Transcript ap ch 15 powerpoint

Chromosomal Basis of Inheritance
Chapter 15
I. Mendel’s Work was Cool!!
A. Mendel’s Laws
1. Segregation
The two alleles for each gene
separate during gamete formation
2. Independent Assortment
Each sex cell gets one of each
chromosome (mom’s or dad’s)
I. Mendel’s Work was Cool!!
B. Chromosomal Theory of Inheritance
1. Mendelian genes have specific
loci on chromosomes
2. Chromosomes undergo
segregation and independent
assortment.
II. Another Geneticist....Mr. Morgan
A. Morgan’s Experiments
1. Used Fruit Flies
• Breed Fast, only four pair of chromosomes,
cute
• Wild Type....Most common
Red-eyes, normal wings, gray
body
• Mutant Forms....Alternates to wild type
QuickTime™ and a
decompressor
are needed to see this picture.
A. Morgan’s Experiments
2. Mated a red-eyed female to a whiteeyed male
• Resulted in .....All red-eyed flies
3. Selfed these and got....
• 3 Red-eyed to 1 White-eyed
BUT....
• All the girls were red-eyed
• Half the boys were RED-eyed
• Half the boys were WHITE-eyed
B. Morgan’s Conclusions
1. Specific genes are on specific
chromosomes
2. Eye color in flies is linked to sex of the
fly.
3. SEX-LINKED genes: Genes that are
carried on the X sex chromosome
QuickTime™ and a
decompressor
are needed to see this picture.
III. Sex Chromosomes and their Cool Linkage
Issues
• Human Female = XX
• Human Male = XY
• Grasshopper Female =X
• Grasshopper Male =XX
• Bird Females =XY
• Bird Males = XX
QuickTime™ and a
decompressor
are needed to see this picture.
A. Sex-linkage Problems
• Traits travel only on the X!!
• Always cross xx and xy
• Sex-linked recessive disorders to
know....
• Colorblindness
• Duchenne Muscular Dystrophy
• Hemophilia
• 1. A woman is a
hemophiliac, but her
husband is not. What
are the phenotypic
ratios of their potential
children?
XhXh
x
XHy
•Genotype results
•1/2 XHXh, 1/2 XhY
•Phenotype results
•Girls – normal,
boys hemophilia
Xh
XH
Y
Xh
XH Xh
XH Xh
XhY
XhY
• 2. A man is a
hemophiliac, but his
wife is normal and
homozygous for the
trait. What are the
phenotypic ratios of
their potential
children?
XHXH x Xhy
XH
Xh
Y
XH
XH Xh
XH Xh
XH Y
XH Y
•Genotype results
•1/2 XHXh, 1/2 XHY
•Phenotype results
•All normal
B. X-Inactivation
• In humans, one X chromosome usually
becomes inactive during embryo
formation.
• A Methyl group -CH3 attaches to the
chromosome
• This is called a BARR Body.
• The X that becomes the Barr body is
random (could be from mom or dad)
• Stays active in the ovaries of females
IV. The Exception to Mendel’s Laws
• Every chromosome has hundreds or
thousands of genes.
A. Linked Genes
•Genes located on the same
chromosome tend to be inherited
together (chromosome is passed on as
a unit)
B. The Example....
• Fruit Fly Traits:
• b+ Gray (wild)
• vg+ Normal wings (wild)
•b
Black (mutant)
• vg
Vestigial wings (mutant)
• P1 = b+ b+ vg+ vg+ x b b vg vg
• F1 = all b+ b vg+ vg
• Next test cross......
• b+ b vg+ vg x b b vg vg
• According to Mendel, we get....
• Equal Numbers of all genotypes
• Oops....he got....
• 965
• 944
b+ b vg+ vg
b b vg vg
206 b+ b vg vg
185 b b vg+ vg+
C. The Conclusions
1. These genes must be on the same
chromosome and inherited together
2. Why did we get any of the
recombinants then?
•
CROSSING OVER
D. Terms...
Genetic Recombination: General
term for production of offspring with
new combinations of traits inherited
from two parents
Parental Types: Look like the parents
Recombinants: Look different from the
parents
E. Fun Problems
1. Frequency of Recombination
(crossover frequency)
 Take the number of recombinants and
divide by total offspring
 Multiply by 100 to get percent
2. Let’s do our flies from before....
• Total Recombinants…391
• Total Offspring....2300
• Frequency of Recombination.....17%
3. When two genes are on different
chromosomes, they have a 50%
recombination frequency.
F. What is a Linkage Map?
1. Genetic map based on recombination
frequencies.
2. The farther apart two genes are on a
chromosome, the more likely they are to
cross over.
3. Map Units: 1% recombination frequency.
These are NOT physical distances!
4. Genes that are really far apart on
chromosomes show a 50% recombination
frequency just like unlinked genes.
G. Making a Linkage Map
1. Determine the recombination
frequencies of each set of genes
using the equation.
2. Using the frequencies, try to map out
the space on the chromosomes
Example...
• b-cn = 9%
• b-vg = 17 %
• cn-vg = 9.5%
C. Genetic Disorders caused by Chromosome
Problems
1. Nondisjunction:
• Homologous pairs do not separate during
Meiosis I
• Sister Chromatids do not separate during
Meiosis II
2. Aneuploidy: Abnormal chromosome
number
• Trisomic: 3 copies of a chromosome
EX. Down’s Syndrome: Trisomy 21
• Monosomic: 1 copy of a chromosome
EX. Turner’s Syndrome: XO, missing a sex
chromosome.
Problems with non-disjunction in Sex Chromosomes
• XO – Turner’s Syndrome
• 1 in 5000 females
• No mental impairment
• Sterile
• XXY – Klinefelter’s Syndrome
•1 in 2000
•Male sex organs, but small testis
•Sterile
•Have female characteristics (breast
enlargement)
• XYY – no name
•Tend to be taller than average
3. Polyploidy: More than 2 complete
chromosome sets
• (more on this weirdness later)
D. Alterations in Chromosome Structure
1. Deletion:
• Fragment of a chromosome is lost
2. Duplication:
• Fragment joins a homologous
chromosome, duplicate genes result
(ugh!)
3. Inversion:
• Fragment reattaches in reverse order
4. Translocation:
• Fragment joins a non homologous
chromosome
E. Genetic Imprinting...
A. A gene on one chromosome is
silenced, while the other allele is left
to be expressed.
B. The same alleles may have different
effects on offspring, depending on
which parent they came from
E. Genetic Imprinting...
• Example...Fragile X Syndrome:
• -Abnormal X chromosome
• -Worse if you get the abnormal X from
mom
• -Worse in boys because they can only
get their X from mom.
C. Extranuclear Genes
1. Remember....MITOCHONDRIA and
CHLOROPLASTS have DNA.
2. This DNA gets passed on to children,
but not by Mendel’s laws
3. Mom’s egg has the cytoplasm!! Lots
of it!! Dad’s sperm was shaved to be
fluid dynamic.
4. You get your mom’s mitochondrial
DNA
Practice Problems
1. A man with hemophilia (a recessive,
sex-linked condition) has a daughter
of normal phenotype. She marries a
man who is normal for the trait.
• What is the probability that a
daughter of this mating will be a
hemophiliac?
• What is the probability that a son of
this mating will be a hemophiliac?
•
Show your work:
Cross = XHXh x XHY
XH
Genotype Results
XH
1/4 XHXH, 1/4 XHXh
1/4 XHY, 1/4 XhY
Phenotype Results
Girls – all normal
Boys – 1/2 normal, 1/2 hemophilia
Y
Xh
XH XH
XH Xh
XHY
XhY
• If the couple has four sons, what is the
probability that all four will be born
with hemophilia?
• 1/2 x 1/2 x 1/2 x 1/2 = 1/16
More Practice Problems
2. Pseudohypertophic muscular dystrophy is a
disorder that causes gradual deterioration
of the muscles. It is seen only in boys born
to apparently normal parents and usually
results in death in the early teens. Is this
disorder caused by a dominant or a
recessive allele? Explain.
• Is the inheritance autosomal or sex-linked?
Explain.
• Why is this disorder seen only in boys and
never in girls?
• Recessive – if it were dominant, at least
one parent would be affected
• Sex-linked – if it were autosomal, girls
would be impacted as well
• For a girl to get it, she would have to
inherit from both parents, but boys
don’t live past teen years
Even More Practice 
3. Red-green color blindness is caused
by a sex-linked recessive allele. A
color-blind man marries a woman
with normal vision whose father was
color-blind. What are the
phenotypic ratios of the potential
children. Don’t forget to specify the
sex. Show your work below.
XbY x XBXb
XB
Xb
•Genotype results
•1/4 XBXb, 1/4XbXb,
•1/4 XBY, 1/4 XbY
Y
Xb
XB Xb
XbXb
XB Y
XbY
Phenotype results
•1/2 girls normal, 1/2 girls colorblind
•1/2 boys normal, 1/2 boys colorblind
• Using the information above, what is
the probability of having a colorblind
daughter?
• 1/2 chance of girl, 1/2 chance of
colorblind = 1/4
• What is the probability that their FIRST
son will be colorblind?
• 1/2 chance boy is colorblind
• 4. A wild-type fruit fly (heterozygous for gray
body color and normal wings) is mated with
a black fly with vestigial wings. The offspring
have the following phenotypic distribution....
Wild Type...778
Black Vestigial...785
Black Normal...158
Gray Vestigial...162
•
What is the recombination frequency
between these genes for body color and
wing type? Show your work.
• 158 + 162 = 320 recombinants
• 1883 total offspring
• 320/1883
• 17%
• 5. What pattern of inheritance would
lead a geneticist to suspect that an
inherited disorder of cell metabolism is
due to a defective mitochondrial
gene?
• If the disorder is always inherited from
the mother
• 6. Determine the sequence of genes
along a chromosome based on the
following recombination frequencies:
• A-B....8 map units
• A-C...28 map units
• A-D...25 map units
• B-C...20 map units
• B-D...33 map units
•
28
33
• C----------B----A---------------------D
20
8
25
• 7. Assume that genes A and B are
linked and are 50 map units apart. An
animal heterozygous at both loci is
crossed with one that is homozygous
recessive at both loci. What
percentage of the offspring will show
phenotypes resulting from crossovers?
If you did not know that genes A and B
were linked, how would you interpret
the results of this cross?
• 50% of the results would be from
crossing over
• This is the same as you would get if the
genes were on different chromosomes
• 8. A space probe discovers a planet
inhabited by creatures who reproduce with
the same hereditary patterns seen in
humans. Three phenotypic characters are
height (T=tall,t=short), head appendages
(A=Antenna, a=no antennae), and nose
morphology (N=upturned snout,
n=downturned snout). Since the creatures
are not “intelligent”, Earth scientists are able
to do some controlled breeding experiments
using various heterozygotes in testcrosses.
Remember...testcrosses are crosses where
one parent is totally recessive.
• For a tall heterozygote with antennae,
the offspring were:
• Tall-antennae: 46
• Short-antennae: 7
• Short-no antennae: 42
• Tall-no antennae: 5
• What is the recombination frequency?
• 7+5 = 12/100 = 12%
• For a heterozygote with antennae and
an upturned snout, the offspring were:
• antennae-upturned: 47
• antennae-downturned: 2
• no antennae-upturned: 3
• no antennae-downturned: 48
• Calculate the recombination
frequencies
• 3+2 = 5/100 = 5%
• 9. Using the information above in #8, a
further testcross is done using a
heterozygote for height and nose
morphology. The offspring are....
• Tall-upturned: 40
• Short-upturned: 9
• Tall-downturned: 9
• Short-downturned: 42
• Calculate the recombination frequency
from these data.
• 9+9 = 18/100 = 18%
• 10. Use your answers from 8 and 9 to
determine the correct sequence of
the three linked genes.
12
5
• T-----------------A----N
18