6.2 Mendel and Inheritance Further Questions and Answers

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 tongue-roller.

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:

ashort  short

bshort  heterozygous normal

cheterozygous normal  heterozygous normal.

A5.

Bk Ch6 S6.2 FA5

N: normal wing (dominant); n: short wing (recessive).

ashort wing  short wing: nn  nn

Genotype of each parent is nn; only n gametes are produced.

All offspring have genotype nn and phenotype short wing.

bshort wing  heterozygous normal: 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.

cheterozygous normal  heterozygous normal: 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.

aUse appropriate notation to assign genotypes to the parents in generation I. How can you be certain of their genotypes?

bWhat is the probability of individual

iIII-1, and

iiiV-3

developing the disease? Explain your answer in each case.

cSuggest 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).

aI-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.

biIndividual 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.

iiThe 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.

cIndividuals 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 spiral-tailed 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.

aDraw a family pedigree representing this information. Assign genotypes to each family member in the pedigree, using appropriate notation.

bWhat is the probability of any other children being albino?

A8.

Bk Ch6 S6.2 FA8

aFamily pedigree:

bEach 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).

6.2 Mendel FQApage 1 of 4

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