 3/16 = 30 chocolate, 160 × 4/16 = 40 yellow. χ² = (92 − 90)²/90 + (29 − 30)²/30 + (39 − 40)²/40 = 0.044 + 0.033 + 0.025 = 0.10.

  • Answer: 0.10

5. Using χ² ≈ 0.10 for the 9:3:4 hypothesis, which conclusion is correct at p = 0.05?

  1. df = 2, critical value 5.99; fail to reject, so the counts are consistent with 9:3:4.
  2. df = 3, critical value 7.82; fail to reject, so the counts are consistent with 9:3:4.
  3. df = 2, critical value 5.99; reject, so the counts do not fit 9:3:4.
  4. df = 2, critical value 5.99; this proves the 9:3:4 hypothesis is true.
Show the answer

Three classes give df = 3 − 1 = 2, critical value 5.99. 0.10 is far below it, so the small deviations are consistent with chance: fail to reject 9:3:4.

  • Correct: df = 2, critical value 5.99; fail to reject, so the counts are consistent with 9:3:4.: df = 2 and 0.10 < 5.99: the data fit the hypothesis.
  • df = 3, critical value 7.82; fail to reject, so the counts are consistent with 9:3:4.: There are three color classes, so df = 2, not 3.
  • df = 2, critical value 5.99; reject, so the counts do not fit 9:3:4.: A χ² below the critical value means fail to reject, not reject.
  • df = 2, critical value 5.99; this proves the 9:3:4 hypothesis is true.: A test can fail to reject a hypothesis but never proves it true.

Experimental setup

Leaf color in four o'clock plants

Some four o'clock plants have branches of three kinds: green, white (no working chloroplasts), and variegated (patches of green and white). Each flower makes both eggs and pollen, and the plant has no sex chromosomes. A botanist moved pollen from flowers on one kind of branch onto flowers on another kind, then grew the seeds.

Seedlings from crosses between flowers on different branches
Flower giving the eggs was on aFlower giving the pollen was on aSeedlings
Green branchGreen branchGreen
Green branchVariegated branchGreen
Green branchWhite branchGreen
Variegated branchGreen branchA mix of green, variegated and white
White branchGreen branchWhite (die after using up the seed's food)

6. Which explanation best accounts for the results?

  1. The trait depends on chloroplasts, which pass to the seedling in the egg's cytoplasm, not in the pollen.
  2. The trait is caused by a recessive allele in the nucleus, which is hidden when pollen brings a dominant allele.
  3. The trait is X-linked, so it passes from the plant giving the eggs to its offspring.
  4. The pollen's chloroplasts are destroyed by the egg's nucleus at fertilization.
Show the answer

Chloroplasts carry their own DNA and are passed on in the cytoplasm. An egg brings plenty of cytoplasm; the pollen's sperm brings almost none. So the seedling inherits the chloroplasts of the egg's branch: green, white or a mix.

  • Correct: The trait depends on chloroplasts, which pass to the seedling in the egg's cytoplasm, not in the pollen.: Chloroplasts travel with the egg's cytoplasm, so seedlings match the egg-giving branch.
  • The trait is caused by a recessive allele in the nucleus, which is hidden when pollen brings a dominant allele.: A nuclear allele would give the same result whichever plant gave the eggs, and these crosses differ by direction.
  • The trait is X-linked, so it passes from the plant giving the eggs to its offspring.: Four o'clocks have no sex chromosomes; each flower makes both eggs and pollen.
  • The pollen's chloroplasts are destroyed by the egg's nucleus at fertilization.: The nucleus does not destroy chloroplasts; the sperm cell brings little or no chloroplast DNA to begin with.

7. A mother has blood type A and her child has type O. Which man can be ruled out as the child's father on blood type alone?

  1. A man with type AB
  2. A man with type A
  3. A man with type B
  4. A man with type O
Show the answer

A type O child is ii and got an i from each parent. A type AB man is IAIB and has no i to give, so he cannot be the father. Men with type A, B or O can all carry i.

  • Correct: A man with type AB: IAIB carries no i allele, so he cannot give the child one.
  • A man with type A: A type A man can be IAi and pass i.
  • A man with type B: A type B man can be IBi and pass i.
  • A man with type O: A type O man is ii and passes i to every child.

8. A plant's height is controlled by three genes (A, B, C) on different chromosomes. Each capital-letter allele adds 2 cm to a base height, and the effects simply add up. How many different heights are possible among the offspring of AaBbCc × AaBbCc? Give a whole number.

Type a number in heights.

Show the answer

An offspring can have 0, 1, 2, 3, 4, 5 or 6 capital-letter alleles, and height depends only on that number: 7 classes. With more genes the classes multiply and the distribution looks continuous, as for polygenic traits such as human height.

  • Answer: 7 heights

Part 9 · Summary

Summary

Mendel's ratios assume genes on different chromosome pairs, one dominant and one recessive allele, one gene per trait and nuclear genes. When genes are linked, close on one chromosome, parental combinations outnumber recombinants; the recombination frequency (recombinants ÷ total × 100) measures their distance in map units and gives gene order. Incomplete dominance gives an in-between heterozygote and codominance shows both alleles, as in ABO blood types, a gene with multiple alleles. Polygenic traits vary continuously, pleiotropic genes affect several traits, and epistasis lets one gene mask another. Mitochondrial and chloroplast genes are usually inherited from the mother, so reciprocal crosses differ. A chi-square test compares observed counts with those expected under a hypothesis.

Part 10 · Up next

What comes next

Part 11 · Connections

Connections