12.3 Flashcards

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1
Q

Interpret data from testcrosses to infer unknown genotypes.

A

o Cross the individual with an unknown genotype (e.g. P_) with a homozygous recessive individual
o All offspring will be dominant if the unknown genotype was homozygous dominant
o If unknown genotype is heterozygous, then some of offspring will be homozygous recessive

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2
Q

What is polygenic inheritance

A

 Occurs when multiple genes are involved in controlling the phenotype of a trait
 The phenotype is an accumulation of contributions by multiple genes
 These traits show continuous variation and are referred to as quantitative traits

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3
Q

What is pleiotropy

A

 Refers to an allele which has a >1 effect on phenotype
* Effects are difficult to predict because a gene that affects 1 trait often performs other, unknown functions
 Can be seen in human diseases such as cystic fibrosis or sickle cell anemia
* Multiple symptoms can be traced back to one defective allele

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4
Q

Explain how ABO blood type is an example of multiple alleles

A

 A blood type can have a genotype of IAIA or IAi
 B blood type can have a genotype of IBIB or IBi

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5
Q

Explain how ABO blood type is an example of codominance

A

 AB blood type has a genotype of IAIB
 IAIB individuals express both the A carbohydrate and the B carbohydrate

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6
Q

Incomplete dominance

A

 Heterozygote is intermediate in phenotype between the 2 homozygotes
* Crossing black cats to white cats produces grey F1

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7
Q

Codominance

A

 Heterozygote shows some aspect of the phenotypes of both homozygotes
* Crossing black cats to white cats produces spotted F1

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8
Q

Provide examples of environmental influence on gene expression

A

o Coat color in Himalayan rabbits and Siamese cat
 Allele produces an enzyme that allows pigment production only at temperatures below 33ºC

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9
Q

Discuss how epistasis changes Mendelian ratios

A

o Gene at one locus alters the phenotypic expression of another gene at a second locus
 Example
* When one gene depends on another to produce its phenotype
* B=black and b=brown but E=deposits pigment while e=does not deposit pigment so when you have B_ee, pigment is not deposited and you get a yellow dog

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