Download - Genetics [e f]
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OverviewA) Monohybrid InheritanceB) Dihybrid InheritanceC) The test crossD) Autosomal dominant and autosomal recessive
pedigree charts
E) LinkageF) Sex determinationG) Alleles and their interactions – multiple alleles,
incomplete dominance, codominanceH) Gene interactions – polygenic inheritance,
epistasis
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Linkage
Linked genes are situated on the same chromosome
Why do red heads always have freckles?
Genes for red hair & freckles are linked
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All genes on a single chromosome :
form a linkage group (abc)
are inherited together as long as crossing-over does not occur
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Linked genes:
do not conform to Mendel’s law of independent assortment
fail to produce the expected 9:3:3:1 ratio in a breeding situation involving dihybrid inheritance
in these situations a variety of ratios are produced
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Morgan discovered linkage
in the early 1900 when he crossed Drosophila
1891-1945
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Linkage in Drosophila
genes for body colour and wing length are linked
Body colour: grey [G] black [g]
Wing length: long wings [L] vestigial (short) wings [l]
Vestigial wings
G gL l
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F2 Phenotype ratio expected if GGLL mate with ggll:
Parental genotypes: GGLL x ggllGametes: GL x glF1 genotypes: GgLl
If F1 are allowed to interbreed: GgLl x GgLlExpected F2 phenotypes: 9:3:3:1
However, the F2 showed 3:1
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How was a 3:1 ratio obtained?
3 grey body,long wing :
1 black body,vestigial wing
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Explanation of 3:1 ratio due to
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grey - Gblack - glong wings - Lvestigial wings -l
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In practice: this 3:1 ratio is never achieved reason: total linkage is rare[since crossing-over can occur]
In reality: 4 phenotypes are produced
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What phenotypes in the offspring did Morgan expect when crossing the following?
GgLl x ggll
G – grey bodyg – black bodyL – long wing l – vestigial wing
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Morgan expected a 1:1:1:1 ratio
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But he did not get what was expected
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GgLl x ggll
Genotype GgLl Ggll ggLl ggll
Phenotype grey, long wing
grey, vestigial
wing
black, long wing
black, vestigial
wing
Actual nos. 965 185 206 944
Expected 25% 25% 25% 25%
Observed 41.5% 8.5% 8.5% 41.5%
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GgLl x ggll
Genotype GgLl Ggll ggLl ggll
Phenotype grey, long wing
grey, vestigial
wing
black, long wing
black, vestigial
wing
Observed 41.5% 8.5% 8.5% 41.5%
Like parents
Unlike parents
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How did the phenotypes unlike the parents result?
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Let us explain how crossing-over may happen during gamete
formation in:
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Crossing over produces Recombinant Types
G g
L l
Replication
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Crossing over produces Recombinant Types
G g
L l
Replication
G g
L l
G g
L l
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Crossing over produces Recombinant Types
G g
L l
Replication Crossing
over
G g
L l
G g
L l
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Crossing over produces Recombinant Types
G g
L l
Replication Crossing
over
G g
L l
G g
L l
G
L
g
l
G g
Ll
Recombinants
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Grey body,long wing
Grey body,vestigial
wing
Black body,vestigial wing
Black body,long wing
G – grey bodyg – black bodyL – long wing l – vestigial wing Recombinants
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Grey body,long wing
Grey body,vestigial
wing
Black body,vestigial wing
Black body,long wing
Recombinants
41.5% grey body, long wing41.5% black body, vestigial wing
8.5% grey body, vestigial wing8.5% black body, long wing
Parental phenotypes
Recombinants
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Most breeding experiments involving linkage produce:
1.approximately equal numbers of the parental phenotypes
GreyLong wing
Blackvestigial
BlackLong wing
Greyvestigial
965 944 206 185
2. a significantly smaller number of phenotypes showing new combinations of characteristics also in equal numbers, called recombinants
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CROSSING-OVER AND CROSSOVER VALUES (COV)
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Recombinant frequency
Recombinant frequency = Number of individuals showing recombination X 100
Number of offspring
parental phenotypes: grey body, long wing 965black body, vestigial wing 944
recombinant phenotypes:black body, long wing 206grey body, vestigial wing 185
206 185Recombinant frequency 100 17%
965 944 206 185
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Mutant phenotypes
Short aristae
Black body
Cinnabareyes
Vestigialwings
Brown eyes
Long aristae(appendageson head)
Gray body
Redeyes
Normalwings
Redeyes
Wild-type phenotypes
IIY
I
X IVIII
0 48.5 57.5 67.0 104.5
What did the recombinant frequency or crossover value (COV) demonstrate?
Genes are arranged linearly along the
chromosome.
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Question: Getting Started
In Drosophila one gene controls eye colour and another controls body colour. The allele E for red eyes is dominant to e, the allele for purple eyes. The allele B for grey body is dominant to b, the allele for black body.A female Drosophila, heterozygous for both the eye-colour gene and the body-colour gene, was crossed with a male, homozygous recessive for both of these genes. The phenotypes of the offspring are given in the table.
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E - red eyes e - purple eyesB - grey bodyb - black body
heterozygous for both the eye-colour gene and the body-colour gene
homozygous recessive for both of these genes
Phenotype Number of offspring
Red eyes, grey bodyPurple eyes, black bodyRed eyes, black bodyPurple eyes, grey body
82791316
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a) i) For this cross, write your answer on the lines to give the following information: (3)
Phenotype of male parent x Phenotype of female parentpurple eyes, black body red
eyes, grey body
Genotype of male parent x Genotype of female parenteebb EeBb
Genotype(s) of male x Genotype(s) of female gametes gametes
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ii) From the information in (a) (i), explain why there were approximately equal numbers of red-eyed, grey-bodied and purple-eyed, black-bodied offspring. (2)Due to linkage of genes parental phenotypes occur at approximately equal numbers.Phenotype Number of offspring
Red eyes, grey bodyPurple eyes, black bodyRed eyes, black bodyPurple eyes, grey body
82791316
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b) i) Explain why there were relatively few red-eyed, black-bodied and purple-eyed, grey-bodied offspring. (2)
These are the recombinants which resulted from crossing over. Crossing over occurs at a low rate.
Phenotype Number of offspring
Red eyes, grey bodyPurple eyes, black bodyRed eyes, black bodyPurple eyes, grey body
82791316
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ii) Explain why the number of red-eyed, black-bodied offspring is not exactly equal to the number of purple-eyed, grey-bodied offspring. (1)
Due to random fertilisation of gametes.Phenotype Number of
offspringRed eyes, grey bodyPurple eyes, black bodyRed eyes, black bodyPurple eyes, grey body
82791316
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Extent of cross-over is proportional to distance between genes
Most
Least Intermediate
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GENE MAPPING
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Chromosome maps are constructed by directly converting
the :COV between genes into hypothetical distances along the chromosome
C? C?A B
499
A COV of 4% :between genes
A and B: they are 4 units apart
A COV of 9% :for a pair of genes A and C: they are 9 units apart but does not indicate the linear sequence of genes
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consider the following values involving four genes, P, Q, R and S:
P – Q = 24%R – P = 14%R – S = 8%S – P = 6%
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consider the following values involving four genes, P, Q, R and S:
P – Q = 24%R – P = 14%R – S = 8%S – P = 6%
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consider the following values involving four genes, P, Q, R and S:
P – Q = 24%R – P = 14%R – S = 8%S – P = 6%
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consider the following values involving four genes, P, Q, R and S:
P – Q = 24%R – P = 14%R – S = 8%S – P = 6%
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A linkage map (or chromosome map) tells the relative distances between
gene loci on a chromosome
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Questions: Crossover & Epistasis
In maize the genes for coloured seed and full seed are dominant to the gene for colourless seed and shrunken seed. Pure breeding strains of the double dominant variety were crossed with the double recessive variety and a backcross of the F1 generation produced the following results:
coloured, full seed 380coloured, shrunken seed 14colourless, shrunken seed 386colourless, full seed 10
Calculate the distance in units between the genes for coloured seed and seed shape on the chromosomes. (3)
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coloured, full seed 380coloured, shrunken seed 14colourless, shrunken seed 386colourless, full seed 10
Total number of seeds:
380 + 14 + 386 + 10 = 790Number of recombinants:
14 + 10 = 24Recombinant frequency (crossover value): 24 100
3%790
Therefore, distance = 3 units
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OverviewA) Monohybrid InheritanceB) Dihybrid InheritanceC) The test crossD) Autosomal dominant and autosomal recessive
pedigree chartsE) Linkage
F) Sex determinationG) Alleles and their interactions – multiple alleles,
incomplete dominance, codominanceH) Gene interactions – polygenic inheritance,
epistasis
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Sex Chromosomes (heterosomes)
Females XXHomogametic(one type of gamete, X)
X
Y
X X
Males XYHeterogametic (two types of gametes, X or Y)
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Sex Chromosomes in
Drosophila
as in humans
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Sex chromosomes in some organisms differ from those in humans
Birds (including poultry), moths & butterflies :
Female
Female
Male
Male
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Sex chromosomes in some organisms differ from those in humans
In some insects e.g. grasshopper, the Y chromosome is absent :
XO = male XX = female
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Curious Fact:sex is influenced by temperature in
environment in reptiles (turtle, crocodile)
(t <28 °C) (t 28-32 °C) (t >32 °C)
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Question
Scientists analysed the:DNA on the Y chromosome and the DNA in the mitochondria of the Swedish
wolves.
Swedish wolves
Finland RussiaXThey concluded that the Swedish wolf population descended from one male wolf from Finland and one female wolf from Russia.
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Question
a) Explain why DNA on the Y chromosome helped them to reach this conclusion. (1)Y chromosome is inherited from male.
b) Suggest why DNA in the mitochondria helped them to reach this conclusion. (1)Mitochondria in egg. No mitochondria come from male gamete.
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Question: [MAY, 2012]Use your knowledge of biology to describe the significance of the following.All the mitochondria in a human cell are derived from the female parent. (5)
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Barr Bodies
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Barr Bodies
This held for cats and humans
They thought the spot was a tightly condensed X chromosome
Ewart George Bertram & Murray Llewellyn Barr in 1948
1940’s two Canadian scientists noticed a dark staining mass in the nuclei of cat brain cells
Found these dark staining spots in female somatic cells but not males
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The X Chromosomes
One X chromosome: always appears in the
active state
has the normal appearance
If another is present: it is seen in a resting state as a tightly coiled
dark-staining body - the
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Where is the Barr body located?
The Barr body is generally located on the periphery of the nucleus.
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How can the genotypic sex of a person be determined?
By counting the number of in non-dividing cells
Barr bodies = 0 Barr bodies = 1
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How many Barr Bodies?
Barr bodies = 0 Barr bodies = 1 Barr bodies = 3
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Tortoiseshell cats
Calico cats
or
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Calico cats result from random inactivation of an X chromosome
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Zygote
Early embryo
Random inactivation of X chromosome occurs early in
embryonic development
Black inactive
Orange inactive
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Male calico cats are rare
What are the sex chromosomes of a
calico male cat?
XXY
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Calico manx cat
What is unusual?
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X inactivation explains why:
Levels of enzymes or proteins encoded by genes on the X chromosome are the same in both males and females
Even though males have 1 X chromosome and females have 2.
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SEX LINKAGE
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What is the function of the Y
chromosome?function appears to vary according to
species
In humans: it controls the
differentiation of the testes
In most organisms: it does not carry genes
concerned with sex
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Why is the Y chromosome described as genetically inert or genetically empty?
since it carries so few genes
the human Y chromosome contains: only 78 genes the lowest number of known genetic
diseases (44 in total)
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Fig. 6Homologous and non-homologous regions of the sex chromosomes.
Homologous portion of the sex chromosomes
Non-homologous portion of the X chromosome
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Sex-linkage
affects males and females differently
characteristics determined by genes carried on the non-homologous portion of the X chromosome appear in males even if they are recessive
X Y
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Sex-linked Traitsred-green colour
blindnesspremature balding
haemophilia
“3” or “8”??
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Question: [MAY, 2006]Suggest explanations for the following observation.
Colour blindness affects 8% of human males but only 0.7% of females. (2)
Normal vision Colour blind vision
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Colour-blindness is a sex-linked trait which affects more males than females. The allele responsible for this feature is carried on the X chromosome. As males carry only one such chromosome, chances for the disorder to manifest itself is more than in females, where two X chromosomes are present. A heterozygous female is phenotypically normal, as the recessive allele is not expressed in the presence of the dominant one.
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Haemophilia: possible genotypes and phenotypes
Genotype PhenotypeXHXH Normal female
XHXh Normal female (carrier)
XHY Normal male
XhY Haemophiliac male
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Parents
XHY XHXhX
What is the probability for these parents to have a sick child?
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ParentsParental
Gametes
Y
XHY XHXh
XH XH Xh
X
X
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ParentsParental
Gametes
F1 Offspring:
Y
XHY XHXh
XH
Xh
Y
XH
XH XH Xh
X
X
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ParentsParental
Gametes
F1 Offspring:
Y
XHXH XHY
XHXh XhY
XHY XHXh
XH
Xh
Y
XH
XH XH Xh
X
25%
X
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Using this pedigree, how can you tell that:1. colour blindness is a sex-linked disorder? More males than females are affected.2. colour blindness is due to a recessive allele? Affected parents produce normal offspring.
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Eye colour in Drosophila:
Red eye colour:Dominant
is sex-linked
White eye colour:recessive
XR Xr
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What is a ‘reciprocal cross’?
A cross, with the phenotype of each sex reversed as compared with the original cross.
Red-eyed White-eyed X X Red-eyedWhite-eyed
XRY XrXr XRXRXrY
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Why is a ‘reciprocal cross’ done?
To test the role of parental sex on inheritance pattern.
Red-eyed
male
White-eyed female X X Red-eyed
female
White-eyed male
F1 all red-eyedF1 all males white-eyed: all females red-eyed:
To decide whether trait is sex linked compare offspring phenotypes
XRY XrXr XrY XRXR
XrYXRXr XRXr XRY
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Autosomes
If F1 phenotypes are the same from both crosses:
If F1 phenotypes are different both
crosses:
trait is not sex-linkedtrait is sex-linked
N n
Sex chromosomes
N: Normal n: sick
Autosomes
N n
X X
If allele is carried on the
autosomes, males and
females would be normal.
If allele is carried on the sex
chromosomes, males and
females would be different.
Autosomes
n
X Y
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What is the phenotypic ratio in the F2 generation if a white-eyed male fly is mated with a red-eyed female? Eye
colour is sex-linked.
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Question: [MAY, 2006]
Suggest explanations for the following observation.
In the fruit fly, Drosophila, red eyes are dominant to white eyes. When a red-eyed female is crossed with a white-eyed male, the offspring all have red eyes but when the opposite cross was done (a white-eye male with a red-eyed female) [this should read: a white eyed female and a red eyed male – a mistake in exam paper] all the male offspring had white eyes. (2)
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Question: [MAY, 2006]
This is a reciprocal cross carried out to determine if a gene is sex-linked. This result shows that eye colour in Drosophila is sex-linked. A red-eyed female (XRXR) crossed with a white-eyed male (XrY) produce 100% red-eyed offspring (XRXr, XRY). A white-eyed female (XrXr) crossed with a red-eyed male (XRY) produce white-eyed male offspring (XrY).
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3. Duchenne muscular dystrophy is a sex-linked inherited condition which causes degeneration of muscle tissue. It is caused by a recessive allele. The diagram shows the inheritance of muscular dystrophy in one family.
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a) Give evidence from the diagram which suggests that muscular dystrophy is
i) sex-linked; (1)Only seen in males / not in females;
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ii) caused by a recessive allele. (1)Unaffected parents/mother produce a sick child
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b) Using the following symbols,XD = an X chromosome carrying the normal alleleXd = an X chromosome carrying the allele for muscular dystrophyY = a Y chromosomegive all the possible genotypes of each of the following persons. [5, 6, 7, 8] (2)
XdY XDY
XDXd
XDXd or XDXD
Xd XD
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XDXd
Xd
XD
c) A blood test shows that person 14 is a carrier of muscular dystrophy. Person 15 has recently married person 14 but as yet they have had no children. What is the probability that their first child will be a male who develops muscular dystrophy? (1)
XDY
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Person 14: XDXd and Person 15: XDYMale offspring: XDY and XdYThus half of the male offspring develop muscular dystrophyProbability of having male offspring = 0.5Thus probability that their first child will be a male who develops muscular dystrophy: 0.5 x 0.5 = 0.25 / ¼ / 25% / 1 in 4; (NOT ‘1:4’)
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5. Fabry’s disease is a sex-linked recessive genetic disorder that causes mental retardation. A study was carried out into the inheritance of this disorder in a family, and the results are shown in the pedigree below.
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a) Using the symbol A for the dominant allele and a for the recessive allele, state the genotype of person 2
XAXa (1) XA
Xa
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b) Using the evidence from the pedigree, explain why Fabry’s disease is described as a sex-linked recessive genetic disorder. (3)
Only males have the disease. Probably carried on the X chromosome. Male sufferers are produced from unaffected parents. Recessive is only expressed when no dominant allele is present. Females must be carriers. Males have only one locus for the disorder.
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c) Explain why person 3 is unaffected but why one of his children (person 5) has Fabry’s disease.
Person Number 3 is the father XAY.Number 5 inherits Y from the father and the recessive allele from the mother who is a carrier. (3)
XA
Xa
XaY
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XA
Xa
d) What are the chances of the children of persons 6 and 7 having Fabry’s disease? Give reasons for your answer. (4)
XAY XAXa
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Y X a
F2 : No. 6 x No. 7
F2 genotypes: X AYGametes:
F3 genotypes:
F3 phenotypes:
x
x XAX a
X A X A
X AX A
females are all unaffected; half of the males are affected
X AX a X AY X aY
25 % chance of producing a sufferer
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3. The phenotypic expression of a dominant gene in Ayrshire cattle is a notch in the tips of the ears. In the pedigree below, square symbols represent male animals, circles represent female animals and solid symbols represent notched individuals.
I
II
III
1 2
1 2 3 4 5
1 2 3 4 5 6
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3.1 What is pedigree analysis? (2)The study of phenotypes in closely related families. This is done in order to discover the genotype of each individual.
I
II
III
1 2
1 2 3 4 5
1 2 3 4 5 6
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Use the pedigree diagram to determine the probability of notched progeny being produced from each of the following matings. Your working and reasoning must be shown.
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3.2 III 1 x III 3 (2)
I
II
III
1 2
1 2 3 4 5
1 2 3 4 5 6
N
n
nn nn
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3.3 III 2 x III 3 (2)
I
II
III
1 2
1 2 3 4 5
1 2 3 4 5 6
N
n
nn Nn
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3.4 III 3 x III 4 (2)
I
II
III
1 2
1 2 3 4 5
1 2 3 4 5 6
N
n
nn nn
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3.5 III 1 x III 5 (2)
I
II
III
1 2
1 2 3 4 5
1 2 3 4 5 6
N
n
nn Nn
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3.6 III 2 x III 5 (2)
I
II
III
1 2
1 2 3 4 5
1 2 3 4 5 6
N
n
Nn Nn
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HW: GENETICS MATSECNumbers 5, 8,9
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Questions: Getting Started
4.In the fruit fly, Drosophila melanogaster the allele r for white eyes is sex-linked and is recessive to the allele R for the normal red eye colour. Sex-linked genes are carried on the X chromosome. In fruit flies, the sex chromosomes are XX in females and XY in males.
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When a white-eyed female fly is crossed with a red-eyed male, all of the male offspring have white eyes.
i) In the space below, draw diagrams to represent the chromosomes, and alleles present on them, of the parents in this cross. (3)
Female x MaleXr Xr XR Y
XR – red eyeXr – white eye
ii) State the genotype and phenotype of the female offspring. (2)
Genotype: Phenotype:
XRXr
red eyed
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A gene which affects the size of the wings in the fruit flies is not sex-linked (autosomal). The allele for short (dumpy) wings, d, is recessive to the allele, D, for normal wings.A female fly was heterozygous for the genes for eye colour and for wing size. This female was crossed with a white-eyed, short-winged male.
i) Use a genetic diagram to show the expected results of this cross. (4)
D – normal wingd – short wing Parental genotypes: XRXrDd x XrYdd
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XRD XRd XrD XrdXrd XRXrDd XRXrdd XrXrDd Xr XrddYd XRYDd XRYdd XrYDd XrYdd
D – normal wingd – short wing Parental genotypes: XRXrDd x XrYdd
XRD XRd XrD Xrd Gametes: Xrd Yd x
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ii) Using the columns below, list the genotypes and full phenotypes of the male offspring produced. (2)
Genotypes of male offspring Full phenotypes
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ii) Using the columns below, list the genotypes and full phenotypes of the male offspring produced. (2)
Genotypes of male offspring Full phenotypes
XRYDd red eyed, normal wings
XRYdd red eyed, short wings
XrYDd white eyed, normal wings
XrYdd white eyed, short wings
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Calico Cats are heterozygous for X-linked alleles determining coat Colour
Orange: XB XB
Black: XbXb
Calico:XBXb
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Calico cats have orange &
black patches
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What is the probability for offspring to be tortoiseshell (calico) for coat color?
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50% Tortoiseshell (calico) cats
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Questions: SEX LINKAGE (Calico cats)
1a) Explain what is meant by the term sex linkage. (2)Alleles are located on the sex chromosomes. Most probably, the X chromosome.
b) One of the genes which controls coat colour in cats has
its locus on the X chromosome. This sex-linked gene has two alleles, O and B. Female cats have two chromosomes (XX), male cats have one X and one Y chromosome (XY).In female cats, possible genotypes and phenotypes are shown in the table below.
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State two possible genotypes for a male cat and, in each case, indicate the phenotype. (4)
Genotype PhenotypeXOXO Orange
XOXB Orange and black (tortoiseshell)
XBXB Black
Genotype Phenotype
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State two possible genotypes for a male cat and, in each case, indicate the phenotype. (4)
Genotype PhenotypeXOXO Orange
XOXB Orange and black (tortoiseshell)
XBXB Black
Genotype Phenotype
XOY Orange
XBY Black
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c) An orange and black (tortoiseshell) female cat was mated with a black male cat. Draw a genetic diagram to show the possible genotypes and phenotypes of the kittens resulting from this cross. (5)
Parental phenotypes:
XOXB x XBY
Gametes:
tortoiseshell female
xblack male
F1 genotypes: XOXB XOY XBXB XBY
F1 phenotypes: tortoiseshell female
Black male
Orangemale
BlackFemale
XO XB x XB Y
Parental genotypes:
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Calico female with black offspring
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2) A sex-linked gene for fur colour on the X chromosome of cats has two alleles. One allele results in orange fur (XO) and the other in black fur (X+).The heterozygous condition (XOX+) results in ‘tortoiseshell’ cats with irregular patches of orange and black fur.
a) Complete the two genetic crosses shown below.black ♀ x orange ♂ Orange ♀ x black ♂
i) Genotypes of parents
ii) Genotypes of kittens ♀ kittens ♂ kittens ♀ kittens ♂ kittens
iii)Phenotypes of kittens
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2) A sex-linked gene for fur colour on the X chromosome of cats has two alleles. One allele results in orange fur (XO) and the other in black fur (X+).The heterozygous condition (XOX+) results in ‘tortoiseshell’ cats with irregular patches of orange and black fur.
a) Complete the two genetic crosses shown below.black ♀ x orange ♂ Orange ♀ x black ♂
i) Genotypes of parents
X+ X+ x XOY XO XO x X+Y
ii) Genotypes of kittens
XOX+ X+Y♀ kittens ♂ kittens
XOX+ XOY♀ kittens ♂ kittens
iii)Phenotypes of kittens
tortoiseshell blackfemale male
tortoiseshell orangefemale male
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b) Explain why tortoiseshell cats are always female (unless there has been a mutation).Tortoiseshell cats must have two X chromosomes, one being XO, and the other X+. Males have only one X chromosome.
c) Male tortoiseshell cats are very rare and are always sterile with three sex chromosomes. Give these three sex chromosomes with their alleles where appropriate.XOX+Y
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OOH OOH, ME ME ME!!
Any Questions?