Answer:
The correct answer would be 2 in 4.
According to the question Xo and XO show codominance and express themselves completely when present in heterozygous condition. Cats bearing XoXO show patchwork of black and orange fur and are called tortoiseshell cats.
Xo codes for orange color fur and XO codes for black color fur. In addition, Y chromosome does not contain any gene associated with fur color.
Now, genotype of mother cat is XOXO (orange fur). So, the gametes formed would be XO only.
The genotype of father cat is XoY(black fur). So, the gametes would be Xo and Y.
The cross would lead to the formation of two male cats each having XOY as their genotype and two female cats each with XOXo as their genotype.
Hence, both the male cats would show orange fur and both the female cats would show patchwork of orange and black fur.
Therefore, we can conclude that 2 out of 4 would exhibit tortoiseshell coloring.
The correct terms to fill in the blanks are gastric cavity and extracellular. In cnidarians, the mouth of both the medusa and polyp form opens into an internal gastric cavity where extracellular digestion takes place. The polyp and medusa are body forms that can be found in the phylum Cnidaria. The polyp is a nonmotile body form while the medusa is a free swimming body form.
Answer:
Mice and Rabbits
Explanation:
the other ones would be for omnivores
Based solely on the situation given, the most correct answer is (b) DSM requires a recurrent pattern of problematic use. DSM<span>-IV Criteria for </span>Alcohol<span> Abuse states that "</span>alcohol<span> abuse" should be manifested by plenty of factors, including: RECURRING </span>alcohol<span>-related legal problems.</span>