In: Biology
Problem 1
You are doing a genetics experiment with the fruit fly. In the “P” generation, you cross two true-breeding flies. Thefemale parent is brown and wingless and the male parent is black with normal wings. All of the flies in theF1 generation are brown and have normal wings.
Assume the genes are not found on a sex chromosome. Indicate the body color alleles with B and b and thewing size alleles with N and n.
a) The genotypes of the flies in the P generation are: Female:_____BBnn_________ Male:___bbNN___
b) The genotype of the flies in the F1 generation is: __BbNn___________
c) You now take an F1 female and cross her to a true-breeding black, wingless male. This male’s genotype is:_____bbnn_________
d) When the flies in (b) and (c) are mated, you count 1600 offspring in the F2 generation. If the body color geneand wing type genes are not linked, how many flies of each phenotype would you expect? (Be sure to use a Punnett square for a dihybrid cross to show your work.)
Show your work:
Expected flies:
_0_: # of brown, winged flies (Genotype: __BbNn__)
__800__: # of black, winged flies (Genotype: ___Bbnn___) __800___: # of black, wingless flies (Genotype: ____bbNn___) __0_____: # of brown, wingless flies(Genotype: __bbNN___)
e) When you count the F2 generation, you actually get 85 brown, winged flies, 728 black, winged flies, 712 brown,wingless flies, and 75 black, wingless flies. These results indicate that the body color and wing size are linkedgenes.
·What are the phenotypes of the recombinant flies? (Hint: Refer back to the P generation flies in the first paragraph. Recombinant flies do not resemble the original P generation phenotypes.)
·Calculate the recombination frequency (%) between the linked genes for body color and wing size.
Problem 2
You cross a true-breeding yellow-bodied, smooth-winged female fly with a true-breeding red-bodied, crinkle-winged male. The red body phenotype is dominant to the yellow body phenotype and smooth wings are dominant to crinkled wings. Use B or b for body color alleles, and W or w for wing surface alleles.
a) What are the genotypes of the P generation flies?
b) What will be the genotype(s) and phenotype(s) of the F1 offspring?
c) You discover that the genes for body color and wing surface are linked. You perform a dihybrid test crossbetween the F1 flies from part (b) with a true-breeding yellow-bodied, crinkle-winged fly. Use the followingF2 results to determine the recombination frequency (%) between the body color and wing surface genes. (Remember that the recombinants are the ones that do not resemble the parental types from the P generation.)
Body Color |
Wing Surface |
# of Individuals |
red |
smooth |
102 |
yellow |
smooth |
404 |
red |
crinkled |
396 |
yellow |
crinkled |
98 |
You decide to turn your attention to a different gene, one that controls wing length. This gene has two alleles, "L orl" where long wings are dominant to short wings. Remember that the red body phenotype is dominant to the yellow body phenotype. You again mate two true-breeding flies:
P: red-bodied, short wing male X yellow-bodied, long wing female
F1: All red-bodied, long wing
d) You perform a test cross between the F1 flies above with true-breeding yellow-bodied, short-winged flies. You get the following F2 results. What is the recombination frequency (%) between the genes for body color and winglength?
Body Color |
Wing Length |
# of Individuals |
red |
long |
45 |
red |
short |
460 |
yellow |
long |
440 |
yellow |
short |
55 |
e) Based on the information in (c) and (d), what are the two possible arrangements of these three genes: body color,wing surface and wing length? Draw two linkage maps to show the possible arrangements of these genes and the map distance between genes
PROBLEM 1 :-
ANSWER A :-
The parental generation has brown and wingless pure-bred female parent which has genotype 'BBnn' and Black and winged male parent with genotype 'bbNN'
ANSWER B :-
The F1 generation flies will be heterozygotes as they will recieved combination of alleles from both the pure-bred parents. They will have the genotype 'BbNn'.
ANSWER C :-
The male considered for cross with the F1 female is pure-breeding Black and wingless and hence will show the genotype 'bbnn'
ANSWER D :-
If the genes are not linked, the heterozygous ratio will be seen as follows:-
BN | Bn | bN | bn | |
bn | BbNn | Bbnn | bbNn | bbnn |
If the genes are not linked, the ratio would be 1:1:1:1
Brown, wingless : 1 (400)
Brown, winged : 1 (400)
Black, winged : 1 (400)
Black, wingless :1 (400)
If the genes are linked, the ratio would be 1:1:0.1:0.1 which can be seen as the actual results from the experiment.
ANSWER E :-
The recombinants will differ from the parent. They will show the differing characters associated due to combination of alleles from both parents. Hence, the genotypes namely 'BbNn' that is Brown and Winged and 'bbnn' that is Black and wingless will be recombinants as they show features from both parents.
Recombinants frequency for the linked genes = Number of recombinants / Total number of offsprings
= 85 + 75 / 1600
= 160 / 1600
= 0.1 or 10%
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