Question 5 [12 marks /12 minutes] In mouse, the Brachyury mutation (T) shows the

Question

Question 5 [12 marks /12 minutes] In mouse, the Brachyury mutation (T) shows the mutant phenotype when heterozygous with the wildtype allele (T), and is also recessive lethal. Suppose a researcher has identified a new mutant, hairless (h), which is recessive to the wildtype allele of this second locus (H). Both loci are on the same autosome. The researcher wants to estimate the recombination frequency between the T locus and the hairless locus, and so crosses Brachyury mutant phenotype mice with hairless mutant phenotype mice a) Write the genotype(s) and phenotype(s) of the progeny. You must show linkage in the way you write the genotypes (i.e. show the phase) 4 marks] b) What cross should the researcher do next, to produce progeny from which the recombination frequency can be estimated? Show the genotypes and phenotypes of the progeny, indicate recombinants and non-recombinants, and show how the recombination frequency would be [8 marks] calculated

Explanation / Answer

TBT+= Brachyury mutant phenotype

T+T+= Wildtype phenotype

TBTB= Recessive lethal

H=Wild type allele

.h=recessive allele (hh give hairless condition)

Cross between TBT+ and hh phenotype mice.

TBT+ HH x   T+T+ hh will give rise to heterozygous progeny with genotype TBT+ Hh and T+T+ Hh.

Now to know the recombination frequency between T locus and the hairless locus, select the female progeny with heterozygous condition TBT+ Hh and cross them with double recessive males with genotype TBTB hh. Now list the progeny obtained.

Cross between TBT+ Hh (heterozygous females) and TBTB hh (double recessive males) will rise to four combinations of progeny

TBTB hh (parental combinations)

T+TB Hh (parental combination)

TBTB Hh (non parental combination)

T+TB hh (non parental combination)

Now try to count all the progeny

Recombination frequency= (Number of recombinants)/(Total progeny) x 100= answer m.u

The answer will give you the distance between two genes. If the recombination frequency is less than 50%, then the two genes are linked.

If the recombination frequency is more than 50%, then the genes are not linked and they can segregate independently

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