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Answers
1.
The letters in the cells along the top row and left column
reresent a haploid condition. These represent the gamet's genotype.
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2.
The letters at the top of the columns and along the side are haploid becuase they are single letters. The letters is the "cells" of the square are diploid becuse they consist of two letters. The "cells" represent the genotypes of the offspring which may be produced ty these parents. Press here to go back to home page. press
3. Both parents would be long stem and heterozygous.
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4. Using the formula 2n (n = hetrozygotes) 21 = 2 , each parent can produce two types of gametes. Press here to go back to home page. press
5. The different types of gametes would be "L" anad "l". Press here to go back to home page. press
6.
Parents = Ll x Ll
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7.
The genotype ratio would be 1LL to 2 Ll to 1 ll The Phenotyperatio would be 3 long stem to 1 short stem. Press here to go back to home page. press
8. There is a 75% chance of the offspring having long stem and 25% chance of the offspring having short stem. Press here to go back to home page. press
9. Parents
TT x Tt
a. 100 % of the offspring will be tall.
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10. parents =
Tt x Tt
b. The offspring have 25 % chanch of being dwarf. c. The offspring have 50 % chanch of being heterozygous. Press here to go back to home page. press
11. Parents
Tt x tt
b. The offspring have 50 % chanch of being dwarf. c. The offspring have 50 % chanch of being heterozygous. Press here to go back to home page. press
12. paarents = tt x
tt
a. The offspring have 0 % chanch of being tall.
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13. parents Bb
x Bb
a. The offspring have 75 % chanch of being black.
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14. parents = Bb
x bb
a. The offspring have 50 % chanch of being black.
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15. parents =
BB x bb
a. The offspring have 100 % chanch of being black.
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16. parents =
ss x ss
b. The offspring have 100 % chanch of being long haired. c. The offspring have 0 % chanch of being short heterozygous. d. The phenotypic ratio is 0 short : 1 long. e. The genotype ratio is all ss.
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17. parents = Ss
x Ss
a. The offspring have 75 % chanch of being short haired.
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18. parents =
SS x ss
chart a. The offspring have 100 % chanch of being short haired.
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19. parents =
Ss x ss
a. The offspring have 50 % chanch of being short haired.
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20. Draw two lines and label as P1 and P2. Each line reprsents a single set of genes for the parents. P1 ________ x P2 __________ List genotype of the offspring as much as possible. 50 % are heterozygous Ll; therefore, one parent must have a "L"
and the other a "l".
25% are LL; therefore, each parent must have a "L". P1 ___L___ x P2 ___Ll____ 25% are ll; therefore, each parent must have a "l". P1 ___Ll___ x P2 ___Ll____ Now you can perform the Punnett square to check answer.
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21. Draw two lines and label as P1 and P2. Each line reprsents a single set of genes for the parents. P1 ________ x P2 __________ List genotype of the offspring as much as possible. 50 % are short wing ll; therefore, each parent must have a "l".
P1 _l____ x P2 ___l___ 50% are long wing; therefore, they must have at least one large "L". One parent must have a "L". P1 ___Ll___ x P2 ___l____ At this point you may set up a Punnett square using "Ll" and "ll" for P2. Now you can perform the Punnett square to check answer.
Press here to check answer. press 22. Draw two lines and label as P1 and P2. Each line reprsents a single set of genes for the parents. P1 ________ x P2 __________ List genotype of the offspring as much as possible. 50 % are wrinkled ss; therefore, each parent must have a
"s".
P1 _s____ x P2 ___s___ 50% are smooth seed-coat; therefore, they must have at least one large "S". One parent must have a "S". P1 ___Ss___ x P2 ___s__ At this point you may set up a Punnett square using "Ss and "ss" for P2. Now you can perform the Punnett square to check answer.
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23. P1 =
Ss x
P2 = Ss
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24. Draw two lines and label as P1 and P2. Each line reprsents a single set of genes for the parents. P1 ________ x P2 __________ List genotype of the offspring as much as possible. (3 to 1 would indicate that 3 are dominant S_ and 1 are recessive ss.) 25% are wrinkled ss; therefore, each parent must have a
"s".
P1 _s____ x P2 ___s___ 75% are smooth seed-coat; therefore, they must have at least one large "S". One parent must have a "S". P1 ___Ss___ x P2 ___s__ At this point you may set up a Punnett square using "Ss and "ss" for P2. Now you can perform the Punnett square to check answer.
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25. P1 TT x P2 tt
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26. F1 Tt x tall parent TT
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27. F1 Tt
x
dwarf parent tt
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Created by the Center for Learning Technologies, Academic Technology Services. Last modified October 22, 1997. |