Table of contents

1. Introduction to Biology2h 42m
2. Chemistry3h 37m
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 6m
16. Regulation of Expression3h 31m
17. Viruses37m
- Viruses
  37m
18. Biotechnology2h 58m
19. Genomics17m
- Genomes
  17m
20. Development1h 5m
21. Evolution3h 1m
22. Evolution of Populations3h 53m
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
- Soil and Nutrients
  20m
- Nitrogen Fixation
  16m
36. Plant Reproduction47m
- Flowers
  33m
- Seeds
  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 49m
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
- Central and Peripheral Nervous System
  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

12. Meiosis

Genetic Variation During Meiosis

General Biology

12. Meiosis

Genetic Variation During Meiosis

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5:49 minutes

Problem 8

Textbook Question

Two genes of a flower, one controlling blue (B) versus white (b) petals and the other controlling round (R) versus oval (r) stamens, are linked and are 10 map units apart. You cross a homozygous blue oval plant with a homozygous white round plant. The resulting F1 progeny are crossed with homozygous white oval plants, and 1,000 offspring plants are obtained. How many plants of each of the four phenotypes do you expect?

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Identify the genotypes of the parent plants. The homozygous blue oval plant has the genotype BBrr, and the homozygous white round plant has the genotype bbRR.

Determine the genotype of the F1 progeny. Since the genes are linked and the parents are homozygous, the F1 generation will have the genotype BbRr.

Set up the cross for the F1 progeny with the homozygous white oval plants, which have the genotype bbrr. The cross is BbRr x bbrr.

Calculate the expected gamete combinations from the F1 progeny (BbRr). Since the genes are 10 map units apart, the recombination frequency is 10%. This results in 45% non-recombinant gametes (BR and br) and 5% recombinant gametes (Br and bR).

Determine the expected phenotypes and their frequencies among the 1,000 offspring. Non-recombinant offspring (BR x bbrr and br x bbrr) will be 450 blue round (BRbbrr) and 450 white oval (brbbrr). Recombinant offspring (Br x bbrr and bR x bbrr) will be 50 blue oval (Brbbrr) and 50 white round (bRbbrr).

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Key Concepts

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

Genetic Linkage

Genetic linkage refers to the tendency of genes located close to each other on a chromosome to be inherited together during meiosis. The closer the genes are, the less likely they are to be separated by recombination. In this scenario, the two genes controlling petal color and stamen shape are linked, which affects the expected phenotypic ratios in the offspring.

Map Units and Recombination Frequency

Map units, or centimorgans, are a measure of genetic distance based on the frequency of recombination between two genes. One map unit corresponds to a 1% chance of recombination occurring between the genes. In this case, the genes are 10 map units apart, indicating a 10% recombination frequency, which will influence the proportions of phenotypes in the offspring.

Punnett Square and Phenotypic Ratios

A Punnett square is a tool used to predict the genotypic and phenotypic ratios of offspring from genetic crosses. By setting up a Punnett square for the F1 progeny crossed with homozygous white oval plants, one can calculate the expected ratios of the four phenotypes: blue round, blue oval, white round, and white oval, based on the inheritance patterns of the linked genes.

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Related Practice

Textbook Question

A fruit fly somatic cell contains 8 chromosomes. This means that ___________ different combinations of chromosomes are possible in its gametes.

a. 8

b. 16

c. 32

d. 64

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Textbook Question

A wild-type fruit fly (heterozygous for gray body color and red eyes) is mated with a black fruit fly with purple eyes. The offspring are wild-type, 721; black purple, 751; gray purple, 49; black red, 45. What is the recombination frequency between these genes for body color and eye color? Using information from problem 3, what fruit flies (genotypes and phenotypes) would you mate to determine the order of the body color, wing size, and eye color genes on the chromosome?

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Textbook Question

What is the physical basis for the independent assortment of alleles into offspring?

a. There are chromosome divisions during gamete production.

b. Homologous chromosome pairs are separated during gamete production.

c. Sperm and eggs are produced by different sexes.

d. Each gene codes for more than one protein.

e. The instruction manual for producing a human is incomplete.

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Textbook Question

Assume that genes A and B are on the same chromosome and are 50 map units apart. An animal heterozygous at both loci is crossed with one that is homozygous recessive at both loci. What percentage of the offspring will show recombinant phenotypes resulting from crossovers? Without knowing these genes are on the same chromosome, how would you interpret the results of this cross?

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Textbook Question

Triploid (3n) watermelons, which are seedless, are produced by crossing a tetraploid (4n) strain with a diploid (2n) plant. Explain why this mating produces a triploid individual.

941

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Textbook Question

You design Drosophila crosses to provide recombination data for gene a, which is located on the chromosome shown in Figure 15.12. Gene a has recombination frequencies of 14% with the vestigial wing locus and 26% with the brown eye locus. Approximately where is a located along the chromosome?

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Textbook Question

Meiosis results in independent assortment of maternal and paternal chromosomes. If 2n=6 for a given organism, and there is no crossing over, what is the chance that a gamete produced by this diploid organism will receive only paternal chromosomes?

a. 0

b. 1/16

c. 1/8

d. 1/3

1765

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Textbook Question

Why are individuals with an extra chromosome 21, which causes Down syndrome, more numerous than individuals with an extra chromosome 3 or chromosome 16?

a. There are probably more genes on chromosome 21 than on the others.

b. Chromosome 21 is a sex chromosome, and chromosomes 3 and 16 are not.

c. Down syndrome is not more common, just more serious.

d. Extra copies of the other chromosomes are probably fatal.

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