Todays Outline

1
Todays Outline

• Lecture on Inheritance
• Mendels experiments CD activity

2
Early Work in Genetics

• Mendels 1866 paper - based on pea plants
• He manipulated who fertilized whom (P parent)
• looked at the offspring (F1) grandoffspring
  (F2)
• 7 different characteristics, each w/ 2 morphs
• Made sure he had true-breeding varieties - when
  selfed produced only one kind of offspring
• Then made crosses to produce hybrid offspring

3
Monohybrid Cross

• Monohybrid cross - parent plants differ in only 1
  char.
• If cross 2 true-breeding parents, F1 are all
  the same
• But then if cross the F1s, get re-appearance of
  the hidden morph

4
Inheritance

• His observations lead to 4 Hypotheses
• There are alternative forms of genes (the units
  that determine heritable traits) We now call them
  alleles.
• For each inherited character, an individual has
  2 genes. One from each parent. Both genes may be
  the same allele or diff. alleles.
• A sperm or egg carries only one allele. Allele
  pairs segregate from one another at gamete
  production (meiosis).
• When 2 different alleles are present but only one
  is expressed, it is said to be dominant and the
  other recessive.

5
Definitions

• Homozygous both the alleles for a trait are
  same
• True breeding varieties were homozygous
• Parental Purple flowers PP
• Parental White flowers pp
• F1 Purple flowers Pp
• F1 hybrids were heterozygous
• F2 hybrids were a mix
• 1/4 white, 3/4 purple
• 1/4 pp, 1/4 PP, 1/4 Pp, 1/4 pP
• Do Punnett square

6
Definitions

• Expressed physical traits phenotype
• Genetic makeup genotype
• F2 generation phenotypic ratio 31
• F2 generation genotypic ratio 121
• All 7 characters exhibited same pattern
• One parental trait disappeared in F1
  heterozygotes
• Hidden trait reappeared in 1/4 of the F2s
• Principle of Segregation Pairs of genes
  segregate (separate) at gamete formation the
  fusion of gametes at fertilization pairs genes
  once again
• CD activity - Mendels garden

7
Definitions

• Alleles are alternative forms of a gene
• Alleles reside at the same locus on homologous
  chromosomes.

8
Human genetics

• Mendels principles apply
• Characters to trace freckles, hairline, earlobe,
  deafness, sun sneeze
• Cant do crosses at will - rely on family tree
  studies
• Inherited disorders are reason for genetic
  testing
• Recessive disorders are particularly tricky -
  carriers
• Examples deafness, cystic fibrosis, albinism,
  etc.
• Dominant genetic disorders also occur
  Achondroplasia, Huntingtons disease, etc.
• Genetic dominance says nothing about frequency or
  favorableness of a trait

9
Genetic effects

• One single gene may affect one characteristic
• But can also have
• One single gene that affects many characteristics
  pleiotropy
• Many genes that affect one characteristic
  polygenic inheritance

10
Do Practice Questions

11

• 1. Assume tall (T) is completely dominant to
  dwarf (t). If a homozygous dominant individual is
  crossed with a homozygous dwarf, the offspring
  will _____.
• A all be intermediate in height
• B all be tall
• C be 1/2 tall and 1/2 dwarf
• D be 3/4 tall and 1/4 dwarf
• E all be short

12

• 2. In a certain plant, the alleles A, B, and C
  are completely dominant to the alleles a, b, and
  c. A plant with the genotype AABbcc will have the
  same phenotype as a plant with the genotype
  _____.
• A Aabbcc
• B aabbcc
• C AaBBcc
• D AABBCc
• E none of the above

13

• 3 Pea plants are tall if they have the genotype
  TT or Tt, and they are short if they have
  genotype tt. A tall plant is mated with a short
  plant. Which outcome below would indicate that
  the tall plant was heterozygous?
• A All of the offspring are short.
• B All of the offspring are tall.
• C The ratio of tall offspring to short offspring
  is 31.
• D The ratio of tall offspring to short offspring
  is 11.
• E None of the above.

14

• 4 What is indicated when a single-character
  testcross yields offspring that all have the
  dominant phenotype?
• A The parent with the dominant phenotype was
  homozygous.
• B The parent with the dominant phenotype was
  heterozygous.
• C Epistasis has occurred.
• D The alleles are codominant.
• E Both parents are heterozygous.

15

• 5 If a homozygous dominant is crossed with a
  heterozygote for a given trait, the offspring
  will be _____.
• A all of the dominant phenotype
• B 1/4 of the recessive phenotype
• C all homozygous dominant
• D all homozygous recessive
• E present in a 9331 ratio

16

• 6 In Mendel's monohybrid cross of purple-flowered
  and white-flowered peas, all members of the F1
  generation had the _____ phenotype because their
  genotype was _____ at the flower-color locus.
• A white-flowered ... homozygous recessive
• B white-flowered ... heterozygous
• C purple-flowered ... homozygous recessive
• D purple-flowered ... homozygous dominant
• E purple-flowered ... heterozygous

17

• 7 If a heterozygous plant is allowed to
  self-pollinate, what proportion of the offspring
  will also be heterozygous?
• A ¼
• B 1/3
• C ½
• D 2/3
• E all of them

18

• 8 When two average-height parents give birth to a
  child exhibiting achondroplasia, it is most
  likely due to a new mutation. This is because
  _____.
• A the frequency of achondroplasia is unknown
• B achondroplasia is a relatively rare disorder
• C achondroplasia is caused by an allele that is
  always expressed, therefore the parents must not
  have the allele
• D such mutations are statistically predictable
• E none of the above

19

• 9 A man who has type B blood and a woman who has
  type A blood could have children of which of the
  following phenotypes?
• A A or B only
• B AB only
• C AB or O
• D A, B, or O
• E A, B, AB, or O

20

• 10 Which of the following matings cannot produce
  a child with blood type O? The letters refer to
  blood types (phenotypes).
• A A x A
• B A x B
• C O x AB
• D O x O
• E none of the above

21

• 11 A woman with type O blood is expecting a
  child. Her husband is type A. Both the woman's
  father and her husband's father had type B blood.
  What is the probability that the child will have
  type O blood?
• A 100
• B 75
• C 50
• D 25
• E 0

22

• 12 Michelle and Keith are apparently normal, but
  their daughter was born with alkaptonuria, an
  inherited metabolic disorder. If alkaptonuria is
  like most other human hereditary disorders, the
  probability of their next child being born with
  alkaptonuria is _____.
• A 0
• B ¼
• C ½
• D 2/3
• E ¾

23

• 13 Tom's brother suffers from phenylketonuria
  (PKU), a recessive disorder. The brothers'
  parents do not have PKU. What are the chances
  that Tom, who is normal for this trait, is a
  carrier of PKU?
• A ¼
• B 1/3
• C ½
• D 2/3
• E 4/3