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1. Introduction to Biology2h 42m
2. Chemistry3h 40m
3. Water1h 26m
4. Biomolecules2h 23m
5. Cell Components2h 26m
6. The Membrane2h 31m
7. Energy and Metabolism2h 0m
8. Respiration2h 40m
9. Photosynthesis2h 49m
10. Cell Signaling59m
11. Cell Division2h 47m
12. Meiosis2h 0m
13. Mendelian Genetics4h 44m
14. DNA Synthesis2h 27m
15. Gene Expression3h 20m
16. Regulation of Expression3h 31m
17. Viruses37m
- Viruses
  37m
18. Biotechnology2h 58m
19. Genomics17m
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  17m
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21. Evolution3h 1m
22. Evolution of Populations3h 52m
23. Speciation1h 37m
24. History of Life on Earth2h 6m
25. Phylogeny2h 31m
26. Prokaryotes4h 59m
27. Protists1h 12m
28. Plants1h 22m
29. Fungi36m
- Fungi
  19m
- Fungi Reproduction
  17m
30. Overview of Animals34m
- Overview of Animals
  34m
31. Invertebrates1h 2m
32. Vertebrates50m
33. Plant Anatomy1h 3m
34. Vascular Plant Transport1h 2m
- Water Potential
  1h 2m
35. Soil37m
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  20m
- Nitrogen Fixation
  16m
36. Plant Reproduction47m
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  33m
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  14m
37. Plant Sensation and Response1h 9m
38. Animal Form and Function1h 19m
39. Digestive System1h 10m
- Digestion
  1h 3m
- Blood Sugar Homeostasis
  7m
40. Circulatory System1h 57m
41. Immune System1h 12m
42. Osmoregulation and Excretion50m
- Osmoregulation and Excretion
  50m
43. Endocrine System1h 4m
- Endocrine System
  1h 4m
44. Animal Reproduction1h 2m
- Animal Reproduction
  1h 2m
45. Nervous System1h 55m
- Neurons and Action Potentials
  1h 8m
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  46m
46. Sensory Systems46m
- Sensory System
  46m
47. Muscle Systems23m
- Musculoskeletal System
  23m
48. Ecology3h 11m
49. Animal Behavior28m
- Animal Behavior
  28m
50. Population Ecology3h 41m
51. Community Ecology2h 46m
52. Ecosystems2h 36m
53. Conservation Biology24m
- Conservation Biology
  24m

22. Evolution of Populations

The Hardy-Weinberg Principle

General Biology

22. Evolution of Populations

The Hardy-Weinberg Principle

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3:04 minutes

Problem 5`

Textbook Question

A fruit fly population has a gene with two alleles, A1 and A2. Tests show that 70% of the gametes produced in the population contain the A1 allele. If the population is in Hardy-Weinberg equilibrium, what proportion of the flies carry both A1 and A2?
a. 0.7
b. 0.49
c. 0.42
d. 0.21

Verified step by step guidance

First, understand that the Hardy-Weinberg equilibrium principle states that allele and genotype frequencies in a population will remain constant from generation to generation in the absence of other evolutionary influences.

Given that 70% of the gametes contain the A1 allele, the frequency of A1 (p) is 0.7. Consequently, the frequency of the A2 allele (q) can be calculated as 1 - p, which is 1 - 0.7.

Calculate the frequency of the A2 allele: q = 1 - 0.7 = 0.3.

In Hardy-Weinberg equilibrium, the proportion of heterozygous individuals (carrying both A1 and A2 alleles) is given by the formula 2pq, where p is the frequency of A1 and q is the frequency of A2.

Substitute the values of p and q into the formula: 2pq = 2 * 0.7 * 0.3. This will give you the proportion of flies that are heterozygous (A1A2).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Hardy-Weinberg Equilibrium

The Hardy-Weinberg equilibrium is a principle that states that allele and genotype frequencies in a population will remain constant from generation to generation in the absence of evolutionary influences. These influences include mutation, selection, gene flow, genetic drift, and non-random mating. It provides a mathematical model to study genetic variation in populations.

Allele Frequency

Allele frequency refers to how common an allele is in a population. It is calculated as the number of copies of a specific allele divided by the total number of all alleles for that gene in the population. In this question, the allele frequency of A1 is given as 0.7, meaning 70% of the gametes carry the A1 allele.

Genotype Frequency Calculation

In a population in Hardy-Weinberg equilibrium, the genotype frequencies can be calculated using the allele frequencies. For two alleles, A1 and A2, the frequency of the heterozygous genotype (A1A2) is given by 2pq, where p is the frequency of A1 and q is the frequency of A2. Given p = 0.7, q can be calculated as 1 - p = 0.3, leading to a heterozygous frequency of 2(0.7)(0.3) = 0.42.