Assigment Help
Solve the following genetic problems as described. Some of the problems are already solved but need to be explained more In detail as described marked in yellow.
Question Five
If Y/Y . O/O is crossed to y/y . o/o, and the F1 is testcrossed, what percentage of the testcross progeny will be Y/y . O/o if the two genes are
The cross is Y/Y. O/O x y/y. o/o , thus the F1 would be Y/y. O/o.
- unlinked –
If the genes are unlinked, all four progeny classes from the testcross (including y/y ; o/o) would equal 25 %
Show the solution in a punnet square !!!
- completely linked (no crossing-over at all)
With completely linked genes, the F1 would produce only Y O and y o gametes. Therefore, there would be a 50% chance of having y o/y o progeny from a testcross of this F1.
THIS IS CORRECT ANSWER – DOES NOT NEED ANY MORE SOLUTION
- 25 map units apart? Explain your answer.
If the two genes are linked and 25 map units apart, 25% of the testcross progeny should be recombinants. This is because the F1 is Y O/y o, y o is one of the parental classes and it should equal 1/2 of the total parental.
Show the solution in a punnet square !!! Show the calculations !!!
6. A rice breeder obtained a triple heterozygote carrying the three recessive alleles
for albino flowers (al), brown awns (b), and fuzzy leaves (fu), all paired with their
normal wild-‐type alleles. This triple heterozygote was test-‐crossed. The progeny
phenotypes were:
50 wild type
42 albino, brown, fuzzy
520 albino
511 brown, fuzzy
202 fuzzy
190 albino, brown
5 brown
3 albino, fuzzy
- Are any of the genes linked? If so, draw a map labeled with map distances.
(Don ́t bother with a correction for multiple crossovers)
There are always more parental alleles !!! other combinations are “recombinants” . how do the parental alleles in that case?
Question four
If the recessive trait albinism (a) is present in 1/40,000 individuals in a population at equilibrium, calculate the frequency of
(a) The recessive mutant allele (heterozygotes in the population
In assuming the population is in H-W equilibrium, then the frequency of individuals with the albino genotype is the Frequency of (aa) =Q2
Frequency of the recessive mutant allele will be 1/40000=0.000025 Square root= 0.005
Therefore the frequency allele of albinism is 0.005
(C)The frequency of the normal allele
The frequency of the normal allele is equal to 1-q = 1 -0.005= 0.995
CORRECT !!!
(d) mating between heterozygotes
. We’d then predict that the frequency of Hopi who are homozygous normal (genotype AA) is p2 , which is 0.873. In other words, 87.3% of the population, or an estimated 5238 people, should be homozygous normal. The frequency of carriers we’d predict to be 2pq, which is 0.123. So 12.3%, or 737 people, should be carriers of albinism, if the population is in H-W.
CALCULATE THE RISK/CHANCE OF TWO HETEROZYGOUS MATES !!! SHOW CALCULATIONS
Aa x Aa) Aa = 2pq
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