In the cat, two gene loci on an autosomal chromosome are 32 map units from each
ID: 64488 • Letter: I
Question
In the cat, two gene loci on an autosomal chromosome are 32 map units from each other. At locus 1, the E allele for big ears is completely dominant to the e allele for small ears. At locus 2, the G allele for grey hair is completely dominant to allele g for white hair. A strain of cat, true-breeding for big ears and grey hair, is mated to strain 2, true-breeding for small ears and white hair. The F1 generation of that mating is mated with strain 2 (ie., a testcross) and 400 testcross progeny are produced.
_____: What is the expected frequency of gametes of genotype EG produced by the F1-generations cats? Show your calculations.
_____: When the F1-generation cats produce gametes, what is the expected frequency of gametes that have recombinant combinations of linked genes?
_____: In the testcross progeny, how many cats are expected to have small ears & white hair? Show your calculations.
____: In the testcross progeny, how many cats are expected to have small ears & grey hair? Show your calculations.
_____: If the two loci were not linked, and segregated independently from each other, how many cats are expected to have small ears & grey hair in the testcross progeny? Show your calculations.
Explanation / Answer
1.What is the expected frequency of gametes of genotype EG produced by the F1-generations cats:
In F1 EEGG X eegg
F1 : EeGg
all the progeny will be EG type.
therefore expected frequency will be 100 %
2.the expected frequency of gametes that have recombinant combinations of linked genes:
The map distance between two genes is 32. therefore the recombinant frequency will 32 %.
total progeny are 400.
therefore 32% of 400= 128 progeny will have recombinant combinantion of linked genes.
3. cats expected to have small ears & white hair:
This trait is parental type. therefore
Linkage distance is 32 map units.
therefore recombination frequency will be 32%(1 % linkage is 1 map unit.)
this means that the recombinants are 32% and parental types are 100-32= 68%.
therefore EG type progeny will be approx 50% od the parental type. Because in linkage the parental types are almost equal (EG and eg are equal).
therefore 50% of 68% will be 34%
therefore out of 400 34 % will be small ear white hair type.
on calculation, 34% of 400= 34/100*400=136 progeny will be eg or small ear white hair type
4. cats expected to have small ears & grey hair
this trait is a result of linkage. therefore this type of progeny will be approx half of the recombinbants.
total recombinants as from 2. are 128
therefore expected frequency of small ear and grey hair are 16% or, 64 in no.
5. If the two loci were not linked, and segregated independently from each other, then cats expected to have small ears & grey hair in the testcross progeny
the test cross reults into a 1:1:1:1 ratio.
total progeny are 400 therefore the eg will be 1/4 of total.
=1/4*400= 100
means 100/400*100=25% of the offspring will have small ear and grey hair expected.
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