CLASS: 11-12
DATE: 9-20-10
PROFESSOR: Pittler
Scribe: Angi Gullard
Proof: Kratika Pareek
Classical Genetics
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Please…Quell your Cell! [S1]
Classical Genetics — Lecture I [S2]
a. Most of material for the five genetics lectures is from: Lewis 7th Ed. Human Genetics: Concepts and
Applications, Ch 4: Mendelian Inheritance
Inheritance [S3]
a. What is inheritance? How do individuals in families have some resemblance, similar mannerisms? How are
traits inherited?
b. How does inheritance occur? Parents and offspring often share observable traits.
c. Grandparents and grandchildren may share traits that are not seen in parents; it may skip a generation.
d. Why do traits disappear in one generation and reappear in another?
Gregor Mendel: The father of modern genetics [S4]
a. Gregor Mendel was one of the first to ask question of why traits may skip a generation, and address it in
plants
b. Today, we will talk about Mendelian genetics
c. He combined plant breeding with statistics and careful recordkeeping, and described the transmission
processes of those traits.
d. He looked at simple traits - seeds, leaves, color of plants and looked at offspring when cross two different
types of plants
Mendel studied pea traits with two distinct forms [S5]
a. Mendel studied pea traits with two distinct forms
i. You see a dominant and recessive form for each of the seven traits – see slide for pea traits
True breeding plants [S6]
a. Read slide
b. Start with two short plants  all short offspring = true breeding
c. Same for all tall plants: two tall, get all tall plants
Monohybrid cross [S7]
a. In a monohybrid cross, if breed two talls, get all tall
b. What happens when true breeding plants with two distinct forms of a trait are crossed?
i. See only dominant trait, but progeny show only one form
1. The observed trait is called dominant.
2. The masked trait is called recessive.
Test cross [S8]
a. Is a plant (tall) showing the dominant trait true-breeding or not? Find out by doing a test cross…can do same
with fruit flies, mice…
b. Test by crossing with a plant showing the recessive trait (tt).
i. On left, 1 tall versus 1 short (tt) if get all tall offspring, means tall plant is homozygous (TT) and always
passes on tall allele (T), so all offspring plants will be tall (Tt)
1. All tall offspring indicate parent is true-breeding
ii. On right, 1 tall versus 1 short  get some tall and some short, because tall parent plant is heterozygous
(Tt), so the offspring may get one tall allele (T) and one short allele (t) or two short alleles (tt)
1. Mixed offspring indicate parent is hybrid, that there is a recessive allele that is hiding
Crossing hybrids to each other [S9]
a. Read slide
b. The short trait is present there, but only in recessive form does trait show as phenotype
Mendel’s data [S10]
a. Read slide – points 1-3
b. Mendel looked at thousands of plants and looked at seven traits….really big garden!
c. Ratio is 3:1 (dominant:recessive) for each trait
Untitled [S11]
a. Read slide
b. Can do Punnett square for 2 or even 7 traits
c. A 3 : 1 ratio means that there is a 75% chance of the dominant phenotype and a 25% chance of the recessive
phenotype. – this does not always have to occur this way, alleles are not always passed in that manner
Punnett Squares [S12]
a. Each representative allele goes in a box
Punnett Squares [S13]
a. Can look at segregation pattern
CLASS: 11-12
DATE: 9-20-10
PROFESSOR: Pittler
XIV.
XV.
XVI.
XVII.
XVIII.
XIX.
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XXI.
XXII.
XXIII.
Scribe: Angi Gullard
Proof: Kratika Pareek
Classical Genetics
Page 2 of 5
b. In the F1 generation – crossing of true-breeding tall and recessive parents yields all heterozygotes
Gamete Formation (sperm and eggs) [S14]
a. Because homologous pairs separate during meiosis, a gamete has only one allele from each pair of alleles.
b. Read slide
c. Could be disease-causing gene, causing dental/ocular disease, can follow that
One-Trait Crosses and Probability [S15]
a. Laws of probability alone can be used to determine results of a cross.
b. What is the likelihood trait will be passed onto offspring?
c. Read slide
Untitled [S16]
a. Read slide
b. The chance of Tw (should be Tt)
c. Adding all possible genotypes together, have 75% of getting dominant allele and having dominant phenotype
Law of segregation [S17]
a. Why do traits “disappear” in one generation only to reappear in a subsequent generation? Also called
generational skipping
b. Each plant possesses two distinct separable units (alleles), and that together makes up each individual unit of
a chromosome.
c. Together, 2 chromatids make up chromosome
d. Only one version is observed in an individual - this is not always true
i. Can have codominance
e. Assume Mendelian transmission – no allele mixing, only one version is seen in any individual
f. The unit (allele) does not disappear. It may be present but hidden.
a. What is hidden? Just do not see trait in offspring. It’s there, just not seen.
b. The recessive allele is passed on and but the dominant allele takes over.
c. The recessive allele can be passed on in next generation, so it shows up again in that generation – one
way of generational skipping
Gene locus (locus = location) [S18]
a. Shown here are two sister chromatids of homologous chromosomes
b. Can have replication, so chromatids are duplicated which are made during replication
c. Look at particular allele z – have recessive (z) and dominant (Z) forms
d. Homozygous (zz) and Heterozygous (Zz)
Law of segregation [S19]
a. Look at parent 1 Tt
i. Replication of homologous chromosomes during interphase of meiosis
ii. Get diploid genome instead of haploid genome
iii. Get segregation in meiosis
iv. Fertilization leads to combination of gametes, producing offspring with combination that occurs
b. How do you get gene diversity in an individual?
i. For each chromosome, need at least one DNA-exchange event, otherwise have non-functional organism
Alleles [S20]
a. Read slide
b. What makes it an allele?
i. One DNA change, one base difference, or microdeletion makes alleles different
Genotype [S21]
a. Read slide
b. Genotype is the hidden plus observed, dominant plus recessive - all the alleles that make up the trait
c. Not just one allele makes up trait, but usually multiple alleles
d. Homozygous indicates two identical alleles - misnomer because can be homozygous as long as have same
phenotype (with different alleles) because trait passed on is exactly the same
e. Heterozygous alleles differ and lead to slight change in the phenotype, not necessarily deleterious
i. Slightly different combinations of genes makes up appearance….does not mean we are Mutant!
f. Phenotype – can be part environment and genotype itself
g. Genotype is not changed by the environment; genotype = genes
Genotype and phenotype [S22]
a. Read slide
b. Can get same phenotype (tall) from two different possible genotypes (TT or Tt)
Modern terms for Mendel’s crosses [S23]
CLASS: 11-12
DATE: 9-20-10
PROFESSOR: Pittler
Scribe: Angi Gullard
Proof: Kratika Pareek
Classical Genetics
Page 3 of 5
a. Read slide
XXIV. Wildtype [S24]
a. Other terms we use with animals, not usually with humans (humans – sometimes normal, diseased,
challenged)
b. WT (wild-type) allele – not just only normal version, just most common (most common sequence at certain
position in allele of gene)
c. Mutant - anything that differs from WT (usually mutant refers to deleterious change)
d. Read slide
e. This is the basis for linkage analysis – determine if certain gene that causes a disease segregates with the
disease
XXV. Law of segregation: the monohybrid cross [S25]
a. Female and male parents are both heterozygotes – pass on one of two possible gametes
b. Phenotypic ratio of 3 for dominant and 1 for recessive
c. Read slide
d. 50:50 chance of passing one allele or other, but actually get ¾ phenotypically tall plants
XXVI. Mode of inheritance [S26]
a. Read slide
b. There are many different modes of inheritance
c. For retinitis pigmentosa, all these modes occur except holandric Y-linked
d. Holandric Y-linked is probably very rare in dental diseases
e. Autosomal recessive – there are 22 autosome in humans –
i. For an autosome, when requires that both alleles have disease-causing change in sequence
f. If only one is necessary - autosomal dominant
g. If on X chromosome, can be X-linked recessive (if it’s in pseudo-autosomal regions where X and Y
chromosomes have homologous regions )
h. Can be X-linked dominant
i. Can be Y-linked – excessive hair around ears
j. Can be mitochondrial - Leber hereditary optic neuropathy
XXVII. Autosomal dominant inheritance [S27]
a. Patterns of inheritance – autosomal dominant inheritance
b. Unaffected – normal alleles (WT alleles aa)
c. Affected parent has an affected (A) and a WT (a) allele
d. So, wherever A is passed, have disease because allele is dominant, always exposed (never hidden)
e. Read slide
XXVIII. Autosomal recessive inheritance [S28]
a. For autosomal recessive, see the opposite
i. Parent needs to have both bad alleles to have the disease
b. Here, both parents are heterozygous (carriers), so both have potential to pass on the bad allele, c
c. Get offspring where 1/4 have disease (cc), 3/4 are normal but 2 of those are carriers and could pass the bad
allele on to offspring
i. This is why we do screenings today for diseases – can do genetic counseling to determine risk for disorder
1. Tay-Sachs in Ashkenazi Jewish populations
2. Sickle cell anemia in African populations
d. Read slide
e. Humans: uni-disomal parenting (very rare)
i. One parent passes both bad alleles
1. In cystic fibrosis – if child gets both bad genes from one parent
f. Autosomal recessive inheritance can skip generations
i. One child that is homozygous is less likely
ii. The more children, the more that can be affected
XXIX. Comparison of autosomal dominant and autosomal recessive inheritance [S29]
a. Read slide
XXX. Law of independent assortment [S30]
a. Two genes on different chromosomes segregate their alleles independently
b. This is a critical thing – for two genes on different chromosomes, the likelihood they are passed to offspring is
50%, completely random, no linkage
c. The inheritance of an allele of one gene does not influence which allele is inherited at a second gene, in
general.
CLASS: 11-12
DATE: 9-20-10
PROFESSOR: Pittler
Scribe: Angi Gullard
Proof: Kratika Pareek
Classical Genetics
Page 4 of 5
i. Have one gene on one chromosome and second gene on another chromosome
ii. There are diseases with double heterozygotes – 2 genes involved
1. People with one or other heterozygotic genotype have no disease
2. People with both heterozygote genotypes have disease
iii. If alleles not on same chromosome, segregate independently….important
XXXI. Law of independent assortment [S31]
a. Where does law of independent assortment come from? Look at meiosis
b. Meiosis I – start with diploid gamete and end up with haploid genome
c. Metaphase one – look at the separation and see potential gametes after meiosis – have two alternatives
d. The possibilities segregate independently
XXXII. Independent assortment of two traits [S32]
a. One trait – nothing for it to be independent from
b. Here, for two independent traits in dihybrid cross – look at round, yellow and wrinkled, green
i. Which allele is dominant? Cross and see what offspring are
ii. What you see is what is dominant
c. Read slide
XXXIII. Two traits segregating independently [S33]
a. More complex- look at traits and segregation of traits in next generation
b. P1 (first generation) – round, yellow crosses with wrinkled, green (recessive allele - lowercase)
c. Punnett square for distribution of potential genotypes
d. For 2nd round of crossing, see distribution with 16 possibilities
i. Ratio is 9:3:3:1 for traits in dihybrid cross
e. This is how alleles segregate independently
XXXIV. Pedigrees [S34]
a. Verify is a genetic disease – rule out phenocopy, get family history, collect DNA, put together pedigree
b. Read through most of common symbols on pedigree
i. Female – circle/ male - square
ii. Have disease-causing trait: filled in
iii. Carrier – half-filled
iv. Person is deceased – line-through, no information
v. Do not know sex (as for an unborn child) – diamond
vi. Pregnant – indicate with P
vii. Horizontal double line - consanguineous mating
XXXV. Autosomal dominant inheritance of brachydactyly [S35]
a. Here, heterozygotes have the phenotype
b. Look at 1 and 2 – oldest ones that represent particular family (Generation I)
c. Patient is usually at bottom of pedigree
d. Want to find as many people as you can to determine mode of transmission
e. Look at Generation I: know that 2 and 3 in generation I have disease, 4 does not – all from same parents, so
could be recessive or dominant - do not know yet
f. Look at Generation II – both males and females affected  rule out X-linked
g. See that in every generation, there is an affected individual, so most likely a dominant disease (unless there
are non-Mendelian genetics going on)
XXXVI. Autosomal recessive inheritance of albinism [S36]
a. Probably someone in Generation III was the patient who initiated the analysis
b. Carriers do not have disease – gene passed to two people in Generation 2
c. Normally have bigger pedigree
i. Huntington’s – very large families, but all inter-related, so hard to follow genetics
ii. Where groups do not inter-marry as much – clean genetics
XXXVII. Genetic predictions [S37]
a. Ellen’s brother Michael has sickle cell anemia, an autosomal recessive disease.
b. What is the chance that Ellen’s child has sickle cell anemia? ...Retinitis pigmentosa? Ehler-Danlos syndrome?
i. Get family history, blood drawn, refer to genetic counselor
c. Autosomal recessive disorder, so the question really is : “What is the likelihood I am a carrier?”
d. Parents must be carriers, or you couldn’t have a child with recessive disease
e. Look at Punnett square – parents must have one normal and one recessive allele
i. Ellen is not affected, so she does not have aa genotype
1. Chance that Ellen is carrier is 2/3 (AA, Aa, Aa, aa)
CLASS: 11-12
DATE: 9-20-10
PROFESSOR: Pittler
Classical Genetics
ii. The chance the child inherits sickle cell is ½ (50:50)
1. From Ellen, since she is a carrier (either her A or a allele can be passed on)
iii. There is a chance that Ellen’s husband is a carrier also
iv. The overall chance the child has sickle cell: 2/3 x 1/2 = 1/3
1. Can do direct testing to see if have allele or not
Scribe: Angi Gullard
Proof: Kratika Pareek
Page 5 of 5
[End 41:52 mins]
Question from 9-21-10: How do you get 1/3? Is the father included?
Answer: Probability of father being a carrier is dependent upon population genetics.
- This question is just asking whether the mother will pass on the mutant allele: 2/3 because mother at best is a
carrier (she has no disease, it is recessive, allele from parents).
- The mother can only pass on one allele, so it is 50:50  1/2 x 2/3 = 1/3