Table of contents

1. Introduction to Genetics51m
2. Mendel's Laws of Inheritance3h 37m
3. Extensions to Mendelian Inheritance2h 41m
4. Genetic Mapping and Linkage2h 28m
5. Genetics of Bacteria and Viruses1h 21m
6. Chromosomal Variation1h 48m
7. DNA and Chromosome Structure56m
8. DNA Replication1h 10m
9. Mitosis and Meiosis1h 34m
10. Transcription1h 0m
11. Translation58m
12. Gene Regulation in Prokaryotes1h 19m
13. Gene Regulation in Eukaryotes44m
14. Genetic Control of Development44m
15. Genomes and Genomics1h 50m
16. Transposable Elements47m
17. Mutation, Repair, and Recombination1h 6m
18. Molecular Genetic Tools19m
- Genetic Cloning
  9m
- Methods for Analyzing DNA
  9m
19. Cancer Genetics29m
- Overview of Cancer
  21m
- Cancer Mutations
  7m
20. Quantitative Genetics1h 26m
21. Population Genetics50m
- Hardy Weinberg
  19m
- Allelic Frequency Changes
  31m
22. Evolutionary Genetics29m

2. Mendel's Laws of Inheritance

Mendel's Experiments and Laws

Genetics

2. Mendel's Laws of Inheritance

Mendel's Experiments and Laws

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Multiple Choice

Suppose that in unicorns, two autosomal loci, A and B, assort independently. If a unicorn with genotype AaBb is crossed with aabb, what proportion of the offspring are expected to have the genotype aabb?

3/4

1/4

1/8

1/2

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Verified step by step guidance

Identify the genotypes of the parents: one parent is AaBb (heterozygous at both loci) and the other is aabb (homozygous recessive at both loci).

Since the loci A and B assort independently, consider the inheritance of each locus separately using a Punnett square or probability rules.

For locus A: the Aa parent can pass on either A or a allele with equal probability (1/2 each), and the aabb parent can only pass on the a allele (probability 1). So, the probability of offspring being aa is the probability that the Aa parent passes a (1/2) times the probability that the other parent passes a (1), which equals 1/2.

For locus B: similarly, the Bb parent can pass B or b with equal probability (1/2 each), and the aabb parent can only pass b (probability 1). So, the probability of offspring being bb is the probability that the Bb parent passes b (1/2) times the probability that the other parent passes b (1), which equals 1/2.

Since the loci assort independently, multiply the probabilities for each locus to find the probability of offspring being aabb: $\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$.