Two-trait crosses with a 4 by 4 square

Grade 9 Science Worksheet - Challenge Me

Circle the best answer.

1. 1.

   Ren test-crosses a round yellow plant (RrYy) with a wrinkled green plant (rryy). What phenotype ratio does the square predict?

   • 1 : 1
   • 1 : 1 : 1 : 1
   • 3 : 1
   • 9 : 3 : 3 : 1

2. Read this first

   Test cross (model data)

   Plant X, a round yellow plant of unknown genotype, crossed with a wrinkled green plant (rryy). Offspring: 26 round yellow, 24 round green, 25 wrinkled yellow, 25 wrinkled green. Total: 100. Table: Test cross counts (model data); columns offspring and number of plants; rows round yellow, round green, wrinkled yellow, wrinkled green, total.

3. 2.

   Jun reads the model counts from a test cross of plant X with a wrinkled green plant (rryy). Which genotype is plant X?

   • RRYY (R: two capital; Y: two capital)
   • RrYy (R: one capital, one small; Y: one capital, one small)
   • RRYy (R: two capital; Y: one capital, one small)
   • RrYY (R: one capital, one small; Y: two capital)

4. Test cross (model data)

   Plant X, a round yellow plant of unknown genotype, crossed with a wrinkled green plant (rryy). Offspring: 51 round yellow, 0 round green, 49 wrinkled yellow, 0 wrinkled green. Total: 100. Table: Test cross counts (model data); columns offspring and number of plants; rows round yellow, round green, wrinkled yellow, wrinkled green, total.

5. 3.

   Kira reads the model counts from another test cross of a round yellow plant X with a wrinkled green plant (rryy). Which genotype is plant X?

   • RRYY (R: two capital; Y: two capital)
   • RRYy (R: two capital; Y: one capital, one small)
   • RrYY (R: one capital, one small; Y: two capital)
   • RrYy (R: one capital, one small; Y: one capital, one small)

6. Use this

   Mendel's F2 seed counts from separate one-trait crosses with garden peas. Seed shape: 5,474 round and 1,850 wrinkled. Seed color: 6,022 yellow and 2,001 green. Table: Mendel's one-trait counts; columns seed trait and number of seeds; rows round, wrinkled, yellow, green.

7. 4.

   Luz reads Mendel's one-trait counts. If seed shape and seed color sort independently, about what fraction of seeds in a two-trait cross are round and yellow?

   • About 9/16, since each share is close to 3/4
   • About 3/2, found by adding the two shares
   • About 1/16, since the recessive shares are close to 1/4
   • About 3/4, the share for either trait alone

8. Model data

   160 plants from RrYy × RrYy. Counted: 92 round yellow, 29 round green, 31 wrinkled yellow, 8 wrinkled green. Expected from 9 : 3 : 3 : 1: 90, 30, 30 and 10. Table: Counted and expected (9 : 3 : 3 : 1 ratio); columns phenotype and counted and expected; rows round yellow, round green, wrinkled yellow, wrinkled green, total.

9. 5.

   Noah grows 160 plants from RrYy × RrYy and records model counts. Which statement fits the counts?

   • The counts match 1 : 1 : 1 : 1, with 40 plants in each class
   • The counts do not fit, because wrinkled green should be 40, which is a quarter of the 160 plants
   • The counts are close to the expected 90, 30, 30 and 10, so they fit independent assortment
   • The counts do not fit, because no count matches its expected number exactly

10. Notebook for RrYy × RrYy

    Line 1: Each parent passes on 4 combinations. Line 2: The square has 4 × 4 = 16 boxes. Line 3: Round green fills 3 boxes. Line 4: The chance of round green is 3/4 + 1/4 = 1. Table: Notebook for RrYy × RrYy; columns line and what the notebook says; rows 1, 2, 3, 4.

11. 6.

    Tala's notebook is about one RrYy × RrYy cross. One line is wrong. Which fix is right?

    • Line 3 should say round green fills 9 boxes
    • Line 4 should multiply: 3/4 × 1/4 = 3/16
    • Line 1 should say each parent passes on 2 combinations
    • Line 2 should say the square has 4 + 4 = 8 boxes

12. Notebook for 480 seeds from RrYy × RrYy

    Line 1: Round yellow fills 9 of 16 boxes. Line 2: Wrinkled green fills 1 of 16 boxes. Line 3: Expected wrinkled green: 480 ÷ 16 = 30 seeds. Line 4: Expected round yellow: 9 × 480 = 4,320 seeds. Table: Notebook for 480 seeds; columns line and what the notebook says; rows 1, 2, 3, 4.

13. 7.

    Ari's notebook scales a model total of 480 seeds from RrYy × RrYy. One line is wrong. Which fix is right?

    • Line 1 should say round yellow fills 3 of 16 boxes
    • Line 2 should say wrinkled green fills 3 of 16 boxes
    • Line 4 should say 9 ÷ 16 × 480 = 270 seeds
    • Line 3 should say 480 ÷ 4 = 120 seeds

14. Claim card

    RrYy × RrYy always gives 9 : 3 : 3 : 1. Both genes show simple dominance (R over r, Y over y). The seed shape gene and the seed color gene can be on different chromosomes or on the same one. Table: Claim card; columns claim and cross; rows always 9 : 3 : 3 : 1.

15. 8.

    Esme claims that two plants with RrYy always give 9 : 3 : 3 : 1, with simple dominance for both genes. Which reply has the correct reason?

    • No, because two genes followed together give 3 : 1
    • Yes, because the dominant alleles stay together when passed on
    • Yes, because the 16 boxes are the same for any two genes
    • No: the ratio needs the two genes to sort independently, as genes on different chromosomes do

16. Model data

    32 plants from RrYy × RrYy. Found: 1 wrinkled green plant. Prediction: wrinkled green fills 1 box of 16, so 32 ÷ 16 = 2 plants are expected. Table: Claim card; columns item and plants; rows plants grown, boxes in the square, wrinkled green found, wrinkled green expected.

17. 9.

    Bo grows 32 plants from RrYy × RrYy (model data) and finds 1 wrinkled green plant. The claim is that the 1/16 prediction is wrong. Which reply is best?

    • The claim is weak, because 18 are expected
    • The claim is right, because the expected number is 8
    • The claim is weak: 2 are expected, and a count near 2 can happen by chance in a small group
    • The claim is right, because exactly 2 must appear
18. 10.

    Claim card: a test cross of RrYy with rryy shows the four allele combinations RrYy can pass on. Jun claims that a test cross of RrYy with rryy shows the four combinations RrYy can pass on. Which reason supports the claim?

    • The rryy plant passes on its own gene pairs, rr and yy, so the offspring copy them
    • The RY combination stays together, so only RY and ry offspring appear
    • The offspring come out 9 : 3 : 3 : 1, which shows the combinations
    • The rryy plant passes on only ry, so each offspring shows one combination from the other plant