Inheritance and HEREDITY = Genetics
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Transcript of Inheritance and HEREDITY = Genetics
Inheritance and HEREDITY = Genetics Father of Genetics: Gregor
Mendel
Austrian Monk Grew Peas and studied their characteristics
(inheritable feature) and their traits (varia- tions of a
character) Flower color: characteristic (gene) Pink or white: Trait
(allele) Mendels Work: Reasons for Studying Peas Easily
distinguished traits
Fast reproduction Self pollinate: develop true breeding varieties -
parents and offspring have the same characteristics from generation
to generation Controlled breeding of plants
Crossed plants with opposing traits to see what would happen P
generation: parent generation - initial cross of a genetic study F1
generation: first filial - first generation of offspring F2
generation: second filial - second generation of offspring
Monohybrid study of one gene Dihybrid study of two genes
Two types of Studies: Monohybrid study of one gene Dihybrid study
of two genes Two types of Organisms: Pure bred = Homozygous - two
of the same allele Hybrid = Heterozygous - two different alleles -
typically one dominant and one recessive Results of Mendels
Work
Law of Dominance: Tall pure bred X Short pure bred found all F1
were Tall one trait would over shadow the other trait Against the
convention of the time blending Took F1 Generation and did a self
cross Tall F1 X Tall F1
Law of Segregation: TookF1 Generation and did a self cross Tall F1
X Tall F1 found 75% of F2 were Tall and 25% of F2 were short traits
were preserved from one generation to the next and separated from
one another during reproduction Law of Segregation Law of
Independent Assortment
Tall Purple flowered pure bred X Short white flowered pure bred F1:
all tall purple flowered - self cross F2: ratio of 9 tall, purple:3
tall, white :3 short, purple :1 short, white traits were inherited
independently of one another and recombined during fertilization
Law of Independent Assortment Modern Terminology Factors = genes
Alternative traits = alleles
Each characteristic controlled by a gene pair or allele pair
Mendels Crosses: How to do genetics Problems
Homozygous Tall Pea plant X Homozygous Short Pea plant Genotype:
genetic make up Phenotype: what it looks like if heterozygous:
dominant trait appears MONOHYBRID CROSS: the Punnett square P
generation: TT x tt
Male Genotype Possible Gametes Female Genotype Possible Gametes
Male Genotype TT Possible Gametes Female Genotype T T t Possible
Gametes tt t Male Genotype TT Possible Gametes Female Genotype T T
t Possible Gametes tt t Male Genotype TT Possible Gametes Female
Genotype T T t Tt Possible Gametes Tt tt t Tt Tt Genotypic Ratio:
Phenotypic Ratio: Male Genotype TT Possible Gametes Female Genotype
T T t Tt Possible Gametes Tt tt t Tt Tt Genotypic Ratio: 4:0
Phenotypic Ratio: 4:0 F1 Generation Cross: Tt X Tt
Male Genotype Possible Gametes Female Genotype Possible Gametes F1
Generation Cross: Tt X Tt
Male Genotype Tt Possible Gametes Female Genotype T t T Possible
Gametes Tt t F1 Generation Cross: Tt X Tt
Male Genotype Tt Possible Gametes Female Genotype T t T Possible
Gametes TT Tt Tt t Tt tt Genotypic Ratio: Phenotypic Ratio: F1
Generation Cross: Tt X Tt
Male Genotype Tt Possible Gametes Female Genotype T t T Possible
Gametes TT Tt Tt t Tt tt Genotypic Ratio: 1:2:1 Phenotypic Ratio:
3:1 Test Cross T___ X tt crossing of a dominant phenotype with an
unknown genotype with a homozygous recessive Possible Outcomes T
___ t t
If some of the offspring show the recessive trait = Possible
Outcomes Must be recessive if tt appears T t t Tt tt t Tt tt
If some of the offspring show the recessive trait = unknown
genotype to be recessive If some of the offspring show the
recessive trait = Possible Outcomes T __ t Tt t Tt If no offspring
show the recessive = Possible Outcomes T T t Tt Tt t Tt Tt
If no offspring show the recessive = most likely that the unknown
genotype is homozygous dominant depends on the number of offspring
produced GENETICS AND PROBABILITY
Coin tosses Formation of gametes represents probability Homozygous
Dominant: probability of getting dominant trait = 100%
Heterozygous: probability of getting dominant trait = 50%
Inheritance: Rule of Multiplication: Probability and dependent
events Determining the probable genotypes of offspring Heterozygous
Parents Probability of getting a gene from mother: Probability of
getting a gene from father: Total probability = x = Probability of
being heterozygous in a heterozygous cross
Rule of Addition: Probability and independent events Probability of
being heterozygous in a heterozygous cross two ways to be
heterozygous: find individual probabilities and add them together +
= TT Tt Tt tt Probability and the Dihybrid Cross
Homozygous Tall Purple X Homozygous short white P Generation TTPP x
ttpp Each gamete must get one of each gene Gametes: TP or TP and tp
or tp F1 Genotypes: TtPp F1 Gametes: formed from TtPp Genes
separate in meiosis
Independent assortment allows for several combinations TtPp OR TtPp
Genes separate in meiosis each gamete gets one of each gene
F1 Genotypes: TtPp Genes separate in meiosis each gamete gets one
of each gene Independent assortment allows for several combinations
TtPp OR TtPp T P t p T p t P Dihybrid Genotype results in 4
possible gametes. Expands the Punnett to a 4x4 grid TP Tp tP tp TP
Tp tP tp TP Tp tP tp TTPP TTPp TtPP TtPp TP TtPp Ttpp TTPp TTpp Tp
TtPP TtPp ttPP ttPp tP TtPp ttpp tp Ttpp ttPp Dihybrid Cross Ratios
Genotypic Ratio: Phenotypic Ratio: Dihybrid Cross Ratios Genotypic
Ratio:
TTPP:TTPp:TtPP:TtPp:TTpp:Ttpp:ttPP:ttPp:ttpp Phenotypic Ratio:
Dihybrid Cross Ratios Genotypic Ratio: 1:2:2:4:1:2:1:2:1
TTPP:TTPp:TtPP:TtPp:TTpp:Ttpp:ttPP:ttPp:ttpp 1:2:2:4:1:2:1:2:1
Phenotypic Ratio: Dihybrid Cross Ratios Genotypic Ratio:
1:2:2:4:1:2:1:2:1
TTPP:TTPp:TtPP:TtPp:TTpp:Ttpp:ttPP:ttPp:ttpp 1:2:2:4:1:2:1:2:1
Phenotypic Ratio: Tall Purple: Tall White: Short Purple: Short
White Dihybrid Cross Ratios Genotypic Ratio:
1:2:2:4:1:2:1:2:1
TTPP:TTPp:TtPP:TtPp:TTpp:Ttpp:ttPP:ttPp:ttpp 1:2:2:4:1:2:1:2:1
Phenotypic Ratio: Tall Purple: Tall White: Short Purple: Short
White 9:3:3:1 Phenotypic Ratio for ANY Dihybrid Cross where both
parents are heterozygous for both traits. Extensions Of Mendels
Ideas
Not everything in genetics is as clear cut as Mendels idea of
dominance.Many traits are controlled by several other types of
inheritance patterns. Incomplete Dominance Blending
neither trait is dominant so both are partially expressed Red
flower X White flower = pink Different Rules: Key: Use a capital
Letter to represent the characteristic Use a capital letter
superscript to represent the trait CR = Red CW = White Sometimes
its just the R and W or just R and r Red X white Genotypic:
Phenotypic: Cross Pink with Pink Codominance Both traits are
equally expressed and DISTINGUISHABLE
Ex: Roan Cattle Red Bull x White Cow ROAN Offspring looking closely
at the fur = #s of red hairs and white hairs KEY: Function the same
way as incomplete dominance CR = red CW = white Codominance Crosses
Crosses Red with White Roan with Roan Multiple Alleles instead of
just two possibilities, there are more
gives multiple factors to consider: which alleles are dominant?
which are codominant? which are incomplete? Ex: Rabbit Fur one of
four genes
Full color dominant over chinchilla Chinchilla dominant over
Himalayan Himalayan dominant over Albino Key for Multiple Allele
Problems
Large Letter for characteristic Superscript for traits - only lower
case is the least recessive Full color - CF Chinchilla - CCh
Himalayan - CH Albino - c Cross a heterozygous full/chinchilla with
a heterozygous Himalayan/albino
Genotypic: Phenotypic: Rabbit Fur Mutations HUMAN BLOOD TYPE
Multiple Alleles and Codominance
Importance: Matching Donors and Recipients - Immunoglobin Responses
Blood Type Interaction Type Type Type Type Epistasis: One gene at a
different locus (location) alters the expression of another
gene
Mice Coloration: Black dominant over brown Color dominant over
colorless BB, Bb black bb brown HOWEVER: Also dependent upon the
gene for color CC, Cc color will be deposited in the hair follicle
cc no color despite BB, Bb, or bb - results in an albino Determine
Colors: BbCc: BBcc: bbCC: bbcc: Cross heterozygous male with a
heterozygous female (not 9:3:3:1) Determine Colors: BbCc: black
BBcc: albino bbCC: brown bbcc: albino Cross heterozygous male with
a heterozygous female (not 9:3:3:1) Labrador Retriever Genetics
Black is dominant to chocolate B or b Yellow is recessive epistatic
(when present, it blocks the expression of the black and chocolate
alleles) E or e Phenotype Possible Genotypes BBEEBbEEBBEeBbEe bbEE
bbEe BBee Bbee bbee Determine the number of chocolate labs produced
from a black female and a yellow male (BbEe x bbee) Polygenic
Traits: one characteristic is controlled by the cumulative effect
of many genes
Ex: Human height and skin color - each controlled by at least three
gene pairs T - tallness t - shortness TTTttt - medium height TtTtTt
- medium height TTTTTT - tall tttttt- short TtTtTt x TtTtTt -
result in many different combinations Human Skin Pigmentation
same basic process - gene codes for activity of melanocytes - all
humans have the same number of pigment producing cells in their
skin - the level of activity determines skin coloration - enhanced
by exposure to UV light - protection against FURTHER UV damage
Pleiotropy -a gene affects an individual in more than one way -
multiple affects sickle cell anemia: mutated gene causes blood
cells to have a sickle cell shape - causes blood clots, weakness,
anemia - PAGE 254 Comparison of Polygenic Inheritance and Pleitropy
Penetrance Degree to which the gene is expressed in the population
of those with the genotype - usually pertains to disease Ex: A
disease with an 85% penetrance means that 85% with the gene has the
disease Differs from expressivity which is variations in the
phenotype of those demonstrating the gene Three individuals with a
trait for green hair could have dark green, light green or medium
green Environmental Impact on Phenotype
page STUDYING INHERITANCE IN HUMANS
must be multigenerational Use Pedigree Analysis: Pedigree Chart to
study one trait at a time and determine the probability of a trait
being passed on to an offspring - usually done to predict the
likelihood of a child inheriting a hereditary disease: EX:
Hemophilia, breast cancer, cystic fibrosis Making a Pedigree
Chart
Square - male Circle - female Shaded - demonstrate trait Unshaded -
do not demonstrate trait Half-shaded: carrier for a homozygous
recessive trait Types of Disorders Recessive Dominant
Multifactoral: based on gene interactions and environmental
influence Chromosomal Inheritance
Thomas MORGAN: worked with fruit flies lots of offspring 2 week
life cycle only 4 pairs of chromosomes mated fruit flies until one
day he noticed a male with white eyes
wild type normal red eyes mutant phenotype - alternatives to the
wild white eyes Key: use mutant to indicate trait and mutant with +
for wild w+ = wild type w = white eyes gray body vs. black body
(mutant) b+ = gray b = black Figure 15.3 Morgan mated white, eyed
male with a red eyed female F1 = all red eyed
Crossed F1 F2 = white eyed - expected UNEXPECTED - white eyed were
only MALE All females (1/2 of population) were red eyed Half of
male (1/4 of population) were white eyed Figure 15.4 EXPERIMENT
Morgan mated a wild-type (red-eyed) female
The F2 generation showed a typical Mendelian 3:1 ratio of red eyes
to white eyes. However, no females displayed the white-eye trait;
they all had red eyes. Half the males had white eyes, and half had
red eyes. Morgan then bred an F1 red-eyed female to an F1 red-eyed
male to produce the F2 generation. RESULTS P Generation F1 X F2
Morgan mated a wild-type (red-eyed) female with a mutant white-eyed
male. The F1 offspring all had red eyes. EXPERIMENT CONCLUSION:
Gene for eye color is on the X chromosome Sex determination in
fruit flies XX = female XY = male X and Y Chromosomes Y chromosome
is structurally different from the X and is missing many of the
alleles that are found on the X Result: it only takes one recessive
on the X to get the recessive phenotype Cross: Wild type female
with mutant type male
F1: All Females: All Males: Cross: Wild type female with mutant
type male
Xw+Xw+xXwY F1: All Females: Xw+Xw All Males: Xw+Y Cross F1 female
with and F1 male
Females: Males: Cross F1 female with and F1 male
Females: Xw+Xw+or Xw+Xw - all red eyed Males: Xw+Y (red) or XwY
(white) CONCLUSION Since all F1 offspring had red eyes, the
mutant
white-eye trait (w) must be recessive to the wild-type red-eye
trait (w+). Since the recessive traitwhite eyeswas expressed only
in males in the F2 generation, Morgan hypothesized that the
eye-color gene is located on the X chromosome and that there is no
corresponding locus on the Y chromosome, as diagrammed here. P
Generation F1 F2 Ova (eggs) Sperm X Y W W+ Implications:
Re-evaluate Mendels Ideas What if traits are located on the same
chromosome? Law of Independent Assortment Revised: applies to genes
on different chromosomes HOWEVER: genes on the same chromosomes
only TEND to be inherited together WHY? Crossing over - rearranges
gene sequences by mixing parental DNA EVIDENCE FOR LINKED
GENES
Heterozygous Wild type (gray body, normal wings) crossed with a
recessive mutant (black body and vestigial wings) Genotypes:
EVIDENCE FOR LINKED GENES
Heterozygous Wild type (gray body, normal wings) crossed with a
recessive mutant (black body and vestigial wings) Genotypes:
b+bv+vXbbvv Recombinant (nonparental-type)
Double mutant (black body, vestigial wings) Wild type (gray body,
normal wings) P Generation (homozygous) b+ b+ vg+ vg+ x b b vg vg
F1 dihybrid (wild type) b+ b vg+ vg TESTCROSS b+vg+ b vg b+ vg b
vg+ b+ b vg vg b b vg+ vg 965 (gray-normal) 944 Black- vestigial
206 Gray- 185 normal Sperm Parental-type offspring Recombinant
(nonparental-type) RESULTS EXPERIMENT Morgan first mated
true-breeding wild-type flies with black, vestigial-winged flies to
produce heterozygous F1 dihybrids, all of which are wild-type in
appearance. He then mated wild-type F1 dihybrid females with black,
vestigial-winged males, producing 2,300 F2 offspring, which he
scored (classified according to phenotype). CONCLUSION If these two
genes were on different chromosomes, the alleles from the F1
dihybrid would sort into gametes independently, and we would expect
to see equal numbers of the four types of offspring. If these two
genes were on the same chromosome, we would expect each allele
combination,B+ vg+ and b vg, to stay together as gametes formed. In
this case, only offspring with parental phenotypes would be
produced. Since most offspring had a parental phenotype, Morgan
concludedthat the genes for body color and wing size are located on
the same chromosome. However, the production of a small number of
offspring with nonparental phenotypes indicated that some mechanism
occasionally breaks the linkage between genes on the same
chromosome. Figure 15.5 b+v+ b+v bv+ bv bv bv bv bv Genotypic
Ratio: Phenotypic Ratio: b+v+ b+v bv+ bv bv b+bv+v b+bvv bbv+v bbvv
bv bv bv Genotypic Ratio: Phenotypic Ratio: b+v+ b+v bv+ bv bv
b+bv+v b+bvv bbv+v bbvv bv bv bv Genotypic Ratio: 4:4:4:4
Phenotypic Ratio: 4:4:4:4 b+v+ b+v bv+ bv bv b+bv+v b+bvv bbv+v
bbvv bv bv bv
Genotypic Ratio: 4:4:4:4 = Parental Type same as parents Phenotypic
Ratio: 4:4:4:4 = Recombinants differ from parents Expected Cross:
Genotypic Ratio: 4:4:4:4 Phenotypic Ratio: 4:4:4:4 If 2300
offspring are produced: Parental Types (like parents) wild type
(b+bv+v) = 575 mutant (bbvv) = 575 Recombinants: (different from
parents) gray, vestigial (b+bvv) = 575 black, normal (bbv+v) = 575
ACTUAL RESULTS: Parental types: wild: 965 mutant: 944 Recombinants:
gray, vestigial: 206 black, normal: 185 HUGE DIFFERENCE: WHY??
Traits Linked on Chromosomes WHY RECOMBINANTS?? Crossing over How
much cross over? RECOMBINATION FREQUENCY - frequency that two genes
of sister chromosomes cross over in prophase I Figure 15.6 =
Testcross parents Gray body, normal wings
(F1 dihybrid) b+ vg+ b vg Replication of chromosomes Meiosis I:
Crossing over between b and vg loci produces new allele
combinations. Meiosis II: Segregation of chromatids produces
recombinant gametes with the new allele Recombinant chromosome
b+vg+ b vg b+ vg bvg+ bvg Sperm Meiosis I and II: Even if crossing
over occurs, no new allele combinations are produced. Ova Gametes
offspring b+ vg+ b vg b+ vg b vg+ 965 Wild type (gray-normal) b vg
b vg+ 944 Black- vestigial 206 Gray- 185 normal Recombination
frequency = 391 recombinants 2,300 total offspring 100 = 17%
Parental-type offspring Recombinant offspring Black body, vestigial
wings (double mutant) Recombination Frequency
RF = (# of recombinants/total offspring) X 100 ( ) Morgans Cross =
_____________ X 100 = 17% (2300) Practice Problems: Pg 291 # 5
Recombination frequencies reflect the distance between genes on a
chromosome.
Allows the calculation of relative distances between chromosomes up
to a recombination frequency of 50%. 50% or greater frequency
eliminates the difference between crossing over and simple
independent chromosomes Once gene frequencies are found, gene maps
can be constructed from the cross over data GENE MAPPING use
recombination frequency to map RELATIVE distances of genes the
farther two genes are from one another on a chromosome the more
likely there is to be crossing over as you look at the frequency of
cross over you can place the genes in order based on the
recombination frequencies translate frequency into map units 1% = 1
map unit Recombination frequencies 9% 9.5% 17% b cn vg Chromosome
The bvg recombination frequency is slightly less than the sum of
the bcn and cnvg frequencies because double crossovers are fairly
likely to occur between b and vg in matings tracking these two
genes. A second crossover would cancel out the first and thus
reduce the observed bvgrecombination frequency. In this example,
the observed recombination frequencies between three Drosophila
gene pairs (bcn 9%, cnvg 9.5%, and bvg 17%) best fit a linear order
in which cn is positioned about halfway between the other two
genes: RESULTS A linkage map shows the relative locations of genes
along a chromosome. APPLICATION TECHNIQUE A linkage map is based on
the assumption that the probability of a crossover between two
genetic loci is proportional to the distance separating the loci.
The recombination frequencies used to construct a linkage map for a
particular chromosome are obtained from experimental crosses, such
as the cross depicted in Figure The distances between genes are
expressedas map units (centimorgans), with one map unit equivalent
to a 1% recombination frequency. Genes are arranged on the
chromosome in the order that best fits the data. Figure 15.7
EXAMPLE: three genes X, Y and Z by doing cross over studies: X is
12 map units from Z Y is 7 map units from Z X is 19 map units from
Y Relative Gene Order: EXAMPLE: three genes X, Y and Z by doing
cross over studies: X is 12 map units from Z Y is 7 map units from
Z X is 19 map units from Y Relative Gene Order: X-Z-Y Four genes
are studied: ABCD
A & B are15 map units apart B & C are 27 map units apart C
& D are 7 map units apart D & A are 5 map units apart A
& C are 12 map units apart B & D are 20 map units apart
ORDER: Four genes are studied: ABCD
A & B are15 map units apart B & C are 27 map units apart C
& D are 7 map units apart D & A are 5 map units apart A
& C are 12 map units apart B & D are 20 map units apart
ORDER: CDAB or BADC Five Genes: ORDER: Gene Pair Recombination
Frequency A & B 5 A & P 2.5
45 B & O 25 B & I 40 B & P O & P 27.5 Five Genes:
ORDER: APBOI Gene Pair Recombination Frequency A & B 5
2.5 A & I 45 B & O 25 B & I 40 B & P O & P 27.5
Sex Determination X-Y System: mammals and fruit flies XY =
male
XX = female X-O System: grasshoppers and crickets one X = male two
X = female Z-W System: birds, some fishes and some insects WW =
Male ZW = female
use different letters to represent chromosomes so not to confuse
the system with the X-Y system Haploid-Diploid System: bees and
ants haploid - male: unfertilized eggs diploid - female: fertilized
eggs therefore, all male bees and ants are bastards (d) The
haplo-diploid system
Figure 15.9bd 22 + XX X 76 + ZZ ZW 16 (Haploid) (Diploid) (b) The
X0 system (c) The ZW system (d) The haplo-diploid system Sex-Linked
Inheritance in Humans
Fathers gametes determine the gender of the offspring passing of
traits from one generation to the next is determined by linkage
X-linked genes: genes found on X chromosome can be passed to
offspring by mother and father - father can only pass X-linked to
daughters Ex: hemophilia, color blindness Y-linked genes: genes
found on Y chromosome - can be passed only to sons by fathers Ex.
Hairy ear rims, SRY (sex determining region of the Y chromosome)
X-linked recessive traits
More common in males only takes one allele - hemizygous condition
Ex: color blindness Can occur in females- need one recessive from
each parent A normal visioned woman whos father was color blind has
children with a color blind man.
A) What is the probability they will have a color blind son? B)
What is the probability they will have a color blind daughter? C)
What is the probability they will have a color blind child? D) What
is the probability their daughter will be color blind? A normal
visioned woman whos father was color blind has children with a
color blind man.
A) What is the probability they will have a color blind son? 1/4 B)
What is the probability they will have a color blind daughter? 1/4
C) What is the probability they will have a color blind child? 1/2
D) What is the probability their daughter will be color blind? 1/2
A certain recessive gene on the X-chromosome causes death to
developing mouse fetuses. If a female carrier mates with a normal
male, what fraction of the offspring will be carriers of the trait?
Answer: A certain recessive gene on the X-chromosome causes death
to developing mouse fetuses. If a female carrier mates with a
normal male, what fraction of the offspring will be carriers of the
trait? Answer: 1/3 - since the gene only need be present on one
chromosome in the males, 1/4 (1/2 of males) will die in utero. Of
the remaining population, 1 out of 3 will be a carrier. X-Linked
Dominant Traits
Seen in males and females almost equally Ex: hypophosphatemia - low
blood phosphate Sex Influenced Traits Traits that are not carried
on the sex chromosomes but are enhanced by the sex hormones Ex:
Male pattern baldness on Chromosome 15 Dominant trait in males
Recessive trait in females Fate of the Double Xs in Female
Cells
Embryonic X-inactivation - formation of the Barr body early stages
of development, one of the Xs condenses and attaches to the inside
of the nuclear envelop - 50/50 chance for either to condense - the
resulting cells from that cell all have the same Barr body - can
cause differences in female phenotype depending on which chromosome
is inactivated Examples: calico cat, sweat glands, partial albinism
Figure 15.11 Two cell populations in adult cat: Active X Early
embryo:
Orange fur Inactive X Early embryo: X chromosomes Allele for black
fur Cell division and X chromosome inactivation Black Figure 15.11
ERRORS IN CHROMOSOMAL INHERITANCE
Separation Problems Non-disjunction: chromosomes or chromatids dont
separate properly Results in additional or missing chromosomes
Aneuploidy: abnormal chromosome # - type depends on which gamete is
received Monosomic: one copy of a chromosome Trisomic: three of one
chromosome Trisomy 21: Downs Syndrome Polyploidy: gamete has 2 full
sets of chromosomes due to complete non-disjunction:resulting
fertilization makes a triploid or tetraploid organism - animals =
death (almost always) - plantscan actually benefit - larger leaves,
fruits and flowers easier to deal with than aneuploidy because
chromosome numbers are equal Nondisjunction of sister chromatids in
meiosis II
Gametes n + 1 n 1 n 1 n 1 n Number of chromosomes Nondisjunction of
homologous chromosomes in meiosis I Nondisjunction of sister
chromatids in meiosis II (a) (b) Alterations in Chromosome
Structure:
errors in replication or crossover 4 types: Deletion: lose part of
the chromosome - can lose important genetic information 2.
Duplication: fragmented chromosome rejoins another chromosome
causing a section to be repeated 3. Inversion: fragmented
chromosome rejoins in reverse position 4. Translocation: fragmented
chromosome rejoins a none homologous chromosome Figure 15.14ad A B
C D E F G H Deletion Duplication M N O P Q R
Inversion Reciprocal translocation (a) A deletion removes a
chromosomal segment. (b) A duplication repeats a segment. (c) An
inversion reverses a segment within a chromosome. (d) A
translocation moves a segment from one chromosome to another,
nonhomologous one. In a reciprocal translocation, the most common
type, nonhomologous chromosomes exchange fragments. Nonreciprocal
translocations also occur, in which a chromosome transfers a
fragment without receiving a fragment in return. Extranuclear Genes
Are genes found in organelles in the cytoplasm
Mitochondrial DNA Trace lineages on mothers side Chloroplasts