WildlifeWorld News

Gorilla Rescued From Plane’s Cargo On Turkish Flight From Nigeria To Thailand

By Micheal Chukwuebuka

A young gorilla was rescued from a plane’s cargo on a Turkish Airlines flight from Nigeria to Thailand and was taken to an Istanbul zoo.

On Sunday, officials said the 5-month-old gorilla, which was discovered in a box last month, was recovering at the zoo, while wildlife officers were considering returning him to his natural habitat.

After a public competition, he has been named Zeytin, or Olive, and is recuperating at Polonezkoy Zoo.

“Of course, what we want and desire is for the baby gorilla to continue its life in its homeland.

“What is important is that an absolutely safe environment is established in the place it goes to, which is extremely important for us,” Fahrettin Ulu, regional director of Istanbul Nature Conservation and National Parks, said Sunday.

In the weeks since he was found, Zeytin has gained weight and is showing signs of recovering from his traumatic journey.

“When he first came, he was very shy, he would stay where we left him.

“He doesn’t have that shyness now. He doesn’t even care about us much. He plays games by himself,” said veterinarian Gulfem Esmen.

Both gorilla species – the western and eastern gorillas, which populate central Africa’s remote forests and mountains, are classified as endangered by the International Union for Conservation of Nature.

As Istanbul emerges as a major air hub between continents, customs officials have increasingly intercepted illegally traded animals. In October, 17 young Nile crocodiles and 10 monitor lizards were found in an Egyptian passenger’s luggage at the city’s Sabiha Gokcen Airport.

What's your reaction?

Excited
0
Happy
0
In Love
0
Not Sure
0
Silly
0
Micheal
Micheal Chukwuebuka is a passionate writer. He is a reporter with STONIX NEWS. Besides writing, he is also a cinematographer.

    Leave a reply

    Your email address will not be published. Required fields are marked *

    More in:Wildlife

    0 %
    $year = date('Y'); return $year;