Question

bilgi: bezeleye bitkisi birbirine zıt birçok karaktere sahiptir ve dışardan gözlenebilen çok fazla çeşit içerir. örneğin bezeleyelerin bir çeşidi; buruşuk tohumluyken başka bir çeşidi yuvarlak tohumludur. bir çeşidi mor renkli çiçeklere sahipken diğer çeşidi beyaz çiçeklere sahiptir. bezeleyelerde bitkinin tüm özellikleri değil, bir özelliği dikkate alınarak çaprazlama yapılır.
aşağıdaki grafiklerde bezeleyelerde çiçek rengi kalıtımı ile ilgili yapılan dört çaprazlama sonucunda oluşan bezeleyelerin genotip oranları verilmiştir.
buna göre, çaprazlanan ata bezeleyeler ile aynı genotipe sahip olmayan bezeleyelerin oluşumu hangi çaprazlamalarda gerçekleşmiştir?
a) yalnız 2
b) yalnız 3
c) 1 ve 2
d) 3 ve 4

Explanation:

Step1: Analyze the first cross

In the first cross (\(Aa\times aa\)), the parents have genotypes \(Aa\) and \(aa\). The offspring genotypes are \(Aa:aa = 1:1\). The parent genotypes are \(Aa\) and \(aa\), and the offspring have both \(Aa\) and \(aa\). So, there are offspring with genotypes different from the parents.

Step2: Analyze the second cross

In the second cross (\(AA\times Aa\)), the parents have genotypes \(AA\) and \(Aa\). The offspring genotypes are \(AA:Aa=1:1\). The parent genotypes are \(AA\) and \(Aa\), and the offspring have both \(AA\) and \(Aa\). So, there are offspring with genotypes different from the parents.

Step3: Analyze the third cross

In the third cross (\(Aa\times Aa\)), if we assume the parents are \(Aa\) (since the offspring is \(100\%\ Aa\) which is not possible for a normal Mendelian cross. But if we consider self - cross of \(Aa\) (assuming some error in the problem's third - cross graph interpretation as \(Aa\times Aa\) gives \(AA:Aa:aa = 1:2:1\), but if we assume the cross is \(AA\times AA\) (not in line with the graph's \(Aa\) label, but for the sake of the problem's logic of parent - offspring genotype comparison), if parents are \(AA\) and the offspring is \(Aa\) (which is wrong for \(AA\times AA\)), but if we consider the cross as \(Aa\times Aa\) (ignoring the graph's \(100\%\ Aa\) for a moment of logical parent - offspring comparison), parents \(Aa\) and offspring \(AA\) and \(aa\) (from correct \(Aa\times Aa\) cross) would have different genotypes. But if we go by the graph's \(100\%\ Aa\) offspring, if parents are \(Aa\) (assuming self - cross), then all offspring have the same genotype as parents.

Step4: Analyze the fourth cross

In the fourth cross (\(Aa\times Aa\)) (assuming normal Mendelian cross as \(AA:Aa:aa=1:2:1\) which is in line with the graph's \(25\%\ AA\), \(50\%\ Aa\), \(25\%\ aa\)). The parents (assuming \(Aa\)) have offspring with \(AA\) and \(aa\) which are different from the parent genotype \(Aa\).

Answer:

C. 1 ve 2