Answer:
The correct answer would be - 25%.
Explanation:
It is given that the Blue flower is dominant over the yellow flower which means Blue is represented by allele B here and allele b for yellow and Bb is a heterozygous case with a blue phenotypic character. Similarly, the Tall plant is dominant over short and represented by T and t respectively.
BbTt is a heterozygous condition and a cross with itself will produce :
Gametes: BT Bt bT and bt
BT Bt bT bt
BT BBTT BBTt BbTT BbTt
Bt BBTt BBtt BbTt Bbtt
bT BbTT BbTt bbTT bbTt
bt BbTt Bbtt bbTt bbtt
Here 4 out of 16 offspring are heterozygous for both traits represented by bold alphabets. Therefore the correct answer is 25%
Answer:
Cross-pollination.............
Answer:
Cystic fibrosis mutation is recessive to normal allele because only one functional or normal allele is enough to produce a functional protein. So, if mutation is present in one allele then also, a normal protein can be made from normal allele. The presence of normal protein prevents the expression of disease.
In addition, mutated allele only results in the loss of function of protein which can be compensated by the expression of normal allele. It does not add any toxic effect to the protein. Consequently, the disease is inherited in autosomal recessive fashion.
In contrast, Huntington mutation not only alters the structure of the functional protein but also adds toxicity to it. The altered protein is enable to interact with 100s of other proteins and inhibit or decrease their function. So, if only one allele is present then also, the mutated protein will be produced and it will result in the phenotype. Consequently, it is inherited as autosomal dominant fashion.
Answer:
I dont know what your experiment is but the control variable is the thing your keeping the same.
Explanation:
