Hereditary disorders
On the site
http://pharmaxchange.info/notes/clinical/genetic_disorders.pdf
you can see examples of disorders caused by failing genes. Read through the leaflet (starting on page 8), noting the disorders. Which ones do you recognise?
http://pharmaxchange.info/notes/clinical/genetic_disorders.pdf
you can see examples of disorders caused by failing genes. Read through the leaflet (starting on page 8), noting the disorders. Which ones do you recognise?
Exam questions
The following questions were taken from previous exams. Try to answer all of them: You'll be able to find the answers underneath the questions!
Paper 1
Humans are in blood group M, N or MN. The alleles for blood group M (M) and blood group N (N) are co-dominant. Humans are also in blood group A, B, AB or O. The alleles controlling these blood groups are IA, IB and i.If two parents have the genotypes ii MM and IA i MN what is the ratio of possible phenotypes of their offspring?
D
What distinguishes an allele from a gene?
A. An allele is made of RNA.
B. An allele is shorter.
C. An allele is a variety of a gene.
D. An allele cannot be transferred during genetic modification.
A. An allele is made of RNA.
B. An allele is shorter.
C. An allele is a variety of a gene.
D. An allele cannot be transferred during genetic modification.
C
Some breeds of dogs are characterized by the presence of a melanistic mask, which is a darkening of the fur near the nose, as shown by the arrow in this photograph.
Which outcome is matched with a valid conclusion if dogs that were pure breeding for melanistic masks were crossed with dogs without melanistic masks?
A. If 0 % of the puppies have a mask, the character is recessive.
B. If 25 % of the puppies have a mask, the character is dominant.
C. If 75 % of the puppies have a mask, the character is dominant.
D. If 100 % of the puppies have a mask, the character is recessive.
Which outcome is matched with a valid conclusion if dogs that were pure breeding for melanistic masks were crossed with dogs without melanistic masks?
A. If 0 % of the puppies have a mask, the character is recessive.
B. If 25 % of the puppies have a mask, the character is dominant.
C. If 75 % of the puppies have a mask, the character is dominant.
D. If 100 % of the puppies have a mask, the character is recessive.
A
What is characteristic of homologous chromosomes?
A. They have an identical DNA sequence.
B. They are of the same length in karyograms.
C. They form pairs in prokaryotes.
D. They carry the same alleles.
A. They have an identical DNA sequence.
B. They are of the same length in karyograms.
C. They form pairs in prokaryotes.
D. They carry the same alleles.
B
A colour blind man and a woman carrier for colour blindness have a son. What is the probability that their son will be colour blind?
A. 25 %
B. 50 %
C. 75 %
D. 100 %
A. 25 %
B. 50 %
C. 75 %
D. 100 %
B
The genetic determination of dogs’ coats can be quite complex, with many different genes acting at the same time.
• The dominant allele E gives brown tones. The recessive allele e results in red tones.
• The colour intensity is due to another gene. The dominant allele B gives a dark colour, whereas the recessive allele b results in a light colour.
What would be the genotype of a light brown dog produced from a cross between a dark brown dog and a light red dog?
A. EEbb
B. EeBb
C. eeBb
D. Eebb
• The dominant allele E gives brown tones. The recessive allele e results in red tones.
• The colour intensity is due to another gene. The dominant allele B gives a dark colour, whereas the recessive allele b results in a light colour.
What would be the genotype of a light brown dog produced from a cross between a dark brown dog and a light red dog?
A. EEbb
B. EeBb
C. eeBb
D. Eebb
D
In a fruit fly experiment, grey body, normal winged (homozygous dominant) fruit flies were mated with black body, short winged (homozygous recessive) fruit flies. The F1 dihybrid females were then used in a test cross. If the genes are always linked and no crossing over occurs, what would be the predicted ratio in the F2 generation?
A. 9 : 3 : 3 : 1
B. 1 : 1 : 1 : 1
C. 3 : 1
D. 1 : 1
A. 9 : 3 : 3 : 1
B. 1 : 1 : 1 : 1
C. 3 : 1
D. 1 : 1
D
Paper 2A
(a) A farmer has rabbits with two particular traits, each controlled by a separate gene. Coat colour brown is completely dominant to white. Tailed is completely dominant to tail-less. A brown, tailed male rabbit that is heterozygous at both loci is crossed with a white, tail-less female rabbit. A large number of offspring is produced with only two phenotypes: brown and tailed, white and tail-less, and the two types are in equal numbers.
(i) Deduce the pattern of inheritance of these traits. (2)
(ii) State both parents’ genotypes and the gametes that are produced by each during the process of meiosis.
Male genotype: ........................................................................................
Female genotype: ........................................................................................
Male gametes: ........................................................................................
Female gametes: ........................................................................................ (2)
(iii) Predict the genotypic and phenotypic ratios of the F2 generation. Show your working. (2)
(b) Outline the biotechnology used to transfer genes from one organism to another. (3)
(Total 9 marks)
(i) Deduce the pattern of inheritance of these traits. (2)
(ii) State both parents’ genotypes and the gametes that are produced by each during the process of meiosis.
Male genotype: ........................................................................................
Female genotype: ........................................................................................
Male gametes: ........................................................................................
Female gametes: ........................................................................................ (2)
(iii) Predict the genotypic and phenotypic ratios of the F2 generation. Show your working. (2)
(b) Outline the biotechnology used to transfer genes from one organism to another. (3)
(Total 9 marks)
(a) (i) autosomal;
linked genes / linkage;
together on same chromosome;
as they did not separate / segregate; 2
(ii) Accept any letters for the alleles of the two genes.
male genotype is BbTt / and female genotype is bbtt / ;
Reject Bb,Tt and bb, tt.
male gametes
: BT and bt / BT and bt female gametes: (all) bt / bt; 2 max(iii) BbTt / and bbtt / ;
1 /
½ / 50% brown tailed : 1 / ½ / 50% white tail-less; 2[9]
(b) desired / specific gene obtained / selected;
mRNA copied with reverse transcriptase;
vector used / needed to get gene into host;
restriction enzymes used to cut DNA / to open plasmid;
sticky ends added / present;
DNA / gene inserted into plasmid;
DNA / gene spliced with DNA ligase;
recombinant plasmid / recombinant DNA mixed with host cells;
use of viral vectors / Agrobacterium used as a vector;
shot gunning / gold / tungsten particles coated in genes and shot into host
cell; 3 maxParagraph. Klik hier om te bewerken.
linked genes / linkage;
together on same chromosome;
as they did not separate / segregate; 2
(ii) Accept any letters for the alleles of the two genes.
male genotype is BbTt / and female genotype is bbtt / ;
Reject Bb,Tt and bb, tt.
male gametes
: BT and bt / BT and bt female gametes: (all) bt / bt; 2 max(iii) BbTt / and bbtt / ;
1 /
½ / 50% brown tailed : 1 / ½ / 50% white tail-less; 2[9]
(b) desired / specific gene obtained / selected;
mRNA copied with reverse transcriptase;
vector used / needed to get gene into host;
restriction enzymes used to cut DNA / to open plasmid;
sticky ends added / present;
DNA / gene inserted into plasmid;
DNA / gene spliced with DNA ligase;
recombinant plasmid / recombinant DNA mixed with host cells;
use of viral vectors / Agrobacterium used as a vector;
shot gunning / gold / tungsten particles coated in genes and shot into host
cell; 3 maxParagraph. Klik hier om te bewerken.
The diploid number of chromosomes in horses (Equus ferus) is 64 and the diploid number in donkeys (Equus africanus) is 62. When a male donkey and a female horse are mated, the result is a mule which has 63 chromosomes.
a) State the haploid number for horses. [1]
b) Explain reasons that mules cannot reproduce. [2]
c) Discuss whether or not horses and donkeys should be placed in the same species. [2]
d) A mule was born at the University of Idaho in the USA with 64 chromosomes. Suggest a mechanism by which this could happen. [1]
a) State the haploid number for horses. [1]
b) Explain reasons that mules cannot reproduce. [2]
c) Discuss whether or not horses and donkeys should be placed in the same species. [2]
d) A mule was born at the University of Idaho in the USA with 64 chromosomes. Suggest a mechanism by which this could happen. [1]
a) 32
b)
- because the chromosome number is not an even number/63
- (so) cannot divide by two during meiosis/cannot perform meiosis/chromosomes cannot pair up during meiosis
- one chromosome has no homologue/WTTE
- because unlikely to/cannot produce viable gametes/sperm/egg cells
c)- to be in same species two organisms must have the same genes arranged on the same chromosomes
OR
- must have the same number of chromosomes
- members of same species produce fertile offspring and a mule is not fertile
d) non-disjunction
Accept description of non-disjunction.
b)
- because the chromosome number is not an even number/63
- (so) cannot divide by two during meiosis/cannot perform meiosis/chromosomes cannot pair up during meiosis
- one chromosome has no homologue/WTTE
- because unlikely to/cannot produce viable gametes/sperm/egg cells
c)- to be in same species two organisms must have the same genes arranged on the same chromosomes
OR
- must have the same number of chromosomes
- members of same species produce fertile offspring and a mule is not fertile
d) non-disjunction
Accept description of non-disjunction.
In the pea plant (Pisum sativum), the allele for tall plants is A and the allele for short plants is a. The allele for green plants is B and the allele for yellow plants is b.
a) Determine the phenotype of Aabb. [1]
b) Compare the information that could be deduced when the genotypes are presented as AaBb or
a) Determine the phenotype of Aabb. [1]
b) Compare the information that could be deduced when the genotypes are presented as AaBb or
[2]
c) Deduce one possible recombinant offspring of individual
c) Deduce one possible recombinant offspring of individual
after a test cross. [1]