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

Probability and Genetics

Genetics

2. Mendel's Laws of Inheritance

Probability and Genetics

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Multiple Choice

In a family where both parents are heterozygous for a recessive genetic disorder (genotype Aa), what is the probability that, when a couple has two children, both children will be affected by the disorder?

1/4

1/8

1/2

1/16

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Verified step by step guidance

Identify the genotypes of the parents: both are heterozygous (Aa) for the recessive disorder, meaning each parent has one dominant (A) and one recessive (a) allele.

Determine the probability that a single child will be affected by the disorder. Since the disorder is recessive, the child must inherit two recessive alleles (aa) to be affected. Use a Punnett square to find this probability, which is the chance of inheriting 'a' from both parents.

Calculate the probability for one child to be affected: the chance of inheriting 'a' from one parent is 1/2, so the probability of 'aa' genotype is $\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$.

Since the two children are independent events, find the probability that both children are affected by multiplying the individual probabilities: $\frac{1}{4} \times \frac{1}{4}$.

Express the final probability as a fraction, which represents the chance that both children will have the recessive disorder.