AP Biology · Free-response skill

Read a cross: does the gene ride on the X?

The move. Predict what a cross gives under each hypothesis, compare the predictions with observed counts using chi-square, and decide which hypothesis the data fit.

How to do it

  1. Write the parents' genotypes for each hypothesis before you predict anything. For a gene on an autosome, use two letters (Cc). For a gene on the X, write the allele on each X (X-C X-c) and give the male one X and a Y (X-C Y).
  2. Predict sons and daughters separately. A son gets his only X from his mother and a Y from his father. A daughter gets one X from each parent. If the two sexes look different in a way an autosomal gene cannot explain, suspect the X.
  3. For chi-square, find the expected count in each class (its share of the ratio times the total), then add (observed - expected) squared / expected over all classes. Degrees of freedom are the number of classes minus 1.
  4. Compare your value with the critical value given. Smaller means the data are consistent with the prediction; larger means reject it. Never say a chi-square test proves a hypothesis.

Practice prompt Try it

In a fictional beetle, females are XX and males are XY. Copper shells are dominant to green shells. A researcher crossed a green-shelled female from a true-breeding green line with a copper-shelled male from a true-breeding copper line. All 124 daughters had copper shells, and all 121 sons had green shells. She then crossed these daughters and sons with each other and counted 400 young, shown in the table. For 3 degrees of freedom at p = 0.05, the critical chi-square value is 7.81.

Shell color in 400 young from crossing the first-generation daughters and sons

YoungNumber
Copper females102
Green females96
Copper males99
Green males103

Using the information and the table, respond to each part.

  1. Suppose the shell-color gene were on an autosome. What shell color would you expect in the first-generation daughters and sons? Give the parents' genotypes to show why.
  2. Now suppose the gene is on the X chromosome. Give the genotypes of the green female and the copper male, and use them to show why every daughter was copper and every son was green.
  3. If the gene is on the X chromosome, how many young would you expect in each of the four groups in the table? Show how you got your numbers.
  4. Work out the chi-square value for the table, comparing it with your expected numbers from part c. Then say whether the counts fit the idea that the gene is on the X chromosome, and why.

Write each part, then check yourself against the checklist. Nothing you type here is saved or graded.

Check yourself and see a model answer

Checklist

Our own checklist of what a full answer does, one line per point. Tick each one your response earns.

0 of 8 points

Model answer, annotated

(a) If the gene were on an autosome, the green female would be cc and the copper male would be CC, because both come from true-breeding lines. Every young beetle would get c from its mother and C from its father, so all would be Cc. Copper is dominant, so every daughter and every son would be copper.

  • Point 1 States the prediction for both sexes: all copper.
  • Point 2 Gives cc and CC for the parents and Cc for every offspring.

(b) On the X, the green female is X-c X-c and the copper male is X-C Y. Each daughter gets one X from each parent: X-c from her mother and X-C from her father, so she is X-C X-c and copper. Each son gets his only X from his mother, which must be X-c, and the Y from his father, so he is X-c Y and green.

  • Point 3 Writes both parents in X notation.
  • Point 4 Daughters get the father's X-C, so they are copper.
  • Point 5 Sons get X-c from the mother and Y from the father, so they are green.

(c) The first-generation daughters are X-C X-c and the sons are X-c Y. The daughters pass X-C or X-c with equal chance, and the sons pass X-c or Y with equal chance, so the young are X-C X-c (copper female), X-c X-c (green female), X-C Y (copper male) and X-c Y (green male) in a 1 : 1 : 1 : 1 ratio. A quarter of 400 is 100, so I expect 100 in each group.

  • Point 6 Shows the cross, the 1 : 1 : 1 : 1 ratio and 100 per group.

(d) Copper females: (102 - 100) squared / 100 = 4 / 100 = 0.04. Green females: (96 - 100) squared / 100 = 16 / 100 = 0.16. Copper males: (99 - 100) squared / 100 = 1 / 100 = 0.01. Green males: (103 - 100) squared / 100 = 9 / 100 = 0.09. Chi-square = 0.04 + 0.16 + 0.01 + 0.09 = 0.30. This is less than the critical value of 7.81, so the small gaps can be explained by chance. The counts are consistent with the gene being on the X chromosome, and I do not reject that idea.

  • Point 7 One term per group, then the total of 0.30.
  • Point 8 Compares 0.30 with 7.81 and concludes the data fit the X-chromosome idea.