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Genetics

Unit 9

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Genetics

The study of heredity Through heredity, living things inherit traits from

their parents

Traits are physical characteristics

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Basic concepts

Dominant - allele represented by uppercase letters(e.g., Bb, BB). The dominant trait is visible inheterozygous individuals.

Recessive - allele represented by lowercase letters(e.g., B b , bb ). The recessive trait is not visible inheterozygous individuals. Two recessive alleles (bb)are needed to express the trait.

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Basic concepts

P generation - the first generation orparents in a genetic cross

F1 generation - the offspring of the parents, orthe first generation of offspring in a geneticcross

F2 generation - the second generation of

offspring in a genetic cross; the children of the F 1 generation, or the grandchildren of the original parental (P) generation.

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Basic concepts

Homozygous individual with identical alleles (e.g.,GG or gg) for a gene

Heterozygous - individual with two different alleles

(e.g., Gg) for a gene Genotype - the genetic content of an individual,

letters used to denote alleles (BB, Pp)

Phenotype - the external appearance of an individual(brown, purple)

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Can you always predict the genetic content

(genotype) of an individual on the basis of itsexternal appearance (phenotype)?

Not always Individuals exhibiting the recessive traits (homozygous recessive) Individual with the dominant trait could be:

either homozygous dominant or heterozygous

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Punnett square

Used to determine the PROBABILITY of having a certain type of offspring given the alleles of the parents

Black rat (BB) and white rat (bb) true-breeding parents

Bb Bb

Bb Bb

BB Bb

Bb bb

All F 1 is heterozygous (Bb), soall rats are black

The ratio of phenotype is 3:1, threeblack rats , one white rat

1 Homozygous dominant, 2

heterozygous, 1 homozygous recessive

B B

b

bB b

B

b

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Monohybrid cross A cross between parents who are heterozygous, for example:

Tt x Tt

The genotypic ratio is one homozygous dominant: two heterozygous: onehomozygous recessive

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It is used to determine if an individual exhibiting adominant trait is homozygous or heterozygous for thattrait.Test crosses involve breeding the individual in question

with another individual that expresses a recessive version of the same trait.

If all offspring display the dominant phenotype, the

individual in question is homozygous dominant; if the offspring display both dominant and recessivephenotypes, then the individual is heterozygous.

Test Cross

http://en.wikipedia.org/wiki/Dominant_traithttp://en.wikipedia.org/wiki/Homozygoushttp://en.wikipedia.org/wiki/Heterozygoushttp://en.wikipedia.org/wiki/Recessivehttp://en.wikipedia.org/wiki/Offspringhttp://en.wikipedia.org/wiki/Homozygoushttp://en.wikipedia.org/wiki/Homozygoushttp://en.wikipedia.org/wiki/Offspringhttp://en.wikipedia.org/wiki/Recessivehttp://en.wikipedia.org/wiki/Heterozygoushttp://en.wikipedia.org/wiki/Homozygoushttp://en.wikipedia.org/wiki/Dominant_trait

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In the fruit fly , black body (b) is recessive to the normal gray body (B ). You are

given a male with a gray bod y . Determine its genotype.

Procedure : The male has one of two genotypes : B B or B b . This male is crossed

to a black-bodied female of genotype bb . There are two possible outcomes:

Test Cross

Outcome 1: Male is Heterozygous (Bb)

B b

b

b

Bb

bbBb

bb

The progeny of this cross were 50%gray body and 50% black body reflecting the type and frequency of their father's gametes.

Outcome 2: Male is Homozygous (BB)

B B

b

b

Bb

BbBb

Bb

The progeny of this cross were 100%gray heterozygotes ( Bb ) reflecting thetype and frequency of their father'sgametes.

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Cystic fibrosis is a recessive trait that causes a build-up of mucusin the lungs. A woman who has cystic fibrosis mates with a manwho does not have the condition. Their first child has cysticfibrosis. What is the probability that their next child will havecystic fibrosis?

Mom is homozygous recessive cc

C cc ccCc 50%

Dad is heterozygous - Cc

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WR Wr wR wr

wR

Dihybrid exampleIn squash, white fruit (W) is dominant to yellow (w) and round

shaped fruit (R) is dominant to flat (r) shaped fruit. If you crossa true-breeding yellow, round squash with one that isheterozygous for both color and shape, what would be theexpected phenotypic ratios of the offspring?

Genotypes: wwRR x WwRr Gametes: wR WR, Wr, wR, wr

WwRR WwRr wwRR wwRr

50% White, Round (WwRR, WwRr)

50% Yellow, Round (wwRR, wwRr)

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In corn snakes, the presence of black pigment (M) is dominant toalbino (m) and blotched patterns (B) are dominant to motleypatterns (b). If an albino motley snake mated with a snakeheterozygous for both traits and they had a total of 20hatchlings, how many hatchlings would be albino and blotched?

mmbb X MmBb

MB Mb mB mb

mb X

1/4= 0.25

0.25 X 20= 5

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Some hints to remember

Monohybrid Cross: When 2 heterozygotes arecrossed, phenotypic ratio of offspring is always3:1 (A_:aa)

Dihybrid Cross: When 2 heterozygotes arecrossed, phenotypic ratio of offspring is always9:3:3:1 (each phenotype is a fraction out of /16)

9/16 A_B_: 3/16 A_bb: 3/16 aaB_: 1/16 aabb

20

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Pedigree analysis

Graphic method to present the inheritance of particular traits in a group of relatedindividuals

Female

Male

Parents

Siblings

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Pedigree analysis

First determine whether the trait is due to adominant or recessive allele

If trait skips a generation, it must be caused by arecessive allele (hidden in the heterozygote)

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Pedigree analysis

Dominant alleles can not be hidden, so they can notskip generations

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Pedigree analysis Look for parents of the same phenotype If they produce offspring of the opposite phenotype, then

parents must be heterozygous If they were homozygous they could not produce an unlike

offspring In such case, if parents have the trait, is due to a dominant

allele If they lack the trait, it is due to a recessive allele

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Tips for the Pedigree:

-if two parents have the trait, but none of their kids have it,the trait MUST be dominant AND both parents MUST beheterozygous

-if you see two parents that do not have the trait, but their kid(s) have it, the trait is recessive AND both parentsMUST be heterozygous

-if a trait shows up in a grandparent and a grandchild, but not the parents, the trait is recessive (skipping ageneration)