Download 6.2 Mendel and inheritance – Further questions and answers Q1. Bk

6.2 Mendel and inheritance – Further questions and answers
Q1.
Bk Ch6 S6.2 FQ1
Explain why fruit-flies, mice, rabbits, corn and wheat are far better subjects for the study of inherited
characteristics than humans are.
A1.
Bk Ch6 S6.2 FA1
Fruit-flies, mice, rabbits, corn and wheat make better subjects for the study of inherited characteristics
for two important reasons. Firstly, they have a much shorter life span than humans do so it is easy to
investigate the behaviour of inherited characteristics over several generations. They also have large
numbers of offspring at a time and this makes the analysis of offspring ratios more convenient. The
greater the number of offspring or individuals in a sample being studied, the more reliable the results
are likely to be.
Q2.
Bk Ch6 S6.2 FQ2
Describe five examples of variations in the phenotype of humans, which are the result of environmental
influence.
A2.
Bk Ch6 S6.2 FA2
Human characteristics that are subject to environmental influence include length of hair, colour of skin
due to exposure to sun, colour of hair if dyed, gap between front teeth (can be reduced by the wearing
of braces) and pierced ears.
Q3.
Bk Ch6 S6.2 FQ3
Explain why two children of the same parents are not usually identical.
A3.
Bk Ch6 S6.2 FA3
Mendel’s Law of Independent Assortment during gamete formation accounts for the differences
between children of the same parents. The parents’ chromosomes bearing different alleles for the same
genes assort independently during gamete formation with a variety of possibilities occurring in the
different gametes produced.
Q4.
Bk Ch6 S6.2 FQ4
The ability to roll the tongue is a dominant characteristic. Is it possible for two non-rollers to have a
child who can roll the tongue? Explain your answer, using appropriate notation and giving
probabilities.
A4.
Bk Ch6 S6.2 FA4
Individuals who cannot roll their tongue must be homozygous recessive for the characteristic. Since the
ability to roll the tongue is dominant and neither parent has the dominant allele, they will not be able to
produce, a child who can tongue-roll. Using the notation T = can roll the tongue (dominant) and t =
cannot roll the tongue (recessive), both parents in this case have the genotype tt and can only produce
gametes containing t. All offspring will have the genotype tt and zero probability of being a tongueroller.
6.2 Mendel FQA
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Q5.
Bk Ch6 S6.2 FQ5
In Drosophila, normal wing is dominant to short wing. Let N = normal wing and n = short wing. Use a
Punnett square to show the expected ratio of genotypes and phenotypes in each of the following
crosses:
a
short  short
b
short  heterozygous normal
c
heterozygous normal  heterozygous normal.
A5.
Bk Ch6 S6.2 FA5
N: normal wing (dominant); n: short wing (recessive).
a
short wing  short wing: nn  nn
Genotype of each parent is nn; only n gametes are produced.
n
n
b
n
nn
nn
All offspring have genotype nn and phenotype short wing.
short wing  heterozygous normal: nn  Nn
n
n
c
n
nn
nn
N
Nn
Nn
n
nn
nn
50% of offspring are expected to have genotype Nn and phenotype normal wing, and 50% of
offspring are expected to have genotype nn and phenotype short wing. That is, ratio of normal
wing to short wing is 1 : 1.
heterozygous normal  heterozygous normal: Nn  Nn
N
n
N
NN
Nn
n
Nn
nn
25% of offspring are expected to have genotype NN and 50% with genotype Nn. Both of these
genotypes will have phenotype normal wing. 25% of offspring expected to have genotype nn and
phenotype short wing. That is, ratio of normal to short wing is 3 : 1.
Q6.
Bk Ch6 S6.2 FQ6
The following pedigree illustrates the inheritance of Huntington’s disease in a family. Huntington’s
disease is a degenerative disease of the nervous system resulting in loss of coordination, mental
breakdown and eventually death. It is a dominant characteristic and generally does not develop until at
least 35 years of age. In this pedigree all individuals in generation II are over 50 years of age.
6.2 Mendel FQA
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a
b
c
Use appropriate notation to assign genotypes to the parents in generation I. How can you be
certain of their genotypes?
What is the probability of individual
i III-1, and
ii IV-3
developing the disease? Explain your answer in each case.
Suggest why no-one in generations III and IV shows the disease.
A6.
Bk Ch6 S6.2 FA6
H: Huntington’s disease (dominant); h: no Huntington’s disease (recessive).
a
I-1: Hh, I-2: hh.
Individual I-1 has Huntington’s disease so must have a dominant allele, but also has a child (II-2)
who doesn’t develop the disease so I-1 must also be carrying a recessive allele to pass on.
Individual I-2 doesn’t have Huntington’s disease so must have only the recessive genotype.
b
i Individual III-1 has zero chance of developing the disease. We can be sure of this because
the mother of this person, II-2, is over 50 years old and if she was carrying the dominant
allele the disease would already have begun to develop. This is not the case. Individual II-2
is unaffected and so we can presume that she doesn’t have the affected allele to pass on.
ii The development of Huntington’s disease in IV-3 depends upon the alleles inherited by her
father from his mother. Individual III-4 has a 1 in 2 or 50% probability of developing the
disease because of the chance of obtaining the affected allele from his mother. Therefore
probability of Huntington’s disease in IV-3 is one-half or 0: one-half + 0 = one-half.
c
Individuals in generations III and IV are too young to develop the disease. However, individuals
III-3 and III-4 have a 1 in 2 or 50% probability of having inherited the dominant allele and
therefore a 1 in 2 chance of developing the disease.
Q7.
Bk Ch6 S6.2 FQ7
The curly wolf (Lupus spirale) is famous for its long, spiral tail. However, straight-tailed curly wolves
are occasionally born. The animal breeder at the Twirly Hill Zoo needs to know if a particular spiraltailed curly wolf is carrying the gene for a straight tail or not; in other words, is it homozygous (AA) or
heterozygous (Aa)? The usual test cross to investigate this is to mate the wolf with a straight-tailed
curly wolf (aa). Explain how the results of such a cross indicate the genotype of the curly wolf in
question.
A7.
Bk Ch6 S6.2 FA7
A: spiral tail (dominant); a: straight tail (recessive).
In the test cross the spiral-tailed wolf in question is mated with a straight-tailed wolf. If there are any
straight-tailed offspring in the litter then the genotype of the spiral-tailed wolf must be Aa. If the litter
contains only cubs with spiral tails, this would indicate a likelihood of the spiral-tailed parent being
homozygous-dominant. (The larger the litter the more reliable the results, because in this case the
parent is presumed to have genotype Aa and when crossed with aa is expected to give rise to a 1 : 1
ratio of cubs with the different tail types.)
Q8.
Bk Ch6 S6.2 FQ8
Lack of pigment (albinism) is recessive to normal pigmentation in all species. An albino man marries a
normal woman and they have one albino child.
a
Draw a family pedigree representing this information. Assign genotypes to each family member
in the pedigree, using appropriate notation.
b
What is the probability of any other children being albino?
6.2 Mendel FQA
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A8.
Bk Ch6 S6.2 FA8
a
Family pedigree:
b
Each child is an independent event so the probability of another child being albino is 1 in 2. We
can show this using a Punnett square.
N: normal pigmentation (dominant); n: albino (recessive).
n
n
N
Nn
Nn
n
nn
nn
6.2 Mendel FQA
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