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Overview, from Ch 5, again
• In matings, precise phenotypic ratios are produced in descendants
as a result of chromosome segregation.
• In heterozygotes, alleles segregate equally into meiotic products.
• Progeny ratios can be predicted from known genotypes of
parents.
• Parental genotypes can be inferred from phenotypes of progeny.
• In many organisms, sex chromosomes determine sex.
• X-linked genes can show different phenotypic ratios in male and
female progeny.
• In humans, single-gene traits can be studied in pedigrees.
• Organelle genes are inherited maternally.
Testcross
• Distinguishes between A/A and A/a genotypes
mated to a/a based on phenotypes of offspring
– if all progeny are dominant phenotype, then unknown is
A/A
– if at least one offspring is recessive phenotype, then
unknown is A/a
If:
genotypic outcome phenotypic outcome
A/A × a/a
A/a
all A
A/a × a/a
1A/a:1a/a
1A:1a
Mendelian inference
P1(green) × P2(yellow)
↓
F1 (yellow)
parental generation
first filial generation
↓
selfing or intercrossing
F2 3 yellow:1 green
second filial generation
The disappearance of green in the F1 and its reappearance
in the F2 means that yellow is dominant. Also, because 3:1
ratios are typical in monohybrid crosses, the F1 must have
been heterozygous.
Autosomes and sex chromosomes
• Sex chromosomes determine sex in most animals
and some plants
– usually only one pair
– one sex has two alike (e.g., XX)
– one sex has two different types (e.g., XY)
• Remaining chromosomes are called autosomes
• X chromosomes and autosomes contain numerous
genes
• Y chromosomes typically have few genes
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Sex chromosome inheritance
in humans
• 46A XX is female, homogametic (only X gametes)
• 46A XY is male, heterogametic (X and Y gametes)
• Segregate equally into gametes at meiosis
Sperm
Gametes
Eggs
50% X
50% Y
50% X
XX
XY
50% X
XX
XY
X-linked inheritance (1)
• Male inherits Y from father and X from mother
• Female inherits one X from father and one X
from mother
• In most mammals, including humans
– Y chromosome has very few genes, but one
important gene is TDF (testis determining factor)
– X chromosome has no corresponding loci on Y
– Males express all recessive genes on X
chromosome (said to be hemizygous)
X-linked inheritance (2)
• Females: 3 possible X-linked genotypes
XAXA
XAXa
XaXa
• Males: 2 possible X-linked genotypes
XAY
XaY
An XAXa × XAY mating yields a 3A:1a phenotypic ratio, but the
recessive phenotype is restricted to 50% of male offspring.
Sperm
Eggs
Gametes
50% XA
50% XA
XAXA
50% Y
XAY
50% Xa
XAXa
XaY
2
Pedigrees (1)
•
•
•
•
Analysis of inheritance in families
Typically small number of offspring
Mendelian ratios rarely observed
Allow inferences concerning genotypes and
predictions concerning phenotypes of offspring
(genetic counseling)
unaffected
male
unaffected
female
affected
male
affected
female
Pedigrees (2)
• Two children, one of each sex, show the trait
• Conclusions:
– must be autosomal recessive trait
– parents must be heterozygous
– 2/3 chance that each unafflicted child is heterozygous
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Pedigrees (3)
• If gene for trait is very rare, this pedigree is most consistent
with X-linked recessive inheritance
• A single affected female would indicate autosomal
Categories of inheritance
• Autosomal recessive
– e.g., PKU, Tay-Sachs, albinism
• Autosomal dominant
– e.g., Huntington disease
• X-linked recessive
– e.g., color-blindness, hemophilia
• X-linked dominant
– e.g., hypophosphatemia
• Y-linked
• Organelle
Organelle inheritance
• Mitochondria and chloroplasts
– small number of genes on circular chromosome
– mostly inherited through maternal lineage via
egg cytoplasm
• Examples
– white green variegation in plants
– poky mutant in Neurospora
– suspected examples in humans
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Calculating probabilities
What is the probability that the
offspring in question will have
the trait?
?
The trait appears to be autosomal recessive and is
assumed to be rare. The probability that each grandmother
is heterozygous is 2/3; the probability that each parent of
the offspring is heterozygous is % x ½. If each is
heterozygous, the probability of an afflicted child is ¼.
Therefore, using the product rule, the probability that the
child will be afflicted is % x ½ x % x ½ x ¼ = 1/36
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