Answers to problems

Chapter 2

1.  A man with red-green colour blindness is married to a girl with normal colour vision.  What is the probability that his children will be colour blind?

 

Answer:  Red-green colour blindness is an X-linked recessive condition.  A ll the offspring of affected males would have normal colour vision, but all  their daughters would be carriers  In the very rare event of his partner being a carrier of red-green  colour blindness,  half their offspring,  whether male or female, would be colour blind, half their daughters would be carriers and half their sons would be normal. 

 

These results are summarized in the two following Punnett squares:

Xb - Gene for X-linked coulor blindness; X = normal gene

 

1. Affected male – genotypically normal female

 

X

X

 

Xb

Xb X

Xb X

All daughters carriers

 Y

XY

XY

All sons normal

 

2.  Affected male – carrier female (Very rare)

 

 Xb

X

 

Xb

Xb Xb

Xb X

Half daughters colour blind; half carriers

 Y

Xb Y

XY

Half sons colour blind; half normal

 

 

2. A boy has Duchenne muscular dystrophy. His maternal uncle had died of muscular dystrophy at the age of 21 years.  What is the probability that this child's sibs will be affected? 

 

Answer: This boy has an X-lined recessive condition.  The family history on the maternal side indicates that his mother is an obligate carrier.  Consequently, half his brothers would be similarly affected and half would be normal.  All his sisters would be unaffected, but half of them would be carriers.   These esults can be observed  in the Punnett square below.

Xd – gene for Duchenne Muscular dystrophy;  X – normal allele

 

1. Carrier female – Normal male

 

Xd

X

 

X

Xd X

X X

 All sisters normal; half of them carriers

 Y

Xd Y

XY

Half his brothers affected; half normal  

 

 

 

 

 

  1. The adjacent pedigree is of a family with haemophilia as shown in Fig.9.4 .  If the gene for haemophilia is represented by Xh and the normal allele by X', write the genotypes of each individual in the pedigree. If the genotype cannot be determined with certainty write the possible alternatives. 

 

Answer:

Begin with affected individual, III:3 .  His genotypes is  Xh Y.  Similarly, his affected uncle II: 5 must have the same genotor.Because of the family history, his mother II:3 is an obligate carrier with genotype Xh X'.  His father (II:4) must be X' Y.

II:1 – X'Y (unaffected);  II:2 and II:6 are X' X'

III:1 - X'Y (unaffected); III:2 - X' X' ;

III:4 – could be either X' X' or Xh X' (carrier)

III:5 - X'Y (unaffected);

III:6 - Xh X' (obligate carrier because her father is affected)

 

  1. Explain, giving reasons, whether the following pedigrees are compatible with autosomal dominant, autosomal recessive or X-linked dominant and X-linked recessive inheritance.  (Note that a pedigree may be compatible with more than one type of inheritance.)


                                                                       

Answers:

 

a.      This pedigree shows dominant inheritance because of linear transmission from parent to offspring.  It could be either autosomal dominant or X-linked dominant

b.      This is very similar to the above  - autosomal dominant or X-linked dominant inheritance

c.       This pedigree is consistent with X-linked recessive inheritance  because:

i.                    only males are affected,

ii.                  transmission is always through a carrier female, and there  is no male to male transmission

Another possibility is autosomal dominant inheritance with decrased penetrance, thus appearing to skip generations.

 

 

 

5. Examine the pedigree from a family with a genetic disease and answer the questions below:

Text Box:

           

 

 

 

 

 

 

 

 

 

 

 

 

 

 

a. Does this pedigree indicate autosomal dominant, recessive or sex-linked type of inheritance?  Give reasons for your choice.

 Answer:   This pedigree is consistent with autosomal dominnat inheritance with reduced penetrance.  X-linked inheritance is excluded because of male to male transmission (II:1 to III:1).

 

b. Assuming that B and b are the normal and mutant alleles  respectively, what would be the genotypes of the individuals: II.1, II.2 and III.3 ?

 

Answer:

II:1 – Bb . This individual, athough unaffected by the diease is obviously transmitting the disease  

II:2 – bb .  This individual is  normal and unrelated to the affected family

III:3 – Bb.  Affected individual.

 

c. Individual II.3 requested genetic counselling.  What is the probability that her child would be affected. Explain why.

 

Answer: 

II:3 is obviously transmitting the disease although she is unaffected.  The chance that her children would be heterozygotes is 50%.  However, only some of these would be affected, depending on the penetrance.  In the aggbove pedigree, three out of five heterozygotes were affected, equal to 60%.  This is only a rough measure of the penetrance because of the small number of individuals involved.  Thus II:3 has a risk of 60% of 50% (=30%) that her offspring would be affected, and 20% risk that her normal offspring are in fact carriers of the abnormal gene.    

 

 

6. A young lady requested pre-marital genetic counselling because her sister had died in infancy of gangliosidosis, an autosomal recessive disease. What is the risk that this young lady has similarly affected offspring?  What advice should be given?

 

Answer:

This young lady is normal. She could be either heterozygous (Gg) carrier  or normal heterozygote (GG).  Here g represents the recessive gene for gangliosidosis and G the normal allele.   Even if she is a carrier (Gg), the chance that her husband is also a carrier would be very small, because this is a rare disease.  Hence, she has a very small risk that her offspring would be affected. 

 

With the availability of genetic testing it is possible to determine whether this young lady is in fact heterozygous or normal, and if she is heterozygous it is possible to test her husband.  If this lady is heterozygous and her husband is normal (GG) she has no risk of having a child with gangliosidosis.  

 

7. A young man requested pre-marital genetic counselling because his brother is an achondroplastic dwarf.  This condition is inherited as an autosomal dominant disorder. What is the risk that his offspring will be similarly affected?  What advice should be given?

 

Answer:

Achondroplasia is autosomal dominant with 100% penetrance.  This young man is normal and therefore does not carry the abnormal gene for achondroplasia.  He has no increased risk of having affected offspring.   His risks is the same as the general population risk that a new mutation will occur, which is in fact negligibly small.

 

 

 

8.  A young couple had their first child affected with cystic fibrosis.  What is the risk that their future children will be similarly affected?

Answer:

 Cystic fibrosis is an autosomal recessive disorder.  This young couple must both be heterozygous carriers.  The risk that their future offspring would be affected with cystic fibrosis is 25% (1 in 4). 

 

 

 

 

Chapter 3

 

 

1.      The diagram shows a pentose sugar.

a) Label the carbon atoms on the pentose ring.

b) Insert a PO4 group in the correct position.

c) Insert an adenine base in the correct position.

d) What is the name of the resulting nucleotide?

e) Insert a whole thymine nucleotide in the correct position to form a dinucleotide..

f) Label the 3' and 5' ends of the dinucleotide formed.


 

 


2.      The following is the sequence of bases on a single strand of DNA that was annealed with its complementary strand to form double stranded DNA.

     

      5’ - G T T A C G G G C G G A A C C G T A A T -3’

 

a)      Write out the sequence of bases on the complementary strand.

 

3’ – C A A T G C C C G C C T T G G C A T T A  -5’

 

b)     Calculate the proportions of the four nucleotides A,T,C and G in the single DNA strand.

Answer:

There are 20 bases  in all

A  = 5  bases = 25%

T  =  4 bases =  20%

C  = 4  bases =  20%

G  =  7  bases = 35%

 

c)      Calculate the proportions of the four nucleotides A,T,C and G in the double-stranded  DNA   formed after annealing.

 

Answer:

In the complementary strand of DNA there are 

A  = 4  bases =  20%

T  =  5 bases =  25%

C  =  7  bases =  35%

G  =  4  bases =  20%

 

Therefore the double stranded DNA contains

A  =  9  bases  =  22.5%

T  =  9 bases  =   22.5%

C  =  11  bases =  27.5%

G  =  11 bases =  27.5%

 

d)     Compare the A:T and C:G ratios of the single and double-stranded DNA. Explain why they are different.

Answer:

The ratios of A,T,C and G are independent of one another in single stranded DNA but T =A and C=G in the double-stranded DNA.  This is because of specific base pairing in the  in double stranded DNA.

e)       Work out the values of (A + G) and (T + C) for the single and double- stranded DNAs.  Explain the results.

 

Answer:

In the given single-stranded DNA                   (A+ G) =12;   (T+C) = 8.

In the complementary strand of DNA             (A+ G) =8 ;    (T+C) = 12.

In the double-stranded  DNA                           (A+ G) =20 ;  (T+C) = 20.

(A+G) on one strand = (T+C) on the complementary strand.

In double-stranded DNA (A+G) = (T+C) = half the total number of bases

 

 

3.      Estimation of the proportion of bases contained in the genetic material of a virus gave the following result: Adenine 36%;  Thymine 18%; Cytosine 26%; Guanine 20%. From these results determine which type of  genetic material was present in this virus. Explain your result.

Answer:

In this specimen   A is not equal to T and C is not equal to G. Therefore it must be single-stranded  nucleic acid.  Since there is thymine it must be single-stranded DNA.

 

A particular stretch of double-stranded DNA contianed  50 thymine and 100 cytosine bases.  What is the number of adenine and guanine bases in this stretch of DNA? 

Answer:

A = T  = 50 bases

G = C = 100 bases

 

4.      DNA estimations on three specimens of nucleic acid gave the following results:

                              A                     T                      C                     G                     U

specimen 1       20%                 20%                 30%                 30%                 -

 

specimen 2       20%                 -                       35%                 15%                 30%

 

specimen 3       30%                 20%                 30%                 20%                 -

 

Identify whether each of the specimens was DNA or RNA and whether single or double stranded. Explain.

Answer:

Specimen 1:    A=T and C=G.  Therefore this is double stranded DNA

Specimen 2:   T is replaced by U and C is not equal to G.  Therefore this is RNA (single-stranded).

Specimen 3: Ais not equal to T and Cis not equal to G.  Therefore, it must be single-stranded DNA.

Chapter 4

1.  The following sequence is on the transcribed strand of DNA:

            3'  G  C  T  A  A  T  C  A G  T  G  C  G  T  A  5'

Write the sequences of:

a.       the coding strand of  DNA

b.       m-RNA

c.        amino acids

 

Answer:

a.   Coding strand

            5`  C  G  A  T  T  A  G  T C  A  C  G  C  A  T  5'

 

b.  m-RNA (replace the Ts in the coding strand by Us)

5`  C  G  A  U  U  A  G  U C  A  C  G  C  A  U  5'

 

c.                 Arg          Phe       Val         Thr       His

 

 

2.  The following is a schematic diagram of DNA being transcribed:

 


a.       DNA transcribed strand

b.      DNA coding strand

c.       m-RNA

The direction of transcription is shown.

Label the 5' and 3' ends of each strand.

Answer:

Transcription always occurs in the 5’ to 3’ direction  on m-RNA.  Start by labelling  m-RNA.  The 5’ and 3’ ends are reversed on the transcribed strand  but are the same as on the coding strand.

 

3.  The following is the sequence of amino acids of the last part of a protein molecule. Using the genetic code write the codons on m-RNA and label the 3' and 5' ends. (Where more than one codon is possible use the first one in the table.)  Write the corresponding codons on the template strand of DNA.

Answer:  You must refer the table of genetic codes.

                        - arg  -  leu  -  asp - phe -   pro -   ile -   glu -   gly -  val -

m-RNA            5’-AGA-CUU-GAU-UUU-CCU-AUU-GAA-GGU-GUU-3-

DNA template strand

                        3’-TCT-GAA-CTA-AAA-GGA-TAA-CTT-CCA-CAA- 5’

Note: do not forget to enter the 5’ and 3’ ends