Download Chapter 14: Mendel and the Gene Idea

Chapter 14: Mendel and the Gene Idea
14.1 Mendel used the scientific approach to identify
two laws of inheritance
14.2 The laws of probability govern Mendelian
inheritance
14.3 Extensions to Mendelian Genetics:
Inheritance patterns are often more complex than
predicted by simple Mendelian genetics
14.4 Many human traits follow Mendelian
patterns of inheritance
PowerPoint TextEdit Art Slides for
Biology, Seventh Edition
Neil Campbell and Jane Reece
Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings
Before we begin: Define these terms:
Gene: functional unit of heredity - a segment of DNA located in a specific site on a chromosome
that gives the instrcutions to make one (or more) enzyme or other protein.
Dive into the chromosome here! !
Phenotype: Observable characteristics of an organism, for example, hair type, eye color, height.
Genotype: The genetic identity of an individual, including alleles that do not show as outward
characteristics.
Locus: The position on a chromosome where a gene, or some other sequence, is located.
Allele: One of two or more alternative forms of a gene; for example, allele "A" may produce a tall
plant, while allele "a" gives a short plant.
Dominant allele: An allele that is fully expressed in the phenotype, regardless of the characteristics
of its counterpart allele.
Recessive allele: An allele which is not expressed in the phenotype unless two copies are present
in the individual, ie, in homozygous recessive condition.
Homozygous: the state of having two identical alleles of a particular gene (eg AA, aa).
Heterozygous: the state of having two different alleles of a particular gene (eg Aa).
Mendel’s Law of Segregation: Homologous chromosomes separate during the formation of
gametes, and are distributed to different gametes, so that every gamete receives only one member
of the pair.
Mendel’s Law of Independent Assortment: Homologous chromosomes are random distributed at
the metaphase plate such that a mixture of maternal and paternal homologous assort to each
gamete.
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Figure 14. 3 and 14.5 Mendel’s Peas
1. Parental Generation (P): Purple flowers x
white flowers: Mendel bred (crosspollinated) two types of pea plants that were
true-breeding for flower color.
2. F1 Generation: When Mendel grew seed
from this First Filial generation (F1) and
looked at their flower color, he saw 100%
purple flowers. Result: In the (F1)
generation, the white trait was masked.
3. F2 Generation Mendel went one step farther,
allowing the F1 to “self”. Result: The white trait
re-appeared in the F2 generation in a ratio of
~3 purple plants to 1 white.
Next Slide: Mendel’s Genius: He realized
that these results were explainable if three
things were true.
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Figure 14.4: Mendel’s Genius: He realized that these results were
explainable if three things were true. He hypothesized that:
1. Every trait (like flower color) is controlled by two
"heritable factors” (genes), one from each parent
-Each form of the gene is called an allele, and
produces a slightly different protein
Alleles controlling
Flower color:
Purple
White
2. If the two alleles differ, one is dominant
(observed in appearance or physiology) and
one is recessive (only observed when
an individual has two copies of the recessive allele).
Dominant traits mask the appearance of recessive traits.
Alleles controlling
Pea shape:
3. Alleles are randomly segregated into eggs and sperm
when the homologues separate during Meiosis I, allowing
all possible combinations of factors to occur in the gametes.
Wrinkled
Round
Homologous Chromosomes
Mendel's Law of Segregation - When gametes (eggs or sperm) are formed,
the two factors (alleles) separate, and only one factor (allele) is present
in each gamete. [ie: the homologues separate during meiosis]
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Table 14.1 The Results of Mendel’s F1 Crosses for
Seven Characters in Pea Plants
Result: For a
monohybrid cross
(single trait),
Mendelian Genetics
predicts that a 3:1 ratio
(Dominant : recessive)
will be seen in the F2
generation
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Figure 14.6 Phenotype versus genotype
Phenotype (Observable characteristics)
Genotype (Gene alleles)
Purple
PP
(homozygous)
3
Purple
1
Pp
(heterozygous)
2
Pp
(heterozygous)
Purple
1
White
pp
(homozygous)
Ratio 3:1
Ratio 1:2:1
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1
Doing a genetic cross (Monohybrid Cross = 1 gene):
(Most of you have done this in lab already)
Geneticists use letters to represent alleles.
*Dominant trait = Capital letter ; recessive trait = a lowercase letter.
*The same letter is used to indicate both alleles.
*Remember, the definition of “Dominant” means…”SEEN IN THE PHENOTYPE”
Examples
= Flower color:
= Seed color:
= Seed shape:
P= purple,
Y= yellow,
W = wrinkled,
p= white
y = green
w = round
In Humans...
= Widow's peak:
= Freckles:
= Earlobes:
= Cystic fibrosis
W = widow's peak,
F = Freckles,
E = unattached,
C = no CF,
w = continuous hairline (which are you?)
f = no freckles (which are you?)
e = attached (which are you?)
c = cystic fibrosis
E-Z steps for doing genetics problems:
1. Indicate the genotype of the parents using letters
2. Determine what the possible gametes are
3. Determine the genotype and phenotype of the children after reproduction. To consider every
type of offspring possible, use a Punnett Square in which all possible types of sperm are lined
up vertically and all types of eggs are lined up horizontally:
4.Copyright
Fill in© 2005
thePearson
squares
by "multiplying" the alleles from mom and dad:
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Human Genetics: Recessively Inherited Disorders
1. Autosomal recessive human disorders: Show up only in the homozygous recessive
person (aa). With two carrier parents, there is a 1/4 chance of an affected child. There
is 2/4 chance of a carrier child (50%)
A.Cystic fibrosis: Mutation in CFTR (Chloride channel)
B.Tay-Sachs: Mutation in Hexaminidase A gene.
(Jewish - Ashkenazi)
C. Sickle-cell disease: Mutation in
hemoglobin gene. The most common
inherited disease of African-Americans
(1:400 affected).
With these parents, what is the chance of
- giving birth to an affected child?
- giving birth to a child who is also a carrier?
Heterozygote advantage; For some
disease alleles, being a heterozygote
(‘carrier’) offers protection against another disease.
Examples:
1. Sickle cell disease: Ss carriers are resistant
to malaria
2. Cystic fibrosis: Cc carriers are resistant to cholera
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unaffected
carrier mother
Heterozygote advantage
One in 10 Africans are carriers
(heterozygotes) for the sickle cell
trait, even though the recessive
condition, SCD, can be fatal. Why
is the incidence so high?
Sadly, over 1 million people / year
(mostly children, pregnant women)
die from Malaria (2,700 per day)….
People who are
HbBHbB = normal hemoglobin, but
may succumb to malaria if they live in
a region with endemic malaria.
P. falciparum malaria
HbSHbS = have sickle cell disease
SCD and are anemic, in pain, and
may suffer strokes or death.
HbBHbS = heterozygotes suffer mild
symptoms of SCD, but if bitten by a
mosquito carrying the malaria
parasite, the parasite life cycle is
interrupted due to the difficult-todigest, irregular HbS
In parts of the world where malaria
is common, the incidence of SCD is
common because there is an
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Inc. publishing as Benjamin Cummings
advantage
to being
a carrier.
Frequency of Hbs allele
1–10%
10–20%
Human Genetics: Dominantly Inherited Disorders
Autosomal dominant human disorders: child will show the phenotype if he / she receives just 1
allele from either parent (OR if a spontaneous mutation appears in the zygote).
A. Achondroplasia (dwarfism): Cause: Mutation in FGFR3 (fibroblast growth factor receptor 3 gene)
severely limits bone growth. AA = Dominant lethal - fatal. Aa = dwarfism. aa = no dwarfism. 99.96% of all
people in the world are (aa). (Fig 14.15)
B. Polydactyly (extra fingers or toes): PP or Pp = extra
digits, aa = 5 digits. 98% of all people in the world are (pp).
C. Huntington's chorea Cause: Mutation of Huntingtin
gene damages nerves and brain. Dominant lethal (HH = fatal).
Heterozygotes (Hh) live to be ~40 or so, then their nervous
system starts to degenerate. (Woody Guthrie).
D. Marfan's syndrome - Cause: Defect in Fibrillin gene involved
in connective tissue; affects skeletal system, eyes, and CV
system. (Pleiotropic). *Abraham Lincoln, Flo Hyman
With these parents, what is the chance of
- giving birth to an affected child?
- giving birth to a child who is also a carrier?
E. Progeria - Cause: Mutation of
Lamin-A Gene. Defective Lamin-A makes the nucleus
unstable, leading to aging, CV disease, stroke (Note:
since children with Progeria don’t live to reproduce, this is
an example of a spontaneous (rather than inherited), Dominant disorder.
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Fig 14.7: Is there a way to distinguish between a purple
heterozygote (Pp) and a purple homozygous dominant (PP)?
Yes = A testcross:
Mate the unknown genotype
(Purple, but PP or Pp?)
with a recessive genotype:
white (pp) in this case.
If offspring are 100% purple:
the unknown genotype was
homozygous dominant (PP),
to begin with; and
100% of the offspring will be
heterozygotes.
If offspring are 50% purple
And 50% white:
the unknown genotype was
a heterozygote (Pp),
To begin with.
So now you know!
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Fig 14.8: Studying Two Traits at once: (Dihybrid Cross)
AA or Aa = Purple; aa = white
BB or Bb = Tall; bb = short
Result: Classic 9:3:3:1 ratio: 9 Purple & Tall, 3 Purple & short, 3 white & Tall, 1 white & short
One really important thing that Mendel noticed from this type of cross was that the two traits
(like flower color, height) are inherited independently - not together as a unit. This has become
known as Mendel's Law of Independent Assortment - Genes for various traits assort into
gametes independently (due to homologues lining up randomly at the metaphase plate).
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(See Fig 14.9) A simpler way to do a Dihybrid Cross or beyond...
(or, how to avoid making anything larger than a simple Punnett square)
Mendelian genetics reflect the laws of probability
For a dihybrid cross - the chance that 2 independent events will occur together is the product
of their chances of occurring separately.
The chance of yellow (YY or Yy) seeds
The chance of round (RR or Rr) seeds
The chance of green (yy) seeds
The chance of wrinkled (rr) seeds
= 3/4 (the dominant trait)
= 3/4 (the dominant trait)
= 1/4 (the recessive trait)
= 1/4 (the recessive trait)
So...
(Remember multiplication with fractions..its easy! Just multiply across the top and the bottom! )
The chance of yellow and round
The chance of yellow and wrinkled
The chance of green and round
The chance of green and wrinkled
= 3/4 x 3/4 = 9/16
= 3/4 x 1/4 = 3/16
= 1/4 x 3/4 = 3/16
= 1/4 x 1/4 = 1/16
9:3:3:1 - Sound familiar? And no Punnett square needed (whew). Therefore, you can avoid
doing a Punnett square if you can reduce the problem to a series of probability statements.
Next, you guessed it, a tri-hybrid cross
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Figure 14.10 Incomplete dominance in snapdragons
Incomplete Dominance:
2 incompletely dominant alleles
(no ‘recessives’) CR and CW
P Generation
Red
C RC R

Gametes CR
The F1 heterozygote (CRCW) is
intermediate in phenotype (pink!)
compared to the two parentals.
Why does this happen?
Just like with Purple:white, this comes
about when one of the alleles lacks the
ability to produce a pigment, but in this
case, the loss IS detectable in the
phenotype.
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CW
Pink
C RC W
F1 Generation
Gametes
Eggs
Note that traits do not ‘blend’: all 4
colors result in the F2.
White
CW CW
F2 Generation
1⁄
2
CR
1⁄
2
Cw
1⁄
2
1⁄
2
CR
CR
1⁄
2
CW
1⁄
W
2C
CR CR CR CW
CR CW CW CW
Sperm
Table 14.2 Determination of ABO Blood Group by
Multiple Alleles (Co-dominance)
ABO Blood Types
3 alleles;
(Each dominant allele
codes for the linkage of a
glycophorin sugar
molecule to the surface of
RBCs.)
2 are dominant
(IA and IB ),
1 is recessive (i)
This results in 4
possible phenotypes
Which do YOU have?
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Figure 14.12 Polygenic inheritance: skin color

AaBbCc
2 or more genes act
additively to produce
a phenotype
AaBbCc
aabbcc Aabbcc AaBbcc AaBbCc AABbCc AABBCc AABBCC
Ex: human height,
skin color
20⁄
64
15⁄
6⁄
64
1⁄
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64
64
Skin pigmentation
Pleiotropy:
(Greek: pleion, more)
How 1 gene can affect
multiple aspects of the
phenotype.
Ex: Sickle Cell Disease,
Cystic Fibrosis,
Marfan’s Disease
Many organ systems
affected, since each
mutation changes a
protein involved in many
bodily functions
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Fig 14.11 Epistasis: Presence of alleles at one locus affect the
expression of alleles at a second locus. (Epi = above)
Other examples: Albinism in humans
[In class] Can use
probability to solve
these; Variations on
a 9:3:3:1 result
(9:3:4; 9:7, 15:1)
Make sure you can
work an epistasis
genetics problem!
Bonus: What
genotype was
the yellow lab
BB, Bb, or bb?
The nose knows!
9 Black :
4 Yellow :
3 Chocolate
(BBEE, BbEE, BbEe)(BBee, Bbee, bbee) (bbEE, bbEe)
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Fig 15-10: Sex-linked disorders in humans: X Chromosome:
Females: Inherit 2 X chromosomes, one from each parent - 2 alleles for all genes (Fig 15.9)
Males: Inherit one X chromosome (from their mother), so only 1 allele for every gene,
Technically, males are hemi-zygous - neither homozygous nor heterozygous at the X!
If that one allele is mutated, the male will be affected (show the trait).
Daughters are almost always unaffected (carrier females) but have a 50% chance of giving birth
to an affected son and a 50% chance of giving birth to a carrier daughter.
Affected females are more rare because they must inherit two recessive alleles.
List the 5 possible genotypes and phenotypes:
1.
Color blindness: sex-linked recessive
Defect in opsin gene;
XA = normal vision, Xa = color blind
2.
Duchenne muscular dystrophy: Affects 1:3500
US. Males. Children lack Dystrophin (X chromosome).
Progressive muscle weakness and loss of coordination
leads to death in early adulthood.
3.
Hemophilia: Affected children
lack the blood clotting factor VIII. Queen Victoria
was a carrier; thus her sons had a 50% chance of
being affected and her daughters had a 50%
chance of being carriers.
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Fig 15.11: Dosage compensation in females:
X-Chromosome Inactivation (XCI)
1
Nucleus
Females normally inactivate one of their 2 Xs
in early embryonic development (XCI). The inactivated
X-chromosome condenses into a Barr Body
This inactivation is a random event, thus females
(this means all K101 females) are Mosaics
50% of our cells have one active X, and
50% have the other active X.
2
2
Barr body
(randomly inactivated chromosome)
Followed by mitosis
Female
calico kitty:
mosaic!
Dosage Compensation = inactivation
of 1 X chromosome per cell that
equalizes the expression of X-linked
genes in males and females
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3
50% of cells
50% of cells
Just a mention about Multifactorial Diseases:
(and see Fig 14.13)
Cancer, Autism and Schizophrenia are multi-factorial - they are not
'caused' by one dominant or recessive gene, but may be the result of several
inherited genetic susceptibilities PLUS environmental factors.
Example: Cancer: Can be an inherited susceptibility - loss of tumor
suppressors (BRCA1, BRCA2, p53), oncogene activation, OR a spontaneous
mutation PLUS environmental factors .
Example: Autism: at least 4 genes linked to this neurodevelopmental
condition, particularly on Chromosomes 2, 3, 7, 15, and X. Some have to do
with language development, some with brain development, some with
producing brain neurotranmitters.
Example: Schizophrenia: several genes linked to the development of this
disorder, including Chromosome 22 plus a combination of family genetics
and environmental factors.
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Two homologous
chromosomes
undergo synapsis
in meiosis
Fig 15.5: “Linked Genes”
Mendel showed that genes are
inherited independently, but in 1910,
Crossing-over
between a pair of
homologous
chromatids
Thomas Hunt Morgan (Nobel Prize
1933) and Alfred Sturtevant showed
that Independent assortment does not
apply if 2 genes are very close
together “Linked” on a chromosome.
Meiosis I
Crossing Over during meiosis
results in recombination
between genes. The closer two genes
are on a chromosome, the more likely
that they are linked and will travel
together. The farther apart they are, the
more likely they will be separated.
Meiosis II
Can use this information to create
maps of gene order on chromosomes
(See Fig 15.7. 15.8 for more info)
Parental
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Recombinant Recombinant
Four haploid cells produced
Parental
Making a physical map of a chromosome based on recombination frequency (crossing
over): Morgan and Sturtevant constructed many “two-point” testcrosses (see Fig 14-7) to get a
estimate of how closely linked genes were on the Drosophila chromosomes.
Recombination frequency: Frec (or % crossing over))
391 recombinants X 100= 17%
2300 offspring
Thus, we can say that these genes are 17 “Centimorgans” or
“map units” apart (cM or m.u.)
Grey,
normal wings
BbVv
Try it: Problem 1
DdEe x ddee
65 Ddee, 25 ddEe, 256 DdEe, 266 ddee
BV
Grey,
normal
Problem 2:
AaEe x aaee
424 Aaee, 430 aaEe, 70 AaEe, 76 aaee
0
50
100
Bv
Black,
Grey,
vestigial vestigial
bV
Black,
normal
bv
Black,
vestigial wings
bbvv
Make a map of these 3 genes,
assuming E is in the middle
bv
Expected results:
independent assortment
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Actual results
BbVv
575
bbvv
Bbvv
575
575
Parentals
965
944
bbVv
575
Recombinants
206
185
Making a physical map of a chromosome based on recombination frequency (crossing
over): Morgan and Sturtevant constructed many “two-point” testcrosses (see Fig 14-7) to get a
estimate of how closely linked genes were on the Drosophila chromosomes.
Recombination frequency: Frec (or % crossing over))
391 recombinants X 100= 17%
2300 offspring
Thus, we can say that these genes are 17 “Centimorgans” or
“map units” apart (cM or m.u.)
Try it: Problem 1
DdEe x ddee
65 Ddee, 25 ddEe, 256 DdEe, 266 ddee
1. Circle the parentals
2. Add the recombinants = 65 + 25 = 90
3. Add all the offspring = 612 total
4. 90 / 612 X 100 = 14.7 D-E are 14.7 m.u. apart
Problem 2:
AaEe x aaee
424 Aaee, 430 aaEe, 70 AaEe, 76 aaee
1. Circle the parentals
bv
2. Add the recombinants = 424 + 430 = 854
3. Add all the offspring = 1000 total
Black,
4. 854 / 1000 X 100 = 85.3
vestigial
wings
E-A are 85.3 m.u. apart
bbvv
Make a map of these 3 genes, assuming E is in the middle
D
E
A
Expected results:
0
50
100
independent assortment
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Actual results
Grey,
normal wings
BbVv
BV
Grey,
normal
BbVv
575
bv
Black,
Grey,
vestigial vestigial
bbvv
Bbvv
575
575
Parentals
965
Bv
944
bV
Black,
normal
bbVv
575
Recombinants
206
185
Hardy-Weinberg ‘Equlilbrium’
The Hardy-Weinberg equation : Basic assumptions
1. Mutation is not occurring
2. Natural selection is not occurring
3. The population is infinitely large
4. All members of the population breed
5. All mating is totally random
6. Everyone produces the same number of offspring
7. There is no migration in or out of the population
(This is never fully true for any population, but
If you know the allele frequency,
Can determine the genotype frequencies
Allele Frequency: Dom + rec = 1 (p + q = 1)
Genotype Frequency: p2 + 2pq + q2 = 1
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Hardy Weinberg: Basic Relations
1. Two alleles at a gene - A and a
2. Frequency of the A allele = p
3. Frequency of the a allele = q
4. Allele frequencies: p + q = 1
5. Genotype frequencies:
p2 (AA) + 2pq (Aa) + q2 (aa) = 1
p
q
p p2
pq
q pq
q2
Problem 1. Allele B is present in a large, random mating
population at a frequency of 54 %
1. What is the frequency of allele b? ______________
2. What proportion of the population is expected to be
-homozygous for B (BB)? __________ ______
-homozygous for b (bb)? __________
-heterozygous? (Bb) ______________
Problem 2. In a large, random-mating population, the frequency of homozygous
recessive (aa) individuals for sickle cell disease is 90 per 1000 individuals.
What is the frequency of
1. the recessive disease aa? ______________
2. the recessive allele a? ______________
3. the dominant allele A? ______________
4. the heterozygous individual Aa? ______________
5. the homozygous dominant individual AA? ______________
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Hardy Weinberg: Basic Relations
1. Two alleles at a gene - A and a
2. Frequency of the A allele = p
3. Frequency of the a allele = q
4. Allele frequencies: p + q = 1
5. Genotype frequencies: p2 (AA) + 2pq (Aa) + q2 (aa) = 1
p
q
p p2
pq
q pq
q2
Problem 1. Allele B is present in a large random mating
population at a frequency of 54% p
1. What is the frequency of allele b? q _0.46________
2. What proportion of the population is expected to be
-homozygous for B (BB)? p2___0.291_____
-homozygous for b (bb)? q2___0.216_____
-heterozygous? (Bb) 2pq_ 2(.54)(.46) = 0.495__
Problem 2. In a large, random-mating population, the frequency of homozygous
recessive (aa) individuals for sickle cell disease is 90 per 1000 individuals.
1. What is the frequency of the recessive disease aa? q290/1000 = 0.09
2. What is the frequency of the recessive allele a? q sq.rt (0.09) = 0.3
3. What is the frequency of the dominant allele A? p1 - 0.3 = 0.7
4. What is the frequency of the heterozygous individual Aa? 2pq2(.3)(.7) = 0.42
5. What is the frequency of the homozygous dominant individual AA? p2 = 0.49
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(Not on exam): EXCITING and SURPRIZING new results in Nature, March 17, 2005
about the Sequence of the X-chromosome and X-chromosome inactivation.
Nature 434, 400 - 404 (17 March 2005); doi:10.1038/nature03479
X-inactivation profile reveals extensive variability in X-linked gene expression
in females
LAURA CARREL AND HUNTINGTON F. WILLARD
In human females, up to 15% of all genes on the ‘inactive’ X escape XCI, and are expressed from both the
active and the inactive X. An additional 10% of X-linked genes show variable patterns of inactivation and are
expressed to different extents in different women.
Nature 434, 400 - 404 (17 March 2005); doi:10.1038/nature03440
The DNA sequence of the Human X chromosome
The International Human Genome Sequence Consortium
We have determined 99.3% of the euchromatic sequence of the human
X-chromosome. We found 1,098 genes in this sequence, of which 99
encode proteins expressed in the testis.
Nature 434, 279 - 280 (17 March 2005); doi:10.1038/434279a
Genome biology: She moves in mysterious ways
CHRIS GUNTER
The human X chromosome is a study in contradictions. The detailed
sequence of the X, and a survey of inactivated genes in females, help to
illuminate this unique 'evolutionary space'…
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Objectives:
1. Define the terms phenotype, genotype, locus, allele, dominant allele, recessive
allele, homozygous, and heterozygous.
2. Describe Mendel’s Laws (principles) of segregation and independent assortment.
3. Solve genetics problems involving monohybrid, dihybrid, (trihybrid) and test
crosses.
4. Use probability (3/4 x 3/4…) to predict the outcomes of genetic crosses.
5. Be able to discuss the difference between autosomal dominant and autosomal
recessive disorders, being able to give specific examples and work genetics
problems.
6. Solve genetics problems involving incomplete dominance, codominance / multiple
alleles, and epistasis;
7. Pleiotropy: discuss how a single gene can affect many features of the organism.
8. Discuss the the inheritance of X-linked genes in mammals.
9. Discuss what it means to be hemizygous, and what is means to be a mosaic.
What is a Barr Body, and what is dosage compensation?
10. Define linkage, and relate it to specific events in meiosis.
11. Show how data from a test cross involving alleles of two loci (‘two point test cross’)
can be used to distinguish between independent assortment and linkage.
12. Know how to calculate allele frequency using the Hardy-Weinberg equation.
13. Explain how prenatal testing, carrier screening, and newborn screening can be
used to detect genetic conditions before and after birth.
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