Download 3D15 – BO0048 Code Questions Answers 1. Write the features of X

 Code
1.
Questions
Write the features of Xlinked
dominant
inheritance and sexlinked
recessive
inheritance.
2.
By taking an example
show that Mendelian
ratios are not always
the rule and are often
modified.
Answers
1. X-linked dominant inheritance: (5 x 1 = 5marks)
a. The trait is never passed from father to son
b. All daughters of an affected male and a normal female are affected.
c. Mating of affected females and normal males produce 50% the sons
affected and 50% the daughters affected
d. In the general population, females are more likely to be affected than
males, even if the disease is not lethal in males.
e. The trait does not skip generations.
2. Sex-linked recessive inheritance: (5 x 1 = 5marks)
a. The trait is never passed from father to son
b. Males are much more likely to be affected than females.
c. All affected males in a family are related through their mothers who are
known to be carriers because they have affected brothers, fathers or
maternal uncles.
d. Affected females come from affected fathers and affected or carrier
mothers.
e. Trait is typically passed from an affected grandfather through his carrier
daughters to half to his grandsons.
1. Experiments show that the ratios stated by Mendel do not occur in all the cases
of inheritance. Often the ratios are modified by various kinds of gene
interactions. Examples include complementary, supplementary and lethal
interactions. (2 marks)
2. In guinea pigs, there are two dominant genes for coat color – namely A and C.
The dominant genotype CCAA produce agouti color. A black color guinea pigs
possess gene C and not A. The gene A is for agouti and C is for black color. (2
marks)
3. If gene for black color is absent agouti is unable to express itself and albino
with genotype ccAA is produced. So the gene C produced black color and
gene A changed its expression to agouti color. This kind of genes are
supplementary genes. (2 marks)
4. Agouti
x
Albino (2 marks)
CCAA
x
ccaa
F1
3.
4.
In man, hemophilia is
sex-linked and
recessive. What
offspring phenotypic
ratio would be
expected from a
marriage between the
following cases?
i.
A hemophilic man
and a carrier
(heterozygous) female:
ii. A normal man and
a carrier woman
Briefly
explain
the
deletion
and
duplication
chromosomal
aberrations.
3D15 – BO0048
CAca (Agouti)
CAca
x CAca
F2
9:3:4
9 Acouti – CCAA, CcAA, CCAa, CcAa (2 marks)
3 Black – Ccaa, Ccaa
4 Albino – ccaa, ccAA, ccAa
1. Hemophilic man (XhY)
x
Carrier woman (XHXh)
F1: Carrier woman
man
XhXH
2.
2.
XhXh
Normal man (XHY) x
F1: Normal woman
XHXH
1.
Hemophilic woman
Normal man
XHY
Hemophilic
XhY
Carrier woman (XHXh)
Carrier woman
XHXh
(2.5 + 2.5 marks)
(2.5 + 2.5 marks)
Normal man Hemophilic man
XHY
XhY
Deletion: The deficiency is the deletion of a chromosomal of a segment
resulting in the loss of genes. Depending on the length of the lost segment, the
genes lost may vary from a single gene to a block containing many genes.
The break in the chromosome may be caused by several agents such as
chemicals, radiations and viruses. The break occurs at random either in both
the chromatids of a chromosome (chromosome break) or only in one chromatid
(chromatid break).
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3.
4.
5.
6.
Deletion
Terminal Deletion: It refers to the loss of segment from one or the other end of
the chromosome. The terminal part fails to survive and causes terminal
deletion. It is caused by a single break in the chromosome.
Intercalary deletion or interstitial deletion: It involves two breaks and an
intermediate segment is deleted followed by the reunion of the terminal
segments.
Duplication: The presence of same block of genes more than once in a haploid
complement is known as duplication and the additional segment is called a
repeat.
7.
Duplication
Tandem duplication: in this type, the added segment has the same genetic
sequence as is present in the original state in the chromosome
9. Reverse tandem duplication: in this type, the sequence of genes aligned in the
attached chromosome piece is just the reverse of the original segment.
10. Displaced duplication: in this type, the chromosomal segment gets attached to
some non-homologous chromosome.
1. Since each individual will have two homologous chromosomes, each carrying a
particular allele, the frequency of M can be calculated by doubling the number
of homozygous M blood group type and adding to it the frequency of
heterozygous MN blood group type.
2. In this manner the frequency of M = (50x2) + 20 = 120
3. The frequency of N = (30x2) + 20 = 80
4. The relative frequency of M = M / (M+N) = 120 / 200 = 0.6
5. The relative frequency of N = N / (M+N) = 80 / 200 = 0.4
8.
5.
Calculate the gene
frequencies of M and N
for a sample population
of 100 individuals with
50 MM, 20 MN and 30
NN blood groups.
6.
Briefly explain the life
cycle of Neurospora.
3D15 – BO0048
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3D15 – BO0048
(4marks)
The mycelium of the Neurospora contains haploid nuclei which multiply as the
hyphae grow and parts of which may give rise to new colonies. (2marks)
Another form of asexual reproduction is by spores or conidia, one class of
which contains a single nucleus, another class several nuclei. These germinate
and reproduce the whole mycelium. (2marks)
Sexual reproduction is by fertilization of nucleus of one mating type by a
nucleus from the conidia or mycelium of the opposite mating type, resulting in a
fusion nucleus which then undergoes two meiotic and one mitotic division to
form eight haploid ascospores. These spores occur two by two in the ascus
and may be dissected out and tested separately. (2marks)
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