Patterns of Inheritance

Pre-AP Biology

Ms. Haut

Modern Theory of Heredity

• Based on Gregor Mendel’s fundamental principles of heredity– Parents pass on discrete inheritable factors

(genes) to their offspring – These factors remain as separate factors from

one generation to the next

Experimental genetics

•Began with Gregor Mendel’s quantitative experiments with pea plants

• These plants are easily manipulated.• These plants can self-fertilize.

Figure 9.3Figure 9.2

Mendel’s Discoveries

• Developed true-breeding lines—populations that always produce offspring with the same traits as the parents when parents are self-fertilized

• Mendel then crossed two different true-breeding varieties.

• Counted his results and kept statistical notes on experimental crosses

Figure 9.4

Figure 9.5

Mendel’s Law of Segregation

• Mendel performed many experiments.

• 1st Law of genetics– The two members of

an allele pair segregate (separate) from each other during the production of gametes.

– Based on Mendel, we’ve developed four hypotheses from the monohybrid cross:

• There are alternative forms of genes, called alleles.• For each characteristic, an organism inherits two

alleles, one from each parent.• If 2 alleles differ, one is fully expressed (dominant

allele); the other is completely masked (recessive allele)

• Gametes carry only one allele for each inherited characteristic.

– 2 alleles for each trait segregate during gamete production

Useful Genetic Vocabulary

• Homozygous—having 2 identical alleles for a given trait (PP or pp)

• Heterozygous—having 2 different alleles for a trait (Pp); ½ gametes carry one allele (P) and ½ gametes carry the other allele (p)

• Phenotype—an organism’s expressed traits (purple or white flowers)

• Genotype—an organism’s genetic makeup (PP, Pp, or pp)

Monohybrid Crosses

x

x

x

x

x

x

x

Ratio3.15:1

3.14:1

3.01:1

2.96:1

2.95:1

2.82:1

2.84:1

3:1

Punnett Square

• Cross a heterozygous tall pea plant with a dwarf pea plant.

• T = tall, t = dwarf

Tt x tt

TtT

t

tt

t t

Tt Tt

tt tt

The Testcross

• The cross of an individual displaying the dominant phenotype to a homozygous recessive parent

• Used to determine if the individual is homozygous dominant or heterozygous

CAUTION:Must perform many, many crosses to be statistically significant Figure 9.10

Genetic Alleles and Homologous Chromosomes

– Homologous chromosomes• Have genes at specific loci.• Have alleles of a gene at the same locus.

Figure 9.7

Mendel’s Law of Independent Assortment

• During gamete formation, the segregation of the alleles of one allelic pair is independent of the segregation of another allelic pair– Law discovered by following segregation of 2

genes (dihybrid cross)

Dihybrid Cross

Figure 9.8

Gamete formation

AaBb

ABAbaBab

AABb ABAbABAb

AaBbCc

ABCABcAbCAbc

aBCaBcabCabc

Mendelian Inheritance Reflects Rules of Probability

• Rule of Multiplication: The probability that independent events will occur simultaneously is the product of their individual probabilities.

What is the probability that you will roll a 6 and a 4?

1/6 x 1/6 = 1/36 chance

• Question: In a Mendelian cross between pea plants that are heterozygous for flower color (Pp), what is the probability that the offspring will be homozygous recessive?

• Answer: Probability that an egg from the F1 (Pp) will

receive a p allele = ½ Probability that a sperm from the F1 will receive

a p allele = ½ Overall probability that 2 recessive alleles will

unite at fertilization: ½ x ½ = ¼

Mendelian Inheritance Reflects Rules of Probability

Mendelian Inheritance Reflects Rules of Probability

• Question: For a dihybrid cross, YyRr x YyRr, what is the probability of an F2 plant having the genotype YYRR?

• Answer: Probability that an egg from a YyRr parent will

receive the Y and R alleles = ½ x ½ = ¼ Probability that a sperm from a YyRr parent will

receive the Y and R alleles = ½ x ½ = ¼ Overall probability of an F2 plant having the

genotype YYRR: ¼ x ¼ = 1/16

Works for Dihybrid Crosses:

Mendelian Inheritance Reflects Rules of Probability

• Rules of Addition: The probability of an event that can occur in two or more independent ways is the sum of the separate probabilities of the different ways.

What is the probability that you will roll a 6 or a 4?

1/6 + 1/6 = 2/6 or 1/3 chance

Mendelian Inheritance Reflects Rules of Probability

• Question: In a Mendelian cross between pea plants that are heterozygous for flower color (Pp), what is the probability that the offspring will being a heterozygote?

• Answer: There are 2 ways in which a heterozygote may

be produced: the dominant allele may be in the egg and the recessive allele in the sperm, or the dominant allele may be in the sperm and the recessive allele in the egg.

Mendelian Inheritance Reflects Rules of Probability

• Probability that the dominant allele will be in the egg with the recessive in the sperm is ½ x ½ = ¼

• Probability that the dominant allele will be in the sperm with the recessive in the egg is ½ x ½ = ¼

• Therefore, the overall probability that a heterozygote offspring will be produced is ¼ + ¼ = ½

Variations to Mendel’s First Law of Genetics

• Incomplete dominance—pattern of inheritance in which one allele is not completely dominant over the other– Heterozygote has a phenotype that is

intermediate between the phenotypes of the homozygous dominant parent and homozygous recessive parent

Incomplete Dominance in Snapdragon Color

Genotypic ratio:

Phenotypic ratio:

1 CRCR: 2 CRCW: 1 CWCW

1 red: 2 pink: 1 white

F2

Figure 9.16

Variations to Mendel’s First Law of Genetics

• Codominance—pattern of inheritance in which both alleles contribute to the phenotype of the heterozygote

• Roan Cattle

In chickens, black feather color (BB) is codominant to white feather color (WW).  Both feather colors show up in a checkered pattern in the heterozygous individual (BW).

Cross a checkered hen with a checkered rooster. What are the genotypic and phenotypic ratios?

Example of Codominance

• Ex: Feather colors in chickens

• Black (BB) x White (WW) = Black and White checkered Chicken

B W

B

W

BB

WWBW

BW

Multiple Alleles

• Some genes may have more than just 2 alternate forms of a gene.

• Example: ABO blood groups– A and B refer to 2 genetically determined

polysaccharides (A and B antigens) which are found on the surface of red blood cells (different from MN blood groups)

• A and B are codominant; O is recessive to A and B

Multiple Alleles for the ABO Blood Groups

3 alleles: IA, IB, i

Figure 9.18

Blood Types• The immune system

produces blood proteins– That may cause

clotting when blood cells of a different type enter the body. Figure 9.19

http://www.biologycorner.com/resources/blood_type.jpg

Example of ABO Blood Groups

• Ex: Feather colors in chickens

• Black (BB) x White (WW) = Black and White checkered Chicken

IA IB

IA

i

IA IA

IB iIA i

IA IB

Pleiotropy

• The ability of a single gene to have multiple phenotypic effects (pleiotropic gene affects more than one phenotype)

Polygenic Traits

• Skin pigmentation in humans--3 genes with the dark-skin allele (A, B, C) contribute one “unit” of darkness to the phenotype.

• These alleles are incompletely dominant over the other alleles (a, b, c)--An AABBCC person would be very dark; an aabbcc person would be very light--An AaBbCc person would have skin of an intermediate shade

Figure 9.21

Polygenic Trait

Chromosome Theory of Inheritance

• Based on Mendel’s observations and genetic studies and cytological evidence– Genes are located at specific positions on

chromosomes.– The behavior of chromosomes during meiosis

and fertilization accounts for inheritance patterns.

Figure 9.23

– Certain genes are linked• They tend to be inherited

together because they reside close together onthe same chromosome

Experiment

Explanation: linked genes

PpLI PpLILong pollen

Observed PredictionPhenotypes offspring (9:3:3:1)

Purple longPurple roundRed longRed round

Parentaldiploid cellPpLI

Most gametes

Mostoffspring Eggs

3 purple long : 1 red roundNot accounted for: purple round and red long

Meiosis

Fertilization

Sperm

284212155

215717124

P I

P L

P L

P L

P L

P L

P I

P L P I

P I

P L

P I

P I

P I

P I

P L

Purple flower

Figure 9.19

Genes on the samechromosome tend tobe inherited together

– Crossing over can separate linked alleles• Producing gametes with recombinant

chromosomes• This happens in Prophase I of Meiosis

Crossing over produces new combinations of alleles

Figure 9.25

•Thomas Hunt Morgan – Performed some of the early studies of crossing over

using the fruit fly Drosophila melanogaster

•Experiments with Drosophila revealed linkage traits. Why Drosophila?

– Easily cultured– Prolific breeders– Short generation times– Only 4 pairs of chromosomes, visible under

microscope

Morgan’s experiments

•Demonstrated the roleof crossing over in inheritance

Figure 9.24

Morgan’s experiments

•Two linked genes– Can give rise to four

different gamete genotypes.

– Can sometimes cross over.

Figure 9.26

– Morgan and his students• Used crossover data to map genes in

Drosophila

Figure 9.21 A

Geneticists use crossover data to map genes

Linkage Map• Alfred Sturtevant

hypothesized that the frequency of recombinants reflected the distances between genes on a chromosome.– The farther apart two

genes are, the higher the chance of crossover between them and therefore a higher recombination frequency.

Copyright © 2002 Pearson Education, Inc., publishing as Benjamin Cummings

Can be used to map the relative positions of genes on chromosomes.

Figure 9.21 B

Mutant phenotypes

Shortaristae

Blackbody(g)

Cinnabareyes(c)

Vestigialwings(l)

Browneyes

Long aristae(appendageson head)

Gray body(G)

Redeyes(C)

Normalwings(L)

Redeyes

Wild-type phenotypes

Chromosomeg c l

9% 9.5%

17%

Recombinationfrequencies

Figure 9.21 C

Recombination frequencies

Copyright © 2002 Pearson Education, Inc., publishing as Benjamin Cummings

Fig. 15.5b

• Sturtevant used the testcross design to map the relative position of three fruit fly genes, body color (b), wing size (vg), and eye color (cn).– The recombination frequency between cn and

b is 9%.– The recombination frequency between cn and

vg is 9.5%.– The recombination frequency between b and

vg is 17%.– The only possible

arrangement of these three genes places the eye color gene between the other two.

Copyright © 2002 Pearson Education, Inc., publishing as Benjamin Cummings

Fig. 15.6

• Sturtevant expressed the distance between genes, the recombination frequency, as map units.– One map unit (sometimes called a

centimorgan) is equivalent to a 1% recombination frequency.

What is the sequence of these three genes on the chromosome?

• A series of matings shows that the recombination frequency between the black-body gene (b) and the gene for short wings (s) is 36%. The recombination frequency between purple eyes (p) and short wings is 41%. The recombination frequency between black-body gene and purple eyes is 6%.

Answer

B 36% SP 41% S

B 6% P

P 6% BB 36% S 6% + 36% = 42%P 41% S

• You may notice that the three recombination frequencies in our mapping example are not quite additive: 9% (b-cn) + 9.5% (cn-vg) > 17% (b-vg).

• This results from multiple crossing over events.– A second crossing over “cancels out” the first

and reduces the observed number of recombinant offspring.

– Genes father apart (for example, b-vg) are more likely to experience multiple crossing over events.

Copyright © 2002 Pearson Education, Inc., publishing as Benjamin Cummings

• Some genes on a chromosome are so far apart that a crossover between them is virtually certain.

• In this case, the frequency of recombination reaches is its maximum value of 50% and the genes act as if found on separate chromosomes and are inherited independently.

Copyright © 2002 Pearson Education, Inc., publishing as Benjamin Cummings

•If the recombination frequency is 50% or greater, the genes are not linked•If the recombination frequency is less than 50%, the genes are linked