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Biology 222 (Genetics)
 Notes for Exam 1

This is exam 1 from 1998.  I thought this might help you prepare for the exam September 24.

Use the following information for questions 1 and 2.

 While on an Amazon river rafting trip you discover two unusual fruit fly phenotypes.  You bring these variants back to your lab at Penn State and perform a series of experiments.  Your studies demonstrated that two alleles of a single gene determine a leg phenotype of long (l) and short (s).  You also found that two alleles  of an independently assorting gene determine eye color of red (r) and green (g).   The results of two different fly matings are shown below.

                                   _____Number of progeny___
                    Mating            s,g       s,r        l,g       l,r __
     1.           s,g  x  s,g         1198   407       0        0
     2.           l,g  x  l,g            272     89      798    267

1. What are the dominant phenotypes?

 A.  long legs, red eyes
 B.  short legs, red eyes
 C.  long legs, green eyes
 D.  short legs, green eyes
 E.  none of the above
 

2. What are the genotypes of the parents in mating 1 if the gene controlling the leg  phenotype is indicated with B (dominant) or b (recessive) and the gene controlling the eye color phenotype is indicated with C (dominant) or c (recessive)?

 A.  bbCC x bbCC
 B.  BbCc x BbCc
 C.  BBCc x BBCc
 D.  Bbcc x Bbcc
 E.  bbCc x bbCc
 

3. In the pentahybrid cross AABbCcddEe x AAbbCcDdEe, approximately how many progeny would you need to examine to find 5 individuals with the genotype AAbbCcDdEE?

 A.    32
 B.    64
 C.  160
 D.  320
 E.  The genotype AAbbCcDdEE is not possible.
 

4. Several matings of gray rats resulted in 46 black rats, 51 white rats and 94 gray rats.  What is the genetic explanation for the gray rats?

 A.  duplicate genes
 B. complementation
 C.  multiple alleles
 D.  incomplete dominance
 E. homozygous lethality

5. The graph below shows the results of data collected from a wild type and two mutant strains of the plant species I. claudius. What does this graph represent?

 A.  developmental noise
 B.  norm of reaction
 C.  simple Mendelian genetics
 D.  X-linkage
 E.  none of the above

Use the following information for questions 6 and 7.

 The resulting F2 progeny from a Drosophila three point test cross is shown below.

  a+   b+   c       907
  a     b     c+     925
  a+   b     c         93
  a     b+   c+     100
  a     b+   c         15
  a+   b     c+       17
  a+   b+   c+     217
  a      b     c      226
                        2500

6. What is the central gene?

 A.  a
 B.  b
 C.  c
 

7. What is the map distance in map units (m.u.) between a and b, b and c, a and c?

 A.  19, 9, 28
 B.  19, 28, 9
 C.  9, 19, 28
 D.  9, 28, 19
 E. 9, 26.7, 19

8. Class   Observed   Expected____
   AB         315          300
   Ab          310          300
   aB           290          300
   ab           285          300

 Calculate c2 = from the table and determine the number of degrees of freedom (df).

 A.  c2 = 1 33; df = 4
 B. c2 = 2.17; df = 4
 C. c2 = 3.00; df = 3
 D.  c2 = 1 33; df = 3
 E.  c2 = 2.17; df = 3
 

9. A pure line of wheat with red seeds is crossed with a pure line of wheat with white seeds.  The F1 generation was all red-seeded.  The F1 generation was selfed and the resulting F2 generation had 714 red-seeded plants and 553 white-seeded plants.  What is the genetic explanation for these results?

 A.  duplicate genes
 B.  complementation
 C.  multiple alleles
 D.  incomplete dominance
 E.  homozygous lethality
 

10. Which of the following statements is false concerning human hemoglobin?

 A.  HbA HbS leads to mild anemia.
 B.  HbA is dominant regarding anemia.
 C.  HbS HbS leads to severe anemia
 D.  HbA is codominant regarding hemoglobin molecules.
 E.  All of the above statements are true.
 

11. The following human pedigree suggests what mode of inheritance?



 

 A.  X-linked recessive
 B.  X-linked dominant.
 C.  autosomal recessive
 D.  autosomal dominant
 E.  Y-linked

12. Fill in the blanks of the following statement.

Mendelís first law can be summarized as the law of ____________ , whereas his second law can be summarized as the law of ____________ .

 A.  unequal segregation; dependent assortment
 B.  unequal segregation; independent assortment
 C.  equal segregation; dependent assortment
 D.  equal segregation; independent assortment
 E.  independent assortment; unequal segregation
 

13. Fill in the blanks of the following statement.

 Synapsis occurs during ____________ of ____________ .

 A.  metaphase; mitosis
 B.  metaphase I; meiosis I
 C.  prophase I; meiosis I
 D.  prophase II; meiosis II
 E.  anaphase I; meiosis I
 

For questions 14 and 15 match the process with the appropriate response.

14. synapsis                                                         A.  anaphase II
                                                                           B.  anaphase I
15. separation of homologous chromosomes        C.  mitosis
                                                                           D.  prophase II
                                                                           E.  prophase I
 

16. The following human pedigree suggests what mode of inheritance?

 A.  X-linked recessive
 B.  X-linked dominant.
 C.  autosomal recessive
 D.  autosomal dominant
 E.  Y-linked
 

17. A man with blood groups AB and M has children with a woman with blood groups A and MN.  Assuming that the gene responsible for the ABO blood groups is on a different chromosome than the gene responsible for the MN blood groups, which of the following blood types cannot occur in their children?

 A.  O and N
 B.  B and MN
 C.  AB and M
 D.  O and MN
 E.  A and N
 

18. Which of the following statements is false concerning X-inactivation in human females.

 A.  Early in development one of the 2 X chromosomes is inactivated.
 B.  If the functional X has a recessive allele, the recessive phenotype is expressed.
 C.  All human females are genetic mosaics.
 D.  The inactive X chromosome observed in a microscope is called a Barr body.
 E.  All of the above statements are true.
 

19. Fill in the blanks of the following statement.

 The observation that a crossover (X-over) in one region of the chromosome  ____________ the likelihood that a X-over will occur in an adjacent region is referred to as ____________.

 A.  increases; interference
 B.  decreases; interference
 C.  increases; coefficient of coincidence
 D.  decreases; coefficient of coincidence
 E.  decreases; non-disjunction
 

20. Fill in the blanks of the following statement.

 ____________ recombination generates 50% parental and 50% recombinant gametes, whereas ____________ recombination generates >50% parental and <50% recombinant gametes.

 A.  mitotic; meiotic
 B.  meiotic; mitotic
 C.  homologous; site-specific
 D.  interchromosomal; intrachromosomal
 E.  intrachromosomal; interchromosomal
 
 

For questions 21 and 22 match the term with the appropriate response.

21. point mutation                    A.  new activity
                                               B.  residual activity
22. auxotrophic mutation         C.  requires a nutrient
                                               D.  any change to the wild type allele
                                               E.  single base pair change
 

23. Which of the following statements is false?

 A.  A second site suppresser can be defined as a mutation in a second gene resulting in the partial phenotypic reversion to wild type.
 B.  Mutations are used to genetically dissect biological functions.
 C.  A selective system is used to separate rare mutant individuals from wild type.
 D.  Sporadic retinoblastoma usually affects both eyes because of rare mitotic crossovers.
 E.  All of the above statements are true.
 

24. Fill in the blanks of the following statement.

 Proto-oncogenes and ____________ genes normally carry out functions involved in regulating ____________.

 A.  tumor suppresser; cell division
 B.  tumor suppresser; metastasis
 C.  tumor activation; cell division
 D.  tumor activation; metastasis
 E.  metastatic; cell division
 

25. Which of the following statements is false?

 A.  Penicillin kills actively growing cells.
 B.  The fluctuation test demonstrated that bacteriophage T1 resistance was due to physiological changes.
 C.  Mutagens are used to increase the mutation rate.
 D.  Prototrophes are nutritionally self-sufficient.
 E.  Penicillin is used to increase the chance of isolating an auxotrophic mutant.
 
 Chapter 1

GENETICS--->Study of Genes--->From molecules of DNA to the gene pool of an entire population.

Molecules of DNA make up genes--->Individuals inherit genes--->Individuals contribute to the gene pool of a population.

All organisms contain DNA.

DNA contains ALL of the information for building an organism. This information is encoded in genes.

GENES--->Functional units of DNA.

Interactions between genes and environment determine what organisms are.

GENETIC ANALYSIS--->Requires an observable trait that is easily monitored.

PHENOTYPE--->Characteristics of an organism.
Phenotypes change throughout the life of an organism as its genes interact with the environment.
Phenotype is determined by the genotype.

GENOTYPE--->The set of genes an organism inherits. (Very little change over time)
Geneticists study partial genotypes and partial phenotypes of organisms.
(i.e.) Study the effect of one or a few genes.

Genetically different individuals develop differently in identical environments.

Genetically identical individuals develop differently in different environments.
(i.e.) Environment plays a role in phenotypic development.

Genotypes can overlap in the phenotypes they cause.
(i.e.) A particular phenotype can be caused by more than one genotype.

Genetic analysis is greatly simplified by using an easily followed phenotype in an organism that is practical to study. (e.g.) mice vs. elephants

WILD TYPE--->genotypic or phenotypic characteristics of natural populations or standard laboratory strains.

An individual that is not wild type is a variant or mutant.

Each genotype can give different phenotypes because the environment influences the expression of the genotype.

NORM OF REACTION--->Quantitative set of environment-phenotype relationships of a particular genotype.

Describes the range of phenotypes that the environment can cause for a specific genotype.

(See figures 1-17 and 1-18)

DEVELOPMENTAL NOISE--->Random molecular events in cells of a developing organism that cause variations in phenotype.

THUS: Genotype determines phenotype BUT...

expression of the information in the genotype is influenced by the environment, developmental noise, and other genes.

For Genetic Analysis:

1. A phenotype that is easy to identify.

2. A phenotype that is not highly influenced by the environment.

3. Use conditions that allow determination of the influence of a single gene.

 Chapter 2

MENDELIAN ANALYSIS--->Analyzing hereditary information using Mendelís techniques and principles.

Gregor Mendel--->(1860ís)A monk who conducted quantitative and systematic studies of inheritance.

Mendel first proposed the concept of the gene.

BLENDING INHERITANCE--->Each parent contributes essences of characteristics that are mixed to form a new individual.

Problem--->Suggests progeny have characteristics that are intermediate of the parents.

PARTICULATE INHERITANCE---> Characteristics determined by discrete units that are inherited intact.

Mendel discounted the blending theory and gave rise to the notion of particulate inheritance.

MENDELíS PEAS--->An individual pea plant produces both pollen and eggs. Thus, peas can self-fertilize or cross-fertilize.

Each pea results from a separate fertilization event (separate individual).

SELF-FERTILIZATION--->Pollen from one flower fertilizes an egg from the same flower.

Both parents are the same organism (same genotype).

CROSS-FERTILIZATION--->Pollen from one plant fertilizes an egg from another plant.

Parents have different genotypes.

PURE LINE--->All offspring produced by selfing or crossing individuals within the same line produce the same phenotype. (Flower color)

P--->Parental generation. (Individuals in the first cross of a particular series of experiments)

F1--->First Filial Generation. (Progeny from the first cross)

F2--->Second Filial Generation. (progeny from crosses involving individuals from the F1 generation)

DOMINANT--->The phenotype that is expressed in the F1 progeny when two pure lines are crossed.

RECESSIVE--->The phenotype that is covered up by the presence of the dominant phenotype.
                            (Reappears in the F2 generation)

Mendel used reciprocal crosses (Figs 2.4 and 2.5).

Experiment 1
P     pure breeding      X      pure breeding
       purple flowers              white flowers
                       (cross-fertilization)
 
F1                  All purple flowers

No blending.
Reciprocal crosses gave the same result.
Purple is dominant and white is recessive.

                        F1         X         F1 (Self-fertilization)
                                    
F2                    3:1 ratio of purple:white

Recessive phenotype remained as a separate inheritable trait.

Mendel studied 7 distinct characteristics (Table 2-1).

Experiment 2
P  yellow X green (cross)
               
F All yellow   (self)
             
F 3:1 yellow:green (self)
             
F3       ???
Selfing F2 plants resulted in a 1:2:1 ratio.

1 yellow F2 always give rise to yellow.
2 yellow F2 give 3:1 yellow:green ratio.
1 green F2 always gives rise to green

Mendel proposed that each gene is actually present twice (gene pair).

Experiment 2 Again

P     YY (yellow)     X     yy (green)  (cross)
 
F1                All Yy (yellow)              (self)
                                
F2          1 YY (yellow)
              2 Yy (yellow)
              1 yy (green)

TEST CROSS-->Crossing to a recessive individual.

F1      Yy  X  yy
                 
                1:1

 (Yy) Dominant : Recessive (yy)
 

ALLELES--->Different forms of the same gene
 (e.g.) Y and y

HOMOZYGOUS--->Both alleles of a gene pair are the same. YY (dominant) or yy (recessive)

HETEROZYGOUS--->Different alleles of a gene pair. Yy (dominant)

Thus, YY and Yy have different genotypes but the same phenotype.

Underlying the 3:1 F2 phenotypic ratio is a 1:2:1 genotypic ratio of YY:Yy:yy

MONOHYBRID CROSS--->Analyzing one characteristic (one gene) by crossing.

DIHYBRID CROSS--->Analyzing two characteristics (genes) simultaneously by crossing.
 

TRIHYBRID CROSS--->etc...

MENDELíS FIRST LAW--->The two members of a gene pair segregate from each other into gametes, so that one half of the gametes carry one member of the gene pair and the other half of the gametes carry the other member of the gene pair.

DIHYBRID CROSS--->Usually has 4 possible phenotypes.

1) Both dominant

2) One dominant, one recessive

3) One recessive, other dominant

4) Both recessive

Phenotypic ratio of F2 progeny--->9:3:3:1 (Fig. 2-9)

INDEPENDENT ASSORTMENT--->Different genes assort independently from one another.

Tells you about the physical relationship of genes to each other.

MENDELíS SECOND LAW--->During gamete formation the segregation of alleles of one gene is independent of the segregation of alleles of another gene.

PROBABILITY--->The number of times an event is expected to happen divided by the number of opportunities for an event to happen. (i.e.) p=x/y

(e.g.) Dice---> p(of a 3)=1/6
(e.g.) Gametes--->if heterozygous Aa
                        p(A)=1/2
                        p(a)=1/2

PRODUCT RULE--->The probability that two independent events will occur simultaneously is the product of their respective probabilities.

(e.g.)
p(of two 4s)=(1/6)(1/6)=1/36
RrYy; p(gamete with 2 recessive)=(1/2)(1/2)=1/4

SUM RULE--->The probability of either one of two independent (mutually exclusive) events is the sum of the probabilities.
(e.g.)

p(of two 4s or two 5s)=1/36 + 1/36=1/18
p(of 4 or 6)=1/6 + 1/6=1/3

Probability or chance governs transmission of genes.

PUNNETT SQUARE--->(Fig 2-10)

1) Possible gametes from one parent on one side, possible gametes from other parent on the other side.

2) Fill in gametic combinations in the squares.

3) Determine genetic ratios (phenotypic or genotypic) by counting squares.

Too unwieldy to use routinely.

 
BRANCH DIAGRAM--->

1) List probabilities for one event.

2) List probabilities of second event next to those of the first event.

3) Use the product rule to determine genetic ratios.
 
 
P
RRYY
X
rrYY
(cross)
 
 
 
 
 
R=round peas
F1
 
RrYy (self)
 
 
r=wrinkled peas
 
 
 
 
Y=yellow peas
F2
 
?
 
 
y=green peas
 

                    Composition of F2                                                      Product rule result
 
 3/4 (Y-) 9/16 (R-Y-) round, yellow 
3/4 (R-) 
(RR,Rr,rR) 1/4 (yy) 1/4 (yy) 3/16 (R-yy) round, green 
3/4 (Y-) 3/16 (rrY-) wrinkled, yellow 
1/4 (rr) 
1/4 (yy) 1/16 (rryy) wrinkled, green 
   
F2 phenotypic ratio of a dihybrid cross is 9:3:3:1.

Use product rule directly for trihybrid, etc...
 (e.g.) F1  AaBbCcDdEe X AaBbCcDdEe

p(AAbbCcDDEe)=(1/4)(1/4)(1/2)(1/4)(1/2)=1/256
p(A-bbC-D-E-)=(3/4)(1/4)(3/4)(3/4)(3/4)=81/1024
PEDIGREE ANALYSIS
Used in human genetics (Family tree)

Symbols used in pedigree analysis (See fig. 2-12).

Recessive disorders (Fig. 2-13)--->In pedigrees, simple Mendelian inheritance of a recessive disorder is revealed by the appearance of the phenotype in the male (    ) and female (    ) progeny of unaffected individuals.

(e.g.) PKU, cystic fibrosis, albinism

Dominant disorders (Fig. 2-16)--->Pedigrees of Mendelian dominant disorders show affected    and    in each generation and also show affected    and    transmitting the condition to sons and daughters in equal proportions.

(e.g.) achondroplasia (dwarfism)
  D- (dwarfism)
  dd (normal)
Interferes with bone growth during development.
 
Chapter 3

MITOSIS AND MEIOSIS--->Division of the nucleus and its contents (chromosomes).

Eukaryotic cells only

CELL CYCLE--->4 stages

1) Mitosis (M)

2) Gap1 (G1)

3) DNA synthesis (S)

4) Gap2 (G2)

Cells with two chromosome sets are diploid (2n).

Cells with one chromosome set are haploid (1n).

MITOSIS--->(Fig. 3-2)

1) Somatic cells (non-gametes).

2) Produces genetically identical daughter cells from one progenitor.

3) Single division (2 cells).

4) Chromosomes duplicated during interphase

(time between mitoses).

5) 4 continuous stages (Fig. 3-2).

A) Prophase

B) Metaphase

C) Anaphase

D) Telophase

PROPHASE--->Chromosomes consisting of 2 sister chromatids condense.

Chromatids still joined at the centromere.

Nucleoli disappear (RNA synthesis).

Nuclear membrane breaks down.

Nuclear spindle starts to form.

METAPHASE--->

Nuclear spindle becomes prominent.

Chromosomes move to equatorial plane.

Centromere of each chromosome attaches to spindle fibers.

ANAPHASE--->Pairs of sister chromatids separate and migrate to opposite poles (V-shaped structures).

TELOPHASE--->

Nuclear membrane reforms around each daughter nucleus containing one complete set of chromosomes.

Nucleoli reappear; spindle disappears.

Cytoplasm divided in two by a new cell membrane.

MEIOSIS--->(Fig. 3-3)

1) Production of genetically different gametes.

2) Two cell divisions.

3) Produces 4 meiotic products.

MEIOSIS I (MI)--->Homologous chromosomes separate. The two members of each homologous pair are called homologs. (Continuous)

PROPHASE I--->Chromosomes become visible, homologous chromosomes pair (synapsis).

Synapsis requires the synaptonemal complex composed of protein, DNA, and RNA.

Number of homologous pairs of chromosomes is 1n.

Crossing over (X-over) occurs which is an exchange of DNA between homologous chromosomes. Visualized by cross-shaped structures (chiasmata).

METAPHASE I--->Each pair of homologs are positioned at the equatorial plane.

The two centromeres of a homologous pair attach to spindle fibers of opposite poles.

ANAPHASE I--->Members of each homologous pair separate and move to opposite poles.

Centromeres DO NOT divide in Meiosis I.

TELOPHASE I--->2 daughter cells form. Nuclear membrane may or may not reform.

INTERPHASE--->NO DNA synthesis.

Number of chromosomes in each cell has been reduced by 1/2 (2n--->1n)

MEIOSIS II (MII)--->Similar to mitosis where the number of chromosomes in the original and product cells are the same.

Centromeres divide resulting in the separation of sister chromatids.

Net result of meiosis is the generation of 4 haploid gametes (1n) (Fig. 3-5)

Mendelís laws are explained by meiosis.

Chromosome Theory of Inheritance--->Genes are on chromosomes (Sutton & Boveri).

Mendelís particles precisely paralleled the behavior of chromosomes (Fig. 3-6).

Proof of the chromosome theory required:

1) Correlation between differences in phenotypes and chromosomes in microscope.

2) Demonstration of independent assortment of chromosomes.

Thomas Hunt Morgan--->Nobel prize (1934)

SEX LINKAGE--->Pattern of inheritance shown by genes on the sex chromosomes.

In many organisms the gender is determined by the combination of sex chromosomes.

Sex chromosomes also carry genes unrelated to male and female development.

X-linked recessive inheritance--->Many more than .

Drosophila eye color phenotype correlated with chromosome differences in microscope.

(Supported chromosome theory)

Experiment 1--->Drosophila eye color (Fig. 3-14)
 
P red (W.T.) x white-eyed 
F1 All & red  red (dominant)
white (recessive)
F2 3:1 red:white  BUT:  red  white
2 :1 all 
Experiment 2--->Test Cross
 
F1 Red-eyed (Xw+Xw)
x
white-eyed (XwY) 
1:1:1:1 red:            white: red:          white
(Xw+Xw) (XwXw) (Xw+ Y)  (XwY)
Experiment 3-->reciprocal cross of expt.1 (Fig. 3-14)
 
 
P XwXw (white )
 x
Xw+Y (red )
F1
1:1
   red       :     white 
Xw+Xw           Xw
F2 1:1:1:1     red :         white:     red :          white 
(Xw+Xw)   (XwXw)   (Xw+ Y)    (XwY) 
Morganís Explanation

(XX)--->2 chromosomes with eye color genes.

(XY)--->1 chromosome with eye color genes.

Chromosome Theory Proof--->Bridges & Morgan

Experiment 3 revisited
 
P XwXw x Xw+Y
(white) (red)
F1  Xw+Xw & XwY
(red) (white) 
PLUS: rare exceptional progeny (1/2000)

XwXwY            &         Xw+0
(white)                           (red)

The exceptional chromosome arrangements were visualized in a microscope.

NON-DISJUNCTION--->Failure of chromosomes or sister chromatids to separate in meiosis.

Exceptional progeny arise due to non-disjunction during meiosis. (See Fig. 3-17)

The number of X chromosomes dictates the gender of Drosophila. (Different than in humans)

2 Xs---> 1X--->

Xw+0--->sterile because the Y chromosome is necessary for fertility.

X-Linked Dominant Inheritance--->Affected transmit phenotype to all daughters, never to sons.

AUTOSOMES--->All chromosomes other than the sex chromosomes.

Human Sex Linkage--->Pedigree Analysis.

The presence of the Y chromosome determines maleness in humans. (See Table 3-1)

X-Linked Recessive (Humans)

1) Many more than show the recessive phenotype.
(If rare, almost all affected people are )

2) If rare, none of the offspring of an affected are affected, but all daughters are carriers.

3) carriers will pass the condition to 1/2 of her sons and 1/2 of her daughters will be carriers.
(e.g.) color blindness, hemophilia, Duchenneís muscular dystrophy (fatal X-linked)

X-Linked Dominant (Humans)

1) Affected pass condition to all daughters,
no sons.

2) Affected pass condition to 1/2 of sons and 1/2 of daughters.
(e.g.)hypophosphatemia (Vitamin D deficient rickets)

Y-Linked Inheritance (Humans)

No human phenotype other than maleness has been proven to be Y-linked.

(Hairy ear rims?,See Fig. 3-29)

X-INACTIVATION---> mammals inherit 2 X chromosomes. Early in development one of the 2 Xs is inactivated. If the functional X has a recessive allele, the recessive phenotype is expressed.

BARR BODY--->The inactive X (microscope).

Since the inactivation process is random, all are genetic mosaics. (i.e.) a mixture of cells having 2 genotypes corresponding to inactivation alternatives.

(e.g.) sweat glands (See Fig. 3-28)

 HAPLOID ORGANISMS--->(e.g.) yeast

Adults exist as a haploid (1n) cell or cells.

Divide by mitosis to multicelled organism or colony of cells (1n). All recessive phenotypes are seen in 1n cells.

1n cells fuse--->transient diploid goes through meiosis to produce more 1n organisms.

Compare mitosis in a diploid organism (Fig. 3-32) to mitosis in a haploid organism (Fig. 3-34).

Chapter 4

BACKGROUND INFORMATION

Genes specify the structure of proteins.

Different alleles of a gene specify different forms of the corresponding protein.
(e.g.) active, inactive, etc...

Cellular processes occur by pathways in which enzymes act in a series.

Enzymes are proteins encoded by genes.

INCOMPLETE DOMINANCE---> Heterozygotes have intermediate phenotypes of the two homozygotes.

(e.g.) Four-O-Clock Flower Color
 
P C1C1
x
C2C2
Red  White  (Cross) 
F1
C1C2 
(Self) 
Pink
F2
1 C1C1 Red 
2 C1C2 Pink
 1 C2C2 White
CODOMINANCE--->Heterozygote has phenotype of both homozygotes.

(e.g.) Human MN Blood Groups

LMLM Protein M
LNLN Protein N
LMLN Proteins M and N

(e.g.) Human Hemoglobin

HbAHbA Normal Red Blood Cells (RBCs)
HbSHbS Abnormal "sickle" shaped RBCs (severe anemia, often fatal)
HbAHbS No anemia, RBCs form sickles under low oxygen concentrations.

Note: HbA is...

1)Dominant in regard to anemia.
2)Incomplete dominance in regard to cell shape.
3)Codominance in regard to hemoglobin molecules.

MULTIPLE ALLELES-->A gene with more than two alleles.

(e.g.) Human ABO Blood Groups (See Table 4-1)

3 alleles IA, IB, i

IA and IB are fully dominant in IAi and IBi and codominant in IAIB.

LETHAL ALLELES--->Alleles of a gene that kill an organism. May be lethal if homozygous or even heterozygous.

Lethality may occur early during embryonic development, after birth, or later in life.

Many human lethals result in spontaneous abortions.

(e.g.) Mouse coat color

Dark coat--->Normal (W.T.)

Yellow coat--->Variant
 
(pure line) Dark
x
Yellow
1:1 Dark:Yellow
Thus, a single gene determines coat color; the yellow mouse was heterozygous; yellow is dominant to dark.

                      Yellow            x            Yellow
2:1 Yellow:Dark Suggests no homozygous yellow.

Explanation: AYA x AYA
1/4 AA (Dark)
1/2 AYA (Yellow)
1/4 AYAY (Die Before Birth)

COMPLEMENTATION--->Production of the dominant (W.T.) phenotype when two different genotypes determining similar recessive phenotypes come together in the same cell.

Dominant phenotype requires at least one dominant allele of each of two genes.

Homozygosity for the recessive allele of either gene results in the recessive phenotype.

(e.g.) Pea petal color (biochemical pathway)
 
P (AAbb) white line 1
x
white line 2 (aaBB) 
F1
AaBb (purple)
F2 9 A-B- (purple) 
3 A-bb (white) 
3 aaB- (white)
1 aabb (white)
9:7 phenotypic ratio in F2

If the same phenotype is observed in independently isolated organisms, complementation tests can determine if the phenotype is caused by alleles of the same or different genes. (F2 ratios)

DUPLICATE GENES--->Genes present more than once in the genome. One W.T. allele of either gene results in the dominant phenotype.
 
P (Dominant) A1A1A2A2
x
a1a1a2a2 (Recessive) 
F1
A1a1A2a2 
F2 9 A1-A2- (Dominant) 
3 A1-a2 a2 (Dominant)
3 a1a1A2- (Dominant) 
1 a1a1a2a2 (Recessive) 
PLEIOTROPIC--->Allele that causes more than one phenotype.
(e.g.) mouse coat color (Yellow when heterozygous, lethal when homozygous)

SEMILETHAL--->Lethal in some but not all individuals (environment).

 

Chapter 5

GENE LINKAGE

Each chromosome can contain 1000ís of genes.

These genes are linked to each other and segregate together during meiosis. (Fig. 5-1)

Mendelian ratios are NOT observed.

Meiotic recombination generates haploid products (gametes) with genotypes different than both haploid genotypes that made the diploid cell that went through meiosis. (Figs. 5-4, 5-5)

(Interchromosomal Recombination) (Fig. 5-6)

If genes assort independently, crosses between a heterozygote and a tester strain generate:

1) 50% parental type gametes.
2) 50% Recombinant gametes
3) 1:1:1:1 phenotypic ratio

Genes close together on the same chromosome do not assort independently.

 
 
pr+ pr vg+ vg 
x
pr pr vg vg 
 
 
pr+ vg+ 
1339
pr+ --->red eye (normal) 
                          pr vg 
1195
pr --->purple eye 
                          pr+ vg
151
vg+ --->normal wings 
                          pr vg+
154
vg --->vestigial wings 
 
2839 
 
 
PARENTAL COMBINATION--->Original arrangement of alleles on the two chromosomes.
(Two most frequent classes)

INTRACHROMOSOMAL RECOMBINATION---> New combinations generated during meiosis when non-sister chromatids cross over (X-over) between the genes under study. (Figs. 5-2, 5-3, 5-7)

Recombinant classes observed less frequently than parentals. (Fig. 5-8)

X-overs generate two reciprocal products or classes which are about equal in frequency.

Recombinant frequency (RF) significantly <50% suggests linkage.

RF » 50% suggests that the genes are unlinked on separate chromosomes (c2 Test).

LINKAGE MAPPING

To separate linked genes the X-over must occur between them.

The greater the distance between linked genes, the greater the chance of a X-over between them.

The RF between linked genes is used to map their relative distance apart on the chromosome.

RF = 0.01 (1%) = 1 map unit (m.u.)

GENE LOCUS--->Region on a chromosome where alleles of a certain gene are found.

      pr+ vg+ / pr vg                x                   pr vg / pr vg
 
pr vg / pr vg 165 
  pr+ vg+ / pr vg 191 
pr vg+ / pr vg 23
pr+ vg / pr vg 21_
400
44 recombinant x 100 = 11% = 11 m.u.
400 total

11% < 50% pr and vg are linked

Given a genetic distance in m.u. we can predict frequencies of progeny from a test cross.

pr+ vg+ / pr vg     x     pr vg / pr vg

If 11 m.u.

pr vg / pr vg 44.5%
pr+ vg+ / pr vg 44.5%
pr vg+ / pr vg 5.5%
pr+ vg / pr vg 5.5%

 

Three-Point Testcross

Map units are additive, but to order 3 genes we need to perform a three-point testcross. (Fig. 5-10)

Double X-overs--->Smallest class because
two X-overs required.

Compare parental types with double X-overs. The gene that is switched is between the others (middle).
(Figs. 5-12, 5-13)
 
 
P v+ v+ cv cv ct ct 
v v cv+ cv+ ct+ ct+
F1 
v v+ cv cv+ ct ct+ 
(Testcross) 
F2
v cv+ ct+   580 
v+ cv ct     592 
v cv ct+       45 
v+ cv+ ct    40 
v cv ct        89 
v+ cv+ ct+   94 
v cv+ ct        3 
v+ cv ct+        5_ 
                    1448 
1) Consider v and cv (neglect ct for now)
v cv+ & v+ cv parentals; v cv & v+ cv+ recombinants

RF = (45 + 40 + 89 + 94) / 1448 = 0.185 = 18.5%

2) Now consider v and ct (neglect cv)
v ct+ & v+ ct parentals; v ct & v+ ct+ recombinants

RF = (89 + 94 + 3 + 5) / 1448 = 0.132 = 13.2%

3) Now consider cv and ct (neglect v)
cv+ ct+ & cv ct parentals; cv+ ct & cv ct+ recombinants

RF = (45 + 40 + 3 + 5) / 1448 = 0.064 = 6.4%

Now construct the linkage map

v          13.2      ct         6.4     cv

But: 13.2 + 6.4 ? 18.5 WHY?

Because we didnít count the double X-overs between v & cv when we really needed to count them twice.

RF = (45 + 40 + 89 + 94 + 3 + 3 + 5 + 5)/1448
      = 0.196 = 19.6%

INTERFERENCE--->A X-over in one region of the chromosome decreases the likelihood that a
X-over will occur in an adjacent region.

Calculate the frequency and number of double recombinants expected if there is no interference.
(i.e.) use the product rule

Expected frequency = (0.132)(0.064) = 0.0084
Expected number = (0.0084)(1448) = 12

Interference (I)=1-c.o.c.(coefficient of coincidence)

I = 1 - observed number of double X-overs
           expected number of double X-overs

I = 1 - 8/12 = 4/12 = 1/3 = 0.33 = 33%

Obvious departures from 1:1:1:1 ratio are easy to recognize, smaller departures require the c2 test.

c2 Test--->Tells us how often observations deviate from expectations purely on the basis of chance.

1) If linkage is not obvious, hypothesize that the genes are unlinked (Null Hypothesis).

2) Calculate c2.

c2 = total of (O-E)2 for all classes
                       E


(O-E)2 
Class 
O 
E 
(O-E)2 
   E 
AB 150 130 400 3.08
ab 135 130 25 0.19
Ab 110 130 400 3.09
aB 125 130 25 0.19
520 520 c2 = 6.54 
3) Calculate degrees of freedom (df).
df = (number of classes - 1) = 4-1 = 3

4) Find p from table (Table 5-3).
5) Accept null (unlinked) if p > 0.05

Reject null (linked) if p < 0.05

        a+ a b+ b c+ c                 x             a a b b c c
a+ b+c    788 
a b c+     802 
a+ b c       72 
a b+ c+      68 
a+ b+ c+    128 
a b c        122 
a b+ c           9 
  a+ b c+      _11_ 
                    2000 
First determine the middle gene a

1) Consider a and b (neglect c)
a+ b+, a b-->parentals; a+ b, a b+-->recombinants

RF=(72+68+9+11)/2000=160/2000=0.08=8%

2) Consider a and c (neglect b)
a+ c, a c+-->parentals; a+ c+, a c-->recombinants
RF=(128+122+9+11)/2000=270/2000=0.135=13.5%

3) Consider b and c (neglect a)
b+ c, b c+-->parentals; b+ c+, b c-->recombinants
RF=(72+68+128+122+9+9+11+11)/2000=430/2000
     =0.215=21.5%

I=1-(O/E) E=(0.08)(0.135)(2000)=21.6
I=1-(20/21.6)=1-0.926=0.074=7.4%

Chapter 7

Wild type (WT) compared to a mutant or variant.

MUTATION--->Change from one hereditary state to another.

GENE MUTATION--->Mutation in a specific gene resulting in a new allele (point, insertion, or deletion). (Chapter 7)

CHROMOSOME MUTATION--->Hereditary change involving chromosome segments, whole chromosomes, or entire chromosome sets. (Ch.8&9)

FORWARD MUTATION--->Any change from the WT allele.

REVERSE MUTATION--->Any change to the WT allele (true reversion).

SECOND SITE SUPPRESSOR--->Change in the same gene or a second gene resulting in a complete or partial phenotypic reversion to WT.

MUTANT--->An individual or strain carrying a mutation.

Mutations are used to genetically dissect biological functions and to study the process of mutation itself.

LOSS OF FUNCTION MUTATION--->

    A. Null mutation--->No activity.

    B. Leaky mutation--->Some residual activity.

GAIN OF FUNCTION MUTATION---> Mutation results in a new activity

SOMATIC MUTATION--->Mutation in any tissue other than the germinal tissue.

Clone--->Population of identical cells derived from one mutant progenitor (asexual).
(i.e.) Mitosis--->not transmitted to progeny
Often visualized as a sector (See Figs. 7-1 to 7-3).

GERMINAL MUTATION--->Mutation in tissue that forms sex cells (sperm & eggs).

Individual with the "new" germinal mutation will not show the phenotype.

Can be transmitted to progeny. (See Figs. 7-5 & 7-6)

CONDITIONAL MUTATION--->Allele only expresses mutant phenotype under certain environmental conditions.
(e.g.) ts (temperature sensitive)

AUXOTROPHIC MUTATION--->Must be supplied with certain nutrients (amino acids, nucleotides, vitamins).

Commonly used when studying microorganisms.
WT is prototrophic--->nutritionally self-sufficient.

RESISTANCE MUTATION--->Confers the ability to grow in the presence of an inhibitor.
(e.g.) antibiotic or pathogen

POINT MUTATION--->Single base pair change in DNA.

DELETION--->Removal of one or more bases of DNA.

INSERTION--->Addition of one or more bases of DNA.

MUTATION RATE--->

# of mutations     OR     # of mutations
cell division                         gamete

Human Genetics--->Germinal mutations detected by sudden appearance of abnormal phenotype in a pedigree with no previous record of abnormality.

Dominant mutations are easy to detect(see fig. 7-14).

Recessives can go unnoticed for several generations.

X-linked recessive easier to detect than autosomal.
(Queen Victoria-Hemophilia)

Different genes have different mutation rates.
(See Tables 7-1 & 7-2)

MUTANT HUNTS--->Experiments designed to isolate mutants that affect a specific biological function.

SELECTIVE SYSTEM--->Technique designed to separate rare mutant individuals from WT.

Need a selectable phenotype.

MUTAGENS--->Used to increase mutation rates.

(e.g.) Chemicals, radiation

Diploid organism (germinal)

      a+a+b+b+     x     aabb

a+a b+b                 a+abb aab+b
mutant 1                   mutant 2

Diploid organism (somatic)--->look for sectoring in a heterozygote.

Single cell haploid organisms

Advantages:

1) Grow as single cells in liquid culture or as colonies on plates.
2) Easy to examine millions of individuals.
3) Isolated single cells generate a clonal population (colony) of genetically identical cells.
4) Mutants easily identified.(dominant or recessive)

Detection of Reverse Mutations--->auxotrophe

1) Grow culture in minimal medium + supplement.
2) Plate cells on minimal medium without supplement.
3) Survivors are prototrophs.

Penicillin Enrichment--->Auxotrophe selection in bacteria--->Penicillin kills actively growing cells.

1) Grow cells in rich medium.
2) Transfer to minimal medium.
3) Add penicillin (prototrophs die, auxotrophs survive).
4) Plate cells on minimal medium + supplement.

Resistance Mutations--->

1) Grow cells in liquid culture.
2) Plate cells on selective medium (drug or virus)

Fluctuation Test--->Used to determine if resistant mutations were due to random mutations or changes in bacterial physiology.

1) Liquid culture of E. coli.
2) Mix individual cultures with bacteriophage T1.
3). Plate cell-phage mixture.
4) WT cells killed by phage T1.
5) Resistant mutants form colonies.

Fluctuation Test Predictions

A) Random mutation--->Rare mutations could occur early or late in the culture. Predict a large variation in the number of resistant colonies.

B) Physiological change--->Time for physiological adaptation would be relatively constant. Predict small variation in number of colonies.

Answer: Random Mutation
(See Fig. 7-19 & Table 7-3)

Replica Plating--->Demonstration

SOMATIC CELL GENETICS--->Applying mutagenic and selective techniques to animal and plant cell cultures.

Often only identify dominant because diploid.

Mutation and Cancer--->Cancer is a genetic disease caused by mutations in proto-oncogenes (dominant) or tumor suppressor genes (recessive).

Proto-oncogenes and tumor suppressor genes--->Normally carry out functions related to the regulation of cell division.

Mutation in proto-oncogene-->Uncontrolled cell growth (mutant clone)-->tumor (cancer).

Cancer can spread by metastasis.

Genetic Predisposition--->Mutant gene causes an increase in mutation frequency of other genes leading to cancer.

Xeroderma Pigmentosum--->Autosomal recessive predisposition to skin cancer caused by a mutation in a gene involved in repairing UV damaged DNA.

Retinoblastoma--->retinal cancer (Fig. 7-25)

A) Hereditary Predisposition--->Family history

    Inherited as Rr (recessive)
    rr is generated by rare mitotic X-overs.
    Two eyes affected

B) Sporadic--->No previous family history
    RR--->Rr--->rr
    Only one eye affected because requires 2 mutations.

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