Honors Biology Ch. 9 Notes

Honors Biology, Period 5 Chapter 9 Notes Patterns of Inheritance Section 9.1: The Science of Genetics Has Ancient Roots Hippocrates – proposed theory called Pangensis. **key – acquired traits are inherited Aristotle – proposed that we inherit the potential to produce features 19th Century Biologists – Uniformity accepted the Blending Theory of inheritance. **Traits from both parents “mix” and become inseparable. Section 9.2: Experimental Genetics Began in an Abbey Garden Gregor Mendel – ‘The Father of Genetics’ - An Augustinian monk - Lived and worked in an Abbey in Brunn, Austria - Did all of his research on pea plants - Published research paper (in 1866) outlining the mechanism of inheritance Advantages of Pea Plants - Mendel was very familiar with pea plants from rural upbringing - They are very easy to grow - They come in many easily distinguishable varieties - It is easy to strictly control their mating About Pea Plants - They contain male (stamen) and female (carpel) *Figure 9.2B - They tend to self-fertilize in nature - Mendel could promote cross-fertilization if desired *Figure 9.2C Mendel’s Success was based on three key factors 1. His experimental approach (application of mathematics) 2. Choice of organism 3. Limited study to dichotomous traits Key Terms Part I 1. Dichotomous Trait – trait with only 2 distinct alternative forms 2. Self-fertilize – organism’s sperm fertilizes same organism’s egg

3. Cross-fertilize – one organism’s sperm fertilizes a second organism’s egg 4. True-breeding (pure-breeding) – organisms for which self-fertilization produces offspring identical to the parent 5. Hybrid – offspring of two varieties 6. Hybridization (cross) – mating between parents of different verieties 7. Filial – Latin term meaning ‘son’ 8. Phenotype – physical appearance Section 9.3: Mendel’s Law of Segregation Describes the Inheritance of a Single Trait - Figure 9.2D - The 7 pea plant characteristics studied by Mendel - Figure 9.3A – Mendel’s pea plant experiment Mendel’s Observations Resulted in 4 Hypothesis: 1. There are alternative forms of genes that account for variations in inherited characteristics. These versions of a gene are called alleles. 2. For each characteristic, an organism inherits 2 alleles, one from the mother and one from the father. These alleles may be the same or different. o homozygous – when both alleles for a gene are the same o heterozygous – when the two alleles for a gene are different *pure-breeding = homozygous (lol, hot potato) 3. If the two alleles are different, than one determines the organism’s phenotype (appearance) and the other has no noticeable effect on the organism’s appearance. o Dominant – describes the allele that is expressed o Recessive – describes the allele this is not expressed 4. The Law of Segregation – A gamete carries only one allele for each inherited trait because allele pairs separate (segregate) during the production of gametes (meiosis). During what meiotic stage does this happen? – Anaphase I Figure 9.3B – Explanation of Mendel’s observations Using a Punnet Square -

Punnet Square – tool used to calculate genotypes and phenotypes of potential offspring

Dihybrid Cross -

A cross that involves two different traits.

Example #1: A pea plant that is heterozygous for flower color and heterozygous for flower height is allowed to self-fertilize. A. What is the expected phenotypic ratio of the offspring? B. What is the probability that the offspring will be pure-breeding for both purple flowers and dwarf height? Alleles: P – Purple flower color p – White flower color T – tall plants t – dwarf plants Genotypes

Possible Gametes

Male – P p T t

PT

Pt

pT

pt

Female – P p T t

PT

Pt

pT

pt

Punnett Square PT

PT Pt pT

Pt

pT

pt

PPTT

PPTt

PpTT

PpTt

PPTt

PPtt

PpTt

Pptt

PpTT

PpTt

ppTT

ppTt

PpTt

Pptt

ppTt

pptt

pt # Tall Purple: # Tall White: # Dwarf Purple: # Dwarf White:

9/16 3/16 3/16 1/16

Phenotypic ratio: 9 tall purple : 3 tall white : 3 dwarf purple : 1 dwarf white Answers: A. Phenotypic ratio: 9 tall purple : 3 tall white : 3 dwarf purple : 1 dwarf white B. 1/16 Section 9.6: Using A Test Cross to Determine an Unknown Genotype Test cross – mating between an individual of unknown genotype and an individual that is homozygous recessive - When using a test cross, you have to observe the offspring to determine the unknown parental genotype. Example: In Labrador Retrievers, black coat color is dominant to chocolate coat color. You adopt a black lab from Animal Friends. A. What is the genotype of this lab? B. Perform a test cross to determine the unknown genotype. B – black coat color b – chocolate coat color Genotypes Male: B _ Female: bb

Possible Gametes B? bb

Punnett Square Option #1 b b B Bb

Bb

Bb

Bb

B

100% black

Option #2 b b B Bb

Bb

bb

bb

b

1 : 1 ; 50% black, 50% chocolate

Result – when mated, the labs produced a litter of puppies: 5 black, 4 chocolate - Genotype of black lab: Bb Section 9.7: Mendel’s Laws Reflect the Rules of Probability Rule of multiplication – the probability of compound events is the product of the separate Probabilities of the independent events *Also referred to as “AND” problems Example: A pea plant is heterozygous for 3 traits: A, B, and C. This plant is allowed to self-fertilize. Genotypes of parents: Male: AaBbCc Female:AaBbCc 1. What is the chance that the offspring would have the genotype AABBCC? - 1/64 2. What is the chance that the offspring would have the genotype AaBBcc? - 1/32 Rule of addition – probability that an event can occur in two or more separate occasions is the sum. *Also referred to as “OR” problems Example: A pea plant is heterozygous for two traits: A and B. This plant is allowed to self-fertilize. Genotypes: Male: AaBb Female:AaBb 1. What is the probability that the offspring will have a genotype of AABB? - ¼ + ¼ = ½ or 50% Section 9.8: Pedigrees Pedigree – a family tree (unshaded circle) – female (unshaded square) – male (shaded circle showing defect/disease) – female (shaded square showing defect/disease) – male carrier – an individual that is heterozygous for a recessive genetic disorder. (half-shaded

circle or square). They appear normal. Example – in humans, the hearing gene has two alleles. There is a dominant allele produces normal hearing while the recessive allele deafness. D – normal hearing d – deafness **See Figure 9.8B Section 9.9: Many Inherited Disorders Are Controlled By A Single Gene Testing Your Knowledge What is the difference between an infectious disease and a genetic disease? - infectious are caused by living organisms, genetic are inherited Inherited Disorders -

Disorders arising from a defective allele. Inherited disorders can be either dominant or recessive.

Recessive Disorders 1. 2. 3. 4.

Albinism Cystic Fibrosis Sickle Cell Anemia Tay-Sach’s Disease About Cystic Fibrosis: - Most common lethal genetic disease in the United States. - Affects different ethnic groups unequally: 1/17000 African 1/90000 Asian 1/2500 European *about 1/25 Caucasians carry the allele Why does the frequency of the disease vary between cultures? - Intrabreeding About Sickle Cell Anemia - What is it? - Who is at risk? - Carriers are said to have sickle cell trait - Evolutionary connection – the heterozygous advantage

Dominant Disorders 1. Achondroplasia (the homozygous condition is fatal to the embryo) 2. Alzheimer’s Disease 3. Huntington’s Disease 4. Hypercholesterolemia 5. Polydactylism 6. Syndactylism (Christ and Da Vinci) 7. Ectrodactylism (lobster boy) Section 9.10 – 9.11 *read in book Section 9.12: Incomplete Dominance Variations and Exceptions to Mendel’s Laws Exception #1 Incomplete Dominance – A situation where neither allele is dominant. The result is the offspring show phenotypes in between the parental varieties. Example: A pure-breeding RED snapdragon is crossed with a purebreeding WHITE snapdragon R – red flower color r – white flower color P Generation – RR x rr F1 – 100% pink flower F2 – What phenotype ratio do you expect? Actual results: 11 RED : 20 PINK : 9 WHITE Reduced: 1 RED : 2 PINK : 1 WHITE

Section 9.13: Multiple Alleles and Codominance Exception #2 Multiple Alleles – situation where a single gene has more than two possible forms Codominance – situation where more than one allele for a certain gene is dominant Example: Blood type in humans is controlled by a single gene.

This gene has three different alleles Alleles A and allele B are codominant Allele O is recessive Writing alleles: IA – the A allele IB – the B allele i – the O allele Blood Phenotypes and Genotypes: Phenotypes Genotypes Blood type A IA IA or IAi Blood type B IB IB or IBi Blood type AB IA IB Blood type O ii Section 9.14: Pleiotropy Exception #3 Pleiotropy – when a single gene influences more than one trait Example: Sickle Cell Anemia The sickle cell gene can cause: A. Anemia B. Heart failure C. Brain damage D. Spleen damage Exception #4 - **NOT IN TEXTBOOK Epistasis – when multiple genes influence the same trait Section 9.15: Polygenic Inheritance Exception #5 Polygenic Inheritance – the additive effects of two or more genes on a single trait Examples: 1. human skin color 2. human height Example of skin color: **In humans, skin color is controlled by at least 3 different genes, but for this sample problem, use only two different genes.

In humans, skin color is controlled by 2 different genes: A, B. Each allele contributes a unit of darkness to skin color. very light skin light skin medium skin dark skin very dark skin

aabb Aabb or aaBb AAbb or aaBB or AaBb AABb or AaBB AABB

If a man with light skin marries a woman with very light skin, what is the expected phenotypic ratio of their offspring? Very light: aabb Light: Aabb a A

Aa

½ × 4/4 = ½ very light skin, ½ light skin

a

b

Aa X

A aa

aa

b

Bb

Bb

b bb

bb

B

Sections 9.16 – 9.18 *read in book Section 9.19: Linked Genes Exception #6 Linked Genes – genes that are located close together on the same chromosome Example: In sweet peas, Purple flower color (P) is dominant to red (p) flower color and long seeds (L) are dominant to round seeds (l). If two heterozygous plants are crossed, what phenotypic ratio would you expect in the offspring? Expected Phenotypic ratio: 9/16 Purple Long : 3/16 Purple round : 3/16 red Long : 1/16 red round Actual Phenotypic Ratio: 20 Purple Long : 7 red round Ratio – 3 Purple Long : 1 red round How do you explain these results?

- The gene for flower color and the gene for seed shape are on the same chromosome. This means they DO NOT assort independently during Meiosis. *See figure 9.19 Recombination Frequency - percentage of recombinants ( = recombinants/total offspring) Example: Drosophila melanogaster – common fruit fly G – gray body g – black body

L – long wings l – vestigial wings

GgLl (male) x ggll (female)  test cross: one parent is pure-breeding (homozygous) If genes assort independently, what phenotypic ratio is expected? - 1 gray long : 1 gray vestigial : 1 black long : 1 black vestigial If linked: GL GL GGLL

Gl

GgLl

gl GgLl

3 Long Gray : 1 black vestigial expected

ggll

Actual results: (See figure 9.19 C) Using Recombination Frequency to Map Genes Example: G – Gray body g – black body

P – Purple eyes p – red eyes

L – Long wings l – vestigial wings

GPL – linked on the same chromosome gpl – linked on the same chromosome The Recombination Frequency: Gray body and red eyes – 17% Purple eyes and vestigial wings – 31% Gray body and vestigial wings – 14% What is the sequence of genes on this chromosome?

31 map units

P

G

L

14 map 17 map units units Recombination frequency is given below for several gene pairs. Create a linkage map for these genes, showing the map unit distance between loci. j, k – 12%

k, l – 6%

j, m – 9%

9 map units

M

l, m – 15%

12 map units

J 15 map units

L

K 6 map units

21 map units Section 9.22 – 9.24: Sex-Linked Genes Exception #7 Sex-linked genes – genes that are found on either of the sex chromosomes rather than the autosomes Why/How does this change the inheritance pattern of these traits? Chromosomal Make-up Males – XY  males only have one X (hemizygous) Females – XX  females do not have a Y Examples of sex-linked gene disorders: 1. colorblindness (recessive) 2. Hemophilia (recessive) 3. Duchenne Muscular Dystrophy (recessive)

Example Problem: Joe (normal vision) marries Layla (normal vision). They have a son, Chad, who is colorblind. What is the chance their next child will be colorblind?  ¼ If their next child is a boy, what is the chance he will be colorblind?  ½ What is the chance their next child will be a boy with normal vision?  ¼ Alleles – B – normal vision b – colorblindness

Genotypes: Joe - XBY Layla – XBXb Chad - XbY XB XB XB XB B YX Y

Xb XB Xb XbY

Joe x Layla