Question

In: Biology

A purple flowering, smooth seed dihybrid plant (genotype PpSs) is test crossed with a white flowering,...

A purple flowering, smooth seed dihybrid plant (genotype PpSs) is test crossed with a white flowering, wrinkled seed plant (genotype ppss). These produce progeny in the following numbers of four phenotypes: 24:76:74:26 (purple flower + smooth seed coat: purple flower + wrinkled seed coat: white flower + smooth seed coat: white flower + wrinkled seed coat). a) Explain how ratios of progeny show that the two genes are linked. b) How many map units separate the colour and seed coat genes? Show your calculations.

Expert Solution

Answer: The genotype of Purple flowering smooth seed dihybrid plant is PpSs.

Test crossed with white flowering wrinkled seed plant (genotype ppss)

Purple flowering smooth seed X white flowering wrinkled seed

PpSs X ppss

gametes PS, Ps, pS, ps X ps

PS

Ps

pS

ps

PpSs

Purple smooth

Ppss

Purple wrinkled

ppSs

white smooth

ppss

white wrinkled

a) The following numbers of progenies are obtained from the above test cross.

24:76:74:26

24 purple flower + smooth seed coat: 76 purple flower + wrinkled seed coat: 74 white flower + smooth seed coat: 26 white flower + wrinkled seed coat. This ratio indicates that the genes responsible for flower color and sead coat are linked.

b) To find the map units that separate the colour and seed coat genes, we have to calculate the recombinant frequency between the genes P and S.

The total number of progenies = 200

So the recombinant frequency between the genes P and S = 76 + 74 / 200 = 150 / 200 = 0.75 x 100 = 75 map units

The map units that separate the colour and seed coat genes = 75 map units.

gladiator answered 2 years ago

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