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5.0 Linkage
Prepared by Pratheep SandrasaigaranLecturer at Manipal International University
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By the end of this chapter you should be able to:
• Identify Mendel's law of segregation and independent assortment
• Define Chromosomal linkage.• Know the importance of Chromosomal
linkage.• Make genetic map unit .• Make physical map units.
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Diagram adopted fromInternet Sources
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5.1 Mendel's law of segregation vs independent assortment
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• Mendel mated two contrasting, true-breeding varieties, a process called hybridization.
• The true-breeding parents are the P generation.
• The hybrid offspring of the P generation are called the F1 generation.
• When F1 individuals self-pollinate or cross- pollinate with other F1 hybrids, the F2 generation is produced.
Mendel’s Experimental, Quantitative Approach
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Diagram adopted fromCampbell Biology (10th Edition)
Mendel’s Experimental
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Figure 14.2b
RESULTS (F1)
• When pollen from a white flower was transferred to a purple flower, the first-generation hybrids all had purple flowers.
• The result was the same for the reciprocal cross, which involved the transfer of pollen from purple flowers to white flowers
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First filialgenerationoffspring(F1)
5 Examinedoffspring:all purpleflowers
Diagram adopted fromCampbell Biology (10th Edition)
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RESULTS (F2)
• When Mendel crossed the F1 hybrids, many of the F2 plants had purple flowers, but some had white
• Mendel discovered a ratio of about three to one, purple to white flowers, in the F2 generation
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P Generation(true-breeding
parents) Purpleflowers
Whiteflowers
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Diagram adopted fromCampbell Biology (10th Edition)
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P Generation(true-breeding
parents)
F1 Generation(hybrids)
Purpleflowers
Whiteflowers
All plants had purple flowers
Self- or cross-pollination
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Diagram adopted fromCampbell Biology (10th Edition)
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P Generation(true-breeding
parents)
F1 Generation(hybrids)
F2 Generation
Purpleflowers
Whiteflowers
All plants had purple flowers
Self- or cross-pollination
705 purple-flowered
plants
224 whiteflowered
plants
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Diagram adopted fromCampbell Biology (10th Edition)
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• Mendel reasoned that only the purple flower factor was affecting flower color in the F1 hybrids.
• Mendel called the purple flower color a dominant trait and the white flower color a recessive trait.
• The factor for white flowers was not diluted or destroyed because it reappeared in the F2
generation.
• Mendel called it as “heritable factor” is what we now call a gene.
The Law of Segregation
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Figure 14.4
Allele
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Allele for purple flowers
Locus for flower-color gene
Allele for white flowers
Pair ofhomologouschromosomes
Diagram adopted fromCampbell Biology (10th Edition)
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• Alternative versions of genes account for variations in inherited characters, now called alleles.
• For example, the gene for flower colour in pea plants exists in two versions, one for purple flowers and the other for white flowers.
• Each gene resides at a specific locus on a specific chromosome.
The Law of Segregation-Mendel’s First Model:
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• For each character, an organism inherits two alleles, one from each parent
• The two alleles at a particular locus may be identical, as in the true-breeding plants of Mendel’s P generation
• Alternatively, the two alleles at a locus may differ, as in the F1 hybrids
The Law of Segregation-Mendel’s Second Model:
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• If the two alleles at a locus differ
• Then one (the dominant allele) determines the organism’s appearance.
• The other (the recessive allele) has no noticeable effect on appearance
The Law of Segregation-Mendel’s Third Model:
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• law of segregation: the two alleles for a heritable character separate (segregate) during gamete formation and end up in different gametes.
• Thus, an egg or a sperm gets only one of the two alleles that are present in the organism.
• Mendel’s segregation model can be shown using a Punnett square.
The Law of Segregation-Mendel’s Forth Model:
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ARepresents a dominant allele
a Represents a recessive allele
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Figure 14.5-1
P Generation
Appearance:Genetic makeup:
Gametes:
Purple flowers White flowersPP pp
P p
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Diagram adopted fromCampbell Biology (10th Edition)
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Figure 14.5-2
P Generation
F1 Generation
Appearance:Genetic makeup:
Gametes:
Appearance:Genetic makeup:
Gametes:
Purple flowers White flowers
Purple flowersPp
PP pp
P
P
p
p1/21/2
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Diagram adopted fromCampbell Biology (10th Edition)
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Figure 14.5-3
P Generation
F1 Generation
F2 Generation
Appearance:Genetic makeup:
Gametes:
Appearance:Genetic makeup:
Gametes:
Purple flowers White flowers
Purple flowers
Sperm from F1 (Pp) plant
Pp
PP pp
P
P
P
P
p
p
p
p
Eggs from F1 (Pp) plant
PP
ppPp
Pp
1/21/2
3 : 1
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Diagram adopted fromCampbell Biology (10th Edition)
Monohybrid crosses
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TEST YOUR KNOWLEDGE 1
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Define the following terms
•Homozygous An organism with two identical alleles for a character
•HeterozygousAn organism that has two different alleles for a gene
•PhenotypePhysical appearance
•Genotypegenetic makeup
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Phenotype
Purple
Purple
Purple
White
3
1
1
1
2
Ratio 3:1 Ratio 1:2:1
Genotype
PP
Pp
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(homozygous)
(heterozygous)
Pp(heterozygous)
pp(homozygous)
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• Crossing two true-breeding parents differing in two characters produces dihybrids in the F1 generation, heterozygous for both characters
• A dihybrid cross, a cross between F1 dihybrids, can determine whether two characters are transmitted to offspring as a package or independently
The Law of Independent Assortment
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Figure 14.8
P Generation
F1 Generation
Predictions
Gametes
EXPERIMENT
RESULTS
YYRR yyrr
yrYR
YyRr
Hypothesis ofdependent assortment
Hypothesis ofindependent assortment
Predictedoffspring ofF2 generation
Sperm
Spermor
EggsEggs
Phenotypic ratio 3:1
Phenotypic ratio 9:3:3:1
Phenotypic ratio approximately 9:3:3:1315 108 101 32
1/21/2
1/2
1/2
1/41/4
1/41/4
1/4
1/4
1/4
1/4
9/163/16
3/161/16
YR
YR
YR
YRyr
yr
yr
yr
1/43/4
Yr
Yr
yR
yR
YYRR YyRr
YyRr yyrr
YYRR YYRr YyRR YyRr
YYRr YYrr YyRr Yyrr
YyRR YyRr yyRR yyRr
YyRr Yyrr yyRr yyrr
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Diagram adopted fromCampbell Biology (10th Edition)
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• The law of independent assortment states that each pair of alleles segregates independently of each other pair of alleles during gamete formation
• Strictly speaking, this law applies only to genes on different, nonhomologous chromosomes or those far apart on the same chromosome*
• Genes located near each other on the same chromosome tend to be inherited together (linkage)*
law of independent assortment
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Diagram adopted from Human Genetics concepts and Application 9th ed.
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• Linkage refers to the transmission of genes on the same chromosome.
• Linked genes do not assort independently and do not produce Mendelian ratios.
• linkage has been used as a mapping tool in identifying disease-causing genes, and helped pave the way for genome-wide association studies.
Linkage
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Diagram adopted from Human Genetics concepts and Application 9th ed.
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TEST YOUR KNOWLEDGE 2
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1. Imagine that you encounter creatures in outer space whose traits are inherited according to Mendel's laws. In these creatures, purple eyes (E) are dominant to yellow eyes (e). Two purple-eyed creatures mate and produce six offspring (four purple-eyed and two yellow-eyed). What are the genotypes and phenotypes of the parents? What are the genotypes of the offspring?
• Two creatures with purple eye (EE or Ee) mate each other. If the parents were homozygous dominant (EE and EE), the offspring cannot be with yellow eye.
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E eE EE Eee Ee ee
• Genotype of Parent = Ee and Ee• Phenotype of Parents = Purple colored eye• Genotype of offspring = 1:2:1• Phenotype of offspring = 3:1
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2. In fruit flies, long wings (L) are dominant to short wings (l). Two long-winged flies produced 49 short-winged and 148 long-winged offspring. What were the probable genotypes of the parents? About how many of the long-winged offspring should be heterozygous?
• 49 are short wings• 148 are long wings
• 148:49 = 3:1
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• In order to obtain 3:1 phenotype the parents should be heterozygous (Ll and Ll).
• 2/3 of 148 will be heterozygous (from punnet table) = 99
L lL LL Lll Ll ll
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3. In humans, brown eyes (B) are dominant to blue eyes (b). A brown-eyed man marries a blue-eyed woman. They have eight children (all are brown-eyed). What are the possible genotypes of both parents and their offspring?
• For the entire child to be brown colored eye, the father must be homozygous dominant (BB).
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• Hence man (BB), female (bb) while the offspring all will be heterozygous (Bb) and brown eyed.
B Bb Bb Bbb Bb Bb
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4. A black, smooth guinea pig was mated with an albino, rough guinea pig. Their offspring were black, rough and black, smooth. These were the only offspring types produced over a period of several years after multiple mattings. Black color and rough fur are the dominant traits for guinea pigs. What was the probably genotype of each parent?
Parents•Black = B_•White = bb•Rough = R_•Smooth = rr
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In F1 generation
•For the offspring to be constantly black, rough and black, smooth for over the period of several years, Black must be homozygous dominant (BB).•Since the offspring are mixture of smooth and rough, the rough trait must be heterozygous (Rr)•Hence Black-smooth genotype is BBrr while the Albino-rough genotype is bbRr.•Gametes are Br from Black-smooth parent and bR and br for Albino-rough parent.•Offspring will produce BbRr (black, rough) and Bbrr (black, smooth).
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5. In moose, brown coat color (B) is dominant to albino (no pigment) (b) and rough coat (R) is dominant to smooth coat (r). Two moose are selected for breeding and their genotypes are BBRR (female) and bbrr (male). Determine the expected genotypic and phenotypic ratios for the:
a. F1 generation
b. F2 generation
c. Cross between an F1 moose and a moose with the genotype BBRr
• Gametes from BBRR is BR and Gametes from bbrr is br.• F1 generation: Genotype (BbRr) and phenotype (Brown-rough)• F2 generation: Genotype (Punnet table 1) and phenotype (9:3:3:1)• F1 BbRr vs BBRr (Punnet table 2) and phenotype (3:1)
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BR Br bR brBR BBRR BBRr BbRR BbRrBr BBRr BBrr BbRr BbrrbR BbRR BbRr bbRR bbRrbr BbRr Bbrr bbRr bbrr
Punnet table 1
BR Br bR brBR BBRR BBRr BbRR BbRrBr BBRr BBrr BbRr Bbrr
Punnet table 2
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5.2 Chromosomal linkage
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Diagram adopted from Human Genetics concepts and Application 9th ed.
• Observed the unexpected ratios indicating linkage in the early 1900s.
• They crossed true-breeding plants with purple flowers and long pollen grains (genotype PPLL ) to true-breeding plants with red flowers and round pollen grains (genotype ppll ).
• The dihybrid F1 self crossing did not yield the expected 9:3:3:1 phenotypic ratio.
William Bateson and R. C. Punnett’s experiments
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Diagram adopted from Human Genetics concepts and Application 9th ed.
• Those with the parental phenotypes P_L_ and ppll were more abundant than predicted.
• Other two progeny classes, ppL_ and P_ll were less common.
• Bateson and Punnett hypothesized genes are transmitted on the same chromosome and they do not separate during meiosis.
William Bateson and R. C. Punnett’s experiments
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Diagram adopted from Human Genetics concepts and Application 9th ed.
• Why the other two progeny classes, ppL_ and P_ll were still observed even though their number is less?
• If a small but distinct distance exist between gene loci, cross over occurs and can generate few recombinants.
• But many are still parental gametes.
William Bateson and R. C. Punnett’s experiments
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Diagram adopted from Human Genetics concepts and Application 9th ed.
• How to distinguish parental chromosomes from recombinant chromosomes?
• Only if the allele configuration of the two genes is known• Cis (a)• Trans (b).
Allele configuration
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TEST YOUR KNOWLEDGE 3
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Diagram adopted from Human Genetics concepts and Application 9th ed.
• Explain in your own word, what is linkage?
• How do you expect the offspring of F1 generation cross of a linked allele compared to Mendelian hypothesis?
• What is a Cis and Trans allele mean?
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5.3 Linkage Maps
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• Crossing over happens in a way that does not disrupt a particular gene (allele).
• In 1911, Alfred Sturtevant (Thomas Hunt Morgan’s student) from Columbia University made an important hypothesis.
• The farther apart two genes are on a chromosome, the more likely they are to cross over simply because more physical distance separates them.
The frequency of crossover
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Diagram adopted from Human Genetics concepts and Application 9th ed.
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• The correlation between crossover frequency and the distance between genes is used to construct linkage maps, which show the relative positions of genes in chromosomes.
• The distance is represented using “map units” called centimorgans, where 1 centimorgan equals 1 percent recombination.
• These units are used today to construct SNP maps; an estimate of genetic distance along a chromosome.
The frequency of crossover
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Diagram adopted from Human Genetics concepts and Application 9th ed.
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The frequency of crossover
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Diagram adopted from Human Genetics concepts and Application 9th ed.
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Diagram adopted from Human Genetics concepts and Application 9th ed.
• When two genes are widely separated on a chromosome, the likelihood of a crossover is so great
The frequency of crossover
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• The recombinant class may approach up to 50%
• Which may appear to be the result of independent assortment (the result is 1:1:1:1 of the four types).
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• Determining the degree of linkage by percent recombination.
• Lets consider the traits of Rh blood type and a form of anemia called elliptocytosis.
• An Rh + phenotype corresponds to genotypes RR or Rr.
• The anemia corresponds to genotypes EE or Ee.
Solving a Problem- Rh and anemia
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• In 100 one-child families, one parent is Rh negative with no anemia (rree), and the other parent is Rh positive with anemia (RrEe), and alleles are in cis.
• Of the 100 offspring, 96 have parental genotypes ( re/re or RE/re ) and four are recombinants for these two genes ( Re/re or rE/re ).
• Percent recombination is therefore 4 percent, and the two linked genes are 4 centimorgans apart.
Solving a Problem- Rh and anemia
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r r
e e
r
e
R
Evs
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• Nailpatella syndrome is a rare autosomal dominant trait that causes absent or underdeveloped fingernails and toenails, and painful arthritis in the knee and elbow joints.
• The gene is 10 map units from the/ gene that determines the ABO blood type on chromosome 9.
• Geneticists determined the map distance by pooling information from many families to predict genotypes and phenotypes in offspring,.
Solving a Problem- Nail patella syndrome
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Diagram adopted from Internet source.
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TEST YOUR KNOWLEDGE 4
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• Greg and Susan each have nail patella syndrome. Greg has type A blood. Susan has type B blood.
• What is the chance that their child inherits normal nails and knees and type O blood?
Nail patella syndrome
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Diagram adopted from Human Genetics concepts and Application 9th ed.
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• Human blood type is determined by co-dominant alleles.
• Each of us has two ABO blood type alleles, because we each inherit one blood type allele from our biological mother and one from our biological father.
The Human ABO markers: The A, B, and O alleles
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Diagram adopted from Internet source.
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• Greg’s mother has nail-patella syndrome and type A blood.
• His father has normal nails and type O blood.
• Therefore, Greg must have inherited the dominant nail-patella syndrome allele ( N ) and the IA allele from his mother, on the same chromosome.
Analysis for Greg
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n
i
N
IA
Mother Father
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• Susan’s mother has nail-patella syndrome and type O blood, and so Susan inherited N and i on the same chromosome.
• Because her father has normal nails and type B blood, her homolog from him bears alleles n and IB.
• Her alleles are in trans.
Analysis for Susan
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n
IB
N
i
Mother Father
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Nail patella syndrome gene mapping
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Diagram adopted from Human Genetics concepts and Application 9th ed.
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• Determining the probability that Susan and Greg’s child could have normal nails and knees and type O blood.
• The only way this genotype can arise from theirs is if an ni sperm fertilizes an ni oocyte.
• 45% x 5% = 2.25% nnii genotype.
• Calculating other genotypes for their offspring is more complicated For example, a child with nail-patella syndrome and type AB blood?
Greg and Susan offspring gene mapping
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• A man who has type O blood has a child with a woman who has type A blood.
• The woman’s mother has AB blood, and her father, type O.
• What is the probability that the child is of blood type:• O• A• B• AB
The Human ABO blood markers
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