This quiz contains a mixture of SL and HL questions

1. A man suffers from phenylketonuria (PKU), but his wife does not, however her father had the disease. What is the likelihood of their children suffering from the disease?

  • A.   100 %
  • B.   75 %
  • C.   50 %
  • D    0 %
2. What is the pattern of inheritance in this pedigree?

Autosomal_Dominant_Pedigree_Chart

 GAThrawn22
c.c. BY-SA 3.0
  • A.   Sex linked recessive
  • B.   Sex linked dominant
  • C.   Chromosomal
  • D.   Autosomal dominant

3. A woman is a carrier of haemophilia; her husband does not have haemophilia. What are all the possible outcomes for their children?

  Daughters Sons
A. All normal All normal
B. Half carriers Half affected
C. All carriers All affected
D. Half affected Half affected

4. The image shows an Indonesian stamp featuring the four o’clock flower, also known as the marvel of Peru (Mirabilis jalapa).

four oclock flower image

If differently coloured pure breeding homozygous parent plants of this species are crossed, and the resulting F1 generation then self-crossed, what would be the expected phenotypic ratio of the F2 generation?
  • A. 3:1
  • B. 1:1
  • C. 2:2
  • D. 1:2:1

5. The image shows a box and whisker diagram for the heights of a group of children:

box and whisker diagram for the heights of a group of children

Which value would be considered an outlier?

  • A. 93.5 cm
  • B. 90.2 cm
  • C. 108.1 cm
  • D. 106.5 cm

6. The image shows a simplified diagram of a human cell with some gene loci on chromosomes labelled.

human cell with some gene loci on chromosomes labelled

 Aweir03 (adapted)
c.c. BY-SA 4.0
Two individuals with the above genotype are crossed. Which of the following statements is correct?
  • A. The predicted phenotypic ratio should be 1:1:1:1
  • B. Mendel’s law of independent assortment for genes G and H does not apply
  • C. The predicted phenotypic ratio should be 9:3:3:1
  • D. During meiosis each parent will form four haploid gametes cells.

Q7-10. The image below shows part of an illustration of fruit flies (Drosophila melanogaster) from Thomas Hunt Morgan’s 1916 book A Critique of the Theory of Evolution.

illustration of fruit flies - Drosophila

In fruit flies, grey bodies (B) are dominant to black bodies (b) and normal wings (W) are dominant to vestigial wings (w). Flies which are homozygous black bodied with vestigial wings are crossed with individuals who are heterozygous for both traits. The resulting observed phenotypes are below.
Note - this is not the cross shown in the image above.

Observed phenotype  Number of individuals 
Grey bodies and normal wings  630
Grey bodies and vestigial wings 118
Black bodies and normal wings 110
Black bodies and vestigial wings  642
7. Which genotype is correct for the homozygous individuals in this cross?
  • A. BBww
  • B. bbWW
  • C. bbww
  • D. BBWW
8. Which statement applies to the above scenario?
  • A. The predicted ratio of phenotypes is 9:3:3:1
  • B. The likelihood of recombinants from this cross increases if genes for wings and body colour are far apart on the same chromosome
  • C. There is independent assortment of the genes for wings and body colour
  • D. The likelihood of recombinants from these cross decreases if genes for wings and body colour are far apart on the same chromosome

9. What are the possible recombinants for the offspring from this cross?

A. Q9a option
B. Q9b option
C. Q9c option
D. Q9d option
10. Chi squared testing was performed on the observed phenotypes. If the genes are not linked, the predicted phenotypic ratio should be 1:1:1:1. For this test, at the 0.05 level of significance the critical value for chi squared is 7.815. Which of the following should apply based on the results above?
  • A. Chi squared test proves the genes are linked
  • B. The calculated chi squared value should be less than 7.815
  • C. The degrees of freedom = 2
  • D. The expected number of black bodied, normal winged flies is 375