本题目来源于试卷: Continuity and Change D.8 Inheritance，类别为 IB生物学
[单选题]
A dihybrid cross was carried out between two sunflower ld4r)lq tx:gfy o1)-p (dssd+plants (Helianthus gb.5 ifqre+d6 annuus) to see if the genes for seed number and seed shape are linked. The critical value for the chi-squared test at the 0.05 significance level was 7.82. The calculated value for the chi-squagf qd5+b .e6rired test was 10.05; what does this indicate?
A. The null hypothesis can be rejected and the genes are linked
B. The null hypothesis can be accepted and the genes are unlinked
C. The alternative hypothesis can be rejected and the genes are linked
D. The alternative hypothesis can be accepted and the genes are unlinked
参考答案: A
本题详细解析:
If the calculated value (10.05) is above the critical value (7.82), then the null hypothesis is rejected and we can conclude that the traits are linked at the 0.05 significance level. Choice A is correct. Choice B is incorrect as the calculated critical value is above 7.82, so the null hypothesis should be rejected. Choice C is incorrect as although the genes are linked the alternative hypothesis should be accepted. Choice D is incorrect because if the alternative hypothesis is accepted then the genes are linked.
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