PATTERNS OF INHERITANCEChapter 23

23-1

Gregor Mendel 1822 – 1884 Austrian monk Experimented

with garden peas Provided a basis

for understanding heredity

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Mendel cont’d23-3

Published a paper in 1866 stating that parents pass discrete heritable factors on to their offspring Factors retain individuality generation after

generation Identified that each trait is inherited by a pair

of factors, one from each parent One form of a factor may be dominant over an

alternative form Reasoned that each egg and sperm must contain

only 1 copy of a factor for each trait

Mendel cont’d.23-4

Mendel’s law of segregation Each individual has two factors (genes) for

each trait The factors segregate (separate) during the

formation of gametes Each gamete contains only one factor from

each pair of factors Fertilization gives each new individual 2 factors

for each trait

Modern Genetics – Genes23-5

Sections of chromosomes which give instructions for one characteristic or protein.

Located at the same point or locus, on each member of a homologous pair

All together make up the organism’s genome. Controls the physical characteristics of a

species

Modern Genetics - Alleles23-6

Alternative forms of the same gene on each chromosome One allele comes from each parent

Dominant allele Masks other traits present Only 1 dominant allele needs to be present for a

certain trait to be expressed Represented by a capital letter

Recessive Allele 2 copies of the recessive allele need to be present

for trait to be expressed Represented by a lower case letter

Modern Genetics – Alleles cont’d

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Gene locus

Fig. 23.2

Modern Genetics – Alleles cont’d Genotype

Genetic composition of a specific trait Homozygous dominant – 2 dominant alleles Heterozygous – 1 dominant allele and 1 recessive allele Homozygous recessive 2 recessive allele

Phenotype Physical expression of a specific trait

Homozygous dominant or heterozygous Dominant Trait

Homozygous recessive recessive trait

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Single Gene Inheritance23-9

Simplest situation One gene carries all the information

responsible for one trait Widow’s Peak

Alternative forms of alleles for hairline shape

Widow’s peak is dominant to straight hair line W=allele for widow’s peak w= allele for straight hairline

Widow’s peak23-10

Fig. 23.3

Genotype related to phenotype

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Table 23.1

Single Gene Inheritance cont’d. Monohybrid cross

Looks at inheritance of one trait only A punnett squares used to find all possible

combinations of alleles.

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Single Gene Inheritance cont’d. Example 1:

If a homozygous woman with a widow’s peak reproduces with a man with a straight hairline, what kind of hairline will their children have?

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Single Gene Inheritance cont’d. Example:

If two heterozygous parents reproduce what kind of hairline will their children have?

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W wW WW Www Ww ww

Single Gene Inheritance cont’d.

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Genetic Ratios Express ratio’s of possible outcomes

Genotypic ratio Homozygous Dominant: Heterozygous:

Homozygous Recessive Phenotypic ratio

Dominant trait : recessive trait Often expressed as probability The probability resets and is the same for each

pregnancy! Having one child with a trait has no effect on

future children.

Single Gene Inheritance cont’d.

• Look back to example 1– What are the

genotypic and phenotypic ratio’s?

– WW x ww• Genotypic Ratio

– 0 : 4 : 0• Phenotypic Ratio

– 1:0– 100% Widow’s peak

– Ww x Ww• Genotypic Ratio

– 1: 2: 1• Phenotypic Ratio

– 3 : 1– Probability is ¾

Widow’s Peak» 75%

– Probability ¼ Straight» 25%

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Single Gene Inheritance cont’d

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Determining Genotype No way to distinguish between a homozygous

dominant individual and a heterozygous individual just by looking They are phenotypically the same

Test cross may help us determine Used in breeders of plants and animals Cross unknown with a recessive individual

We know one parent genotype, this will help us determine the other genotype

If there are any offspring produced with the recessive phenotype, then the dominant parent must be heterozygous

Single Gene Inheritance cont’d Example :

In rats, large ears is dominant to small ears. A rat breeder has a female rat with large ears, which she breeds with a male rat with small ears. In the first litter, all rats are born with large ears. What is the genotype of the female rat?

In a second litter from the same parents, 4 baby rats have large ears, one has small ears. What is the genotype of the female rat?

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Single Gene Inheritance cont’d

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Practice problems Both a man and a woman are heterozygous for

freckles. Freckles are dominant over no freckles. What is the chance that their child will have freckles?

Both you and your sibling have attached ear lobes, but your parents have unattached lobes. Unattached earlobes (E) are dominant over attached (e). What are the genotypes of your parents?

A father has dimples, the mother of his children does not, and all 5 of their children have dimples. Dimples (D) are dominant over no dimples (d). Give the probable genotypes of all persons concerned.

Homework (WHAT??? It’s BIO!) Bikini Bottom Beach Genetics

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Independent Assortment23-21

Mendel reasoned from the results of his pea plant crosses that each pair of factors assorts independently into gametes Each trait is passed down individually. The

allele you receive for any one gene is not related to any other alleles you receive. We can now explain this through independent

alignment and crossing over in meiosis. (Fig 23.6)

Independent Assortment cont’d

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Fig. 23.6

Independent Assortment cont’d Called the Law of Independent

Assortment Each pairs of factors assorts independently

(without regard to how the others separate) All possible combinations of factors can occur

in the gametes

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Independent Assortment cont’d

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Practice problems For each of the following genotypes, give all

possible gametes WW WWSs Tt Ttgg AaBb

For each of the following, state whether the genotype or a gamete is represented D Ll Pw LlGg

Dihybrid Cross Punnet squares, considering two-trait crosses

at one time. Example:

The traits for hairline and finger length are both single gene traits. As before, widows peak is dominant over straight hairline. Having short fingers is considered dominant over long fingers.

Two parents who are both heterozygous for both traits have children. Determine the genotypic and phenotypic ratios for their

children.

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Dihybrid cross cont’d

Possible Gametes?

Genotypic Ratio?

Phenotypic Ratio?

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Dihybrid Crosses cont’d

Determining Ratio’s Product rule of probability

The chance of 2 or more independent events occurring together is the product of their chance of occurring separately

In our example: Probability of widow’s peak = ¾ Probability of short fingers= ¾ What is the probability of widow’s peak AND

short fingers? ¾ x ¾ = 9/16

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Dihybrid Crosses cont’d23-28

Recall from our single trait crosses Probability of widow’s peak = ¾

Probability of short fingers= ¾ Probability of straight hairline= ¼

Probability of long fingers= ¼

Using the product rule Probability of widow’s peak and short fingers = X =

Probability of widow’s peak and long fingers = X =

3/16 Probability of straight hairline and short fingers = ¼ X

¾ = 3/16 Probability of straight hairline and long fingers = ¼ X ¼

= 1/16

These values are standard for all heterozygous crosses! You don’t need to memorize them, but should be able to figure them out in your head!

Dihybrid Crosses cont’d23-29

Using the product rule Probability of widow’s peak and short fingers

Probability of widow’s peak and long fingers

Probability of straight hairline and short fingers

Probability of straight hairline and long fingers

Dihybrid Crosses cont’d23-30

Two-trait test cross Cross an individual with the dominant

phenotype for each trait with an individual with the recessive phenotype of both traits W?S? x wwss

WS W? ?S ??wws

sWwSs Ws?s ?wSs ?w?s

Dihybrid Cross cont’d23-31

Attached earlobes are recessive, What genotype do children have if one parent is homozygous for earlobes and homozygous dominant for hairline, and the other is homozygous dominant for unattached earlobes and homozygous recessive for hairline?

If an individual from this cross reproduces with another of the same genotype, what are the chances that they will have a child with a straight hairline and attached earlobes?

A child who does not have dimples or freckles is born to a man who has dimples and freckles (both dominant traits) and a woman who does not. What are the genotypes of all persons concerned?

Polygenic Inheritance23-32

Not fully understood by geneticists. Generally:

One trait controlled by 2 or more genes at different loci

The higher the number of dominant alleles you possess, the stronger the expression of the trait.

Result is a continuous range of phenotypes Distribution resembles a bell curve The more gene pairs involved, the more continuous

the pattern of variation Ex: human height, skin pigmentation, eye colour

Polygenic inheritance cont’d Fig. 23.9

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Polygenic Inheritance cont’d

Skin color Controlled by many

gene pairs and many alleles

Let’s assuming a simple model of two alleles at 2 loci A and B If two heterozygous

parents have children, children can range from very light to very dark

Genotype PhenotypeAABB Very Dark SkinAABb or AaBB Dark SkinAaBb, AAbb, or aaBB

Medium brown skin

Aabb, or aaBb Light Skinaabb Very light skin

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Polygenic Inheritance cont’d Eye colour is

controlled by 3 genes we have identified We suspect there

are more Not a clear

dominant and recessive

Brown allele

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Environmental Influences on Inheritance

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Environment can influence gene expression and therefore phenotype Ex: sunlight exposure on skin; coat color in

Himalayan rabbits Human twin studies

Polygenic traits are most influenced “nature vs. nurture” Identical twins separated at birth are studied

If they share a trait in common even though raised in different environments, it is likely genetic

Coat color in Himalayan rabbits

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Fig. 23.10

Incomplete Dominance Incomplete dominance

Heterozygous individual has a phenotype intermediate to the two homozygous individuals

Ex: Curly-haired Caucasian woman and a straight-haired Caucasian man produce wavy-haired children When 2 wavy-haired people have children, the

phenotypic ratio is 1 curly: 2 wavy: 1 straight

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Incomplete dominance23-39

Fig. 23.11

Codominance Multiple allele inheritance

The gene exists in several allelic forms, but each person still has only 2 of the possible alleles

Occurs when both alleles are equally expressed Ex: type AB blood has both A antigens and

B antigens on red blood cells

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Codominance cont’d. ABO blood types

IA = A antigens on RBCs

IB = B antigens on RBCs

i = has neither A nor B antigens on RBCs

Both IA and IB are dominant over I, IA and IB are codominant

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Phenotype

Genotype

A IAIA or IAiB IBIB or Ibi

AB IAIBO ii

Codominance cont’d. Paternity testing- ABO blood groups

often used Can disprove paternity but not prove it

Rh factor- another antigen on RBCs Rh positive people have the antigen Rh negative people lack it

There are multiple alleles for Rh negative, but all are recessive to Rh positive

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Inheritance of blood type23-43

Fig. 23.12

Practice Problems A polygenic trait is controlled by three pairs of alleles.

What are the two extreme genotypes for this trait? What is the genotype of the lightest child that could result

from a mating between two medium-brown individuals? A child with type O blood is born to a mother with type A

blood. What is the genotype of the child? The mother? ‘what are the possible genotypes of the father?

From the following blood types determine which baby belongs to which parents: Baby 1 type O Mrs. Doe type A Mrs. Jones type

A Baby 2 type B Mr. Doe type A Mr. Jones type AB

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Sex-linked inheritance Sex chromosomes

22 pairs of autosomes, 1 pair of sex chromosomes X and y

In females, the sex chromosomes are XX In males, the sex chromosomes are XY

Note that in males the sex chromosomes are not homologous

Traits controlled by genes in the sex chromosomes are called sex-linked traits X chromosome has many genes, the Y chromosome

does not

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Sex-linked inheritance cont’d. X-linked traits

Red-green colorblindness is X-linked The X chromosome

has genes for normal color vision

XB = normal vision Xb –

colorblindness

Genotypes

Phenotypes

XBXB female with normal vision

XBXb carrier female, normal vision

XbXb colorblind femaleXBY male with normal

visionXbY colorblind male

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Cross involving an X-linked allele

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Fig. 23.13

Practice Problems Both the mother and the father of a colorblind male appear to

be normal. From whom did the son inherit the allele for colorblindness? What are the genotypes of the mother, father, and the son?

A woman is colorblind. What are the chances that her son will be colorblind? If she is married to a man with normal vision, what are the chances that her daughters will be colorblind? Will be carriers?

Both the husband and the wife have normal vision. The wife gives birth to a colorblind daughter. Is it more likely the father had normal vision or was colorblind? What does this lead you to deduce about the girl’s parentage?

What is the genotype of a colorblind male with long fingers is s=long fingers? If all his children have normal vision and short fingers, what is the likely genotype of the mother?

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Inheritance of linked genes The sequence of individual genes on a

chromosome is fixed because each allele has a specific locus

All genes on a single chromosome form a linkage group When linkage is complete, a dihybrid produces only 2

types of gametes Any time traits are inherited together, a linkage group

is suspected If very few recombined phenotypes appear in offspring,

linkage is also suspected

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Inheritance of linked genes Crossing over

between 2 alleles of interest, can result in 4 types of gametes

Occurrence of crossing over can indicate the sequence of genes on a chromosome More frequent

between distant genes

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Fig. 23.14

Practice Problems23-51

When AaBb individuals reproduce, the phenotypic ratio is about 3:1. What ratio was expected? What may have caused the observed ratio?

The genes for ABO blood type and for fingernails are on the same homologous pair of chromosomes. In an actual family, 45% of offspring have type B blood and no fingernails, and 45% have type O blood and fingernails; 5% have type B blood and fingernails, and 5% have type O blood and no fingernails. What process accounts for the recombinant phenotypes?