Lecture 20 Basics: Chromosomal inheritance
Laws of segregation and independent assortment
Definition: Law of segregation:
Monohybrid cross:
P
p
P
p
`
Genotypes:
Phenotypes:
Definition: Law of independent assortment:
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Dihybrid crosses:
Genotypes:
Phenotypes:
Sex-linked genes
Book definition:
Female
Male
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Explanations:
A man with hemophilia (a recessive , sex-linked condition) has a daughter of normal phenotype.
She marries a man who is normal for the trait.
1. What is the probability that a daughter of this mating will be a hemophiliac?
2. A son?
3. If the couple has four sons, what is the probability that all four will be born with
hemophilia?
Pedigrees
Fill in key:
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Determining a dominant trait:
Determining a recessive trait:
20.4 Barr bodies
Xa – active X chromosome
Xi – inactive X chromosome
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Two causes of Xi inactivation:
1. Methylation of
2. Activation of Xist gene
Recombination frequencies
Fruit fly genetics:
Crossing over can mix genes
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Recombination common:
Recombination rare:
Calculating a recombination frequency
Few recombinants –
Many recombinants –
Recombination frequency:
Textbook question:
A wild-type fruit fly (heterozygous for gray body color and normal wings was mated with a black
fly with vestigial wings. The offspring had the following phenotypic distribution: wild type, 778;
black-vestigial, 785; black-normal, 158; gray-vestigial, 162.
What is the recombination frequency between these genes for body color and wing type?
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Class Notes:
Learning Goal: Apply principles of chromosomal inheritance by creating sample exam questions.
The following is a list of question starters. Write your group names on the back of an index card,
and create an exam question from the starter you are assigned.
1. Create a dihybrid cross problem: Guinea pigs have short hair that is dominant to long hair
(Ss) and black hair dominant to brown hair (Bb).
2. Create a sex-linked problem using color blindness: Red-green color blindness (Xcb), normal
color vision (XCB)
3. Use this pedigree (or one similar) to create a problem:
4. Create a recombination frequency problem with the following phenotypes, and 100
recombinants. B+ b w+ w (black, winged) female is crossed with a b b w w (brown, wingless)
male. 1000 offspring.
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