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BIOLOGY A ASSIGNMENT 2

GENETICS PROBLEMS

Question 1

The result in F1 generation of crossing a homozygous straight-winged fly with a curved-winged fly would produce genotypically heterozygous flies with straight wings. This is due to the dominant allele for straight-winged trait (L). see table:

The crossing of two F1 flies will result in a F2 generation with the following genotypes: 25% will be homozygous dominant with a straight wings phenotype, 50% will be heterozygous also with a straight wings phenotype and 25% will be homozygous recessive with a curved wings phenotype. see table:

The genotype of the F2 straight-winged flies can be determined by doing a back cross experiment. For this experiment the F2 straight-winged flies are crossed with curved-winged flies (homozygous recessive). The result would determine whether the sample of F2 straight-winged flies are homozygous dominant (LL) or Heterozygous (Ll).

The homozygous dominant crossed with curved-winged flies would produce all straight-winged offsprings and the heterozygous flies crossed with curved-winged flies would produce approximately 50% straight-winged flies and 50% curved-winged flies, a ratio of 1:1

Question 2

Crossing Drosophila A and B resulted in grey-bodied offsprings, therefore at least one of the parents is homozygous dominant.

Crossing A and C resulted in 25% recessive offsprings, therefore both A and C are heterozygous.

Crossing B and C resulted in all grey-bodied offsprings, therefore at least one of the parents is homozygous dominant, in this case the Drosophila B.

Applicable for the crossing of flies A and B or the crossing of flies B and C

Applicable for the crossing of flies A and C

The conclusion is that only the pair of B flies are homozygous dominant and the pairs A and C are heterozygous phenotypically grey-bodied.

When crossing the flies A and C with black-bodied flies the result would be that 50% of their offsprings will be phenotypically black-bodied with a (gg) genotype, and 50% phenotypically grey-bodied with a (Gg) genotype.

Question 3

Both parents have to carry a recessive allele (or recessive trait) of the disease in order to pass on the disease to the children, which is in the order of 25% chance for the child to inherit both recessive alleles from each parent.

In the case of fibrocystic disease, both parents are Heterozygous and phenotypically normal (displaying no symptoms of the disease), a punnet square can illustrate the results.

The probability that their next child will be defective is the same as for the number 1 child which is 25% chance or ratio 3:1, of inheriting the abnormal gene.

Question 4

a)     Both parents need to have a recessive allele in order to pass in on to their child, as both parents of the albino child are phenotypically normal then we can conclude that they both are heterozygous carrying the trait for albinism.

b)     The probability that their next child will be an albino is 25% as the punnet square demonstrate, ratio 3:1