
Discovery of ancient ‘reptile' claw fossils kicks evolution's timeline back by over 35 million years
'I'm stunned,' Per Ahlberg from Uppsala University, who led the study, said in a media release . 'A single track-bearing slab, which one person can lift, calls into question everything we thought we knew about when modern tetrapods evolved.'
A study published in Nature Wednesday dates the fossil tracks to be approximately 355 million years old. It pushes the origin of the species back by 35 to 40 million years from what was earlier thought to be the point when tetrapods evolved from a group of fish that left the sea. This has an implication on the history of human evolution given that we are direct descendants of these tetrapods.
New Delhi: The discovery of ancient fossil footprints of claws in Australia have scientists across the world re-examining evolutionary times of land-based vertebrates.
It was two amateur explorers who discovered the tracks on the banks of the Broken River in Taungurung Country, Victoria, and alerted paleontologists. They were preserved on the upper surface of a loose but fine-grained silty sandstone block. In the Nature study, the authors called it a 'demonstration of the value of citizen science.'
Uppsala University's Ahlberg teamed up with paleontologists from Australia's Flinders University.
'Once we identified this, we realised this is the oldest evidence in the world of reptile-like animals walking around on land, and it pushes their evolution back by 35-to-40 million years older than the previous records in the Northern Hemisphere,' said Professor John Long of Flinders University in a press release.
Their findings are threatening to upend our understanding of evolution of all tetrapods.
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When did the first tetrapods emerge?
As the word suggests, tetrapods include all species that have 'four feet'. They are the first colonists on land and their origin began when fish transitioned from the oceans to adapt to life on land. They are the distant ancestors of all modern amphibians and amniotes that includes vertebrate animals like reptiles, birds and mammals, including humans.
The oldest known tetrapods had primitive fish-like forms, and could barely move on land. The separation of amphibians and amniotes was so far believed to have begun at the start of the Carboniferous period, some 355 million years ago. This separation is known as the tetrapod crown group node.
The new study changes what was previously known by suggesting that the separation dates back to the Devonian period, some 390 million years ago.
'The timeline of these events has seemed clear-cut: the first tetrapods evolved during the Devonian period and the earliest members of the modern groups appeared during the following Carboniferous period,' according to the media release from Uppsala University.
It's the claws that have generated much excitement within the scientific community.
'Claws are present in all early amniotes, but almost never in other groups of tetrapods,' said Ahlberg. 'The combination of the claw scratches and the shape of the feet suggests that the track maker was a primitive reptile.'
The study also suggests that tetrapods originated in Gondwana, the southern supercontinent which Australia was a part of. It also included present-day South America, Africa, Arabia, Madagascar, India, and Antarctica.
The researchers have also found new fossil reptile footprints from Poland, suggesting that tetrapods in the Euramerica—the supercontinent that formed during the Devonian period and included North America, Greenland, northern Europe, and Russia—also originated earlier than previously thought.
So far, researchers have found only fossilised footprints and no fossil bones of ancient tetrapods, but if the new timelines are correct, the fossil footprints findings suggest that the evolution into land-based animals occurred not just earlier but also much quicker than initially thought.
(Edited by Radifah Kabir)
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Indian Express
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Hindustan Times
4 hours ago
- Hindustan Times
Chicken and egg — and duck too
Here at Problematics we usually aim for puzzles that are not the kind you would find in a textbook, but there are exceptions. Some puzzles that can be solved with textbook methods are still interesting because of the way they are packaged or because of their pedigree, with illustrious minds having dwelt on them at some point in history. A prime example of puzzles that are delightful because of both packaging and pedigree are the problems in Bhaskara's Lilavati. While those are widely known, I recently found one that I hadn't come across earlier. It is said to have appeared in a book by the great Euler, and described by the French writer Stendhal before making its way into the writings of the late Russian mathematician Yakov Perelman. To insulate the solution from an internet search, I have added my customary modifications to the version described by Perelman. I have changed the currency to Indian rupees, and tinkered with the prices to bring them within a range that is credible for the story into which I have packaged my adaptation. The story, of course, is entirely my own. #Puzzle 145.1 A family of poultry farmers collects 100 eggs one morning. They are all chicken and duck eggs, the distribution being unequal. Handing the chicken eggs to their son and the duck eggs to their daughter, the farmer parents send them off to the market. The price for each kind is fixed, with the duck eggs being costlier than chicken eggs, as is the case in most places. Each child sells his or her full share of eggs at the respective fixed rates. In the evening, when they compare their earnings, they are thrilled to find that both have made exactly the same amount. I am no farmer, but the internet tells me that hens and ducks lay about one egg daily at the peak of their productive years. It is not surprising, therefore, that the same birds at our farm lay the same number of eggs the following morning. In other words, the family has 100 eggs again, and the unequal distribution of chicken and duck eggs is the same as on the previous day. Mother segregates the produce into a number of baskets, the chicken eggs on one side, the duck eggs on another. Father passes the orders: 'Pick up your respective shares and come back with the same earnings as you did yesterday.' The kids get mixed up, of course (how else would there be a puzzle?) The son picks up the duck eggs by mistake, and the daughter takes the rest. Neither of them notices that his or her count is not the same as on the previous day. At the market, the boy sells the duck eggs at the price for chicken eggs, and his sister sells the chicken eggs at the price for duck eggs. When they compare their earnings in the evening, the boy is alarmed. 'I got only ₹280 today. I don't know how I can explain this to Father,' he says. The girl is equally puzzled about her collection, but pleasantly so. 'I don't know how, but my earnings rose to ₹630 today,' she tells her brother. #Puzzle 145.2 MAILBOX: LAST WEEK'S SOLVERS Hi Kabir, Assuming that the store owner initially bought cat food for 31 cats for N days, or 31N cans. As each cat consumes 1 can/day, the total consumption reduces by 1 can every day. Again, all cans were consumed in one day less than twice the number of days originally planned, or (2N – 1) days. Thus the total number of cans is the sum of an arithmetic progression of (2N – 1) terms starting 31, and with a common difference of –1. The sum of the AP is: [(2N – 1)/2][2*31 + (2N – 1 – 1) (–1)] = 65N – 32 – 2N² Equating the above to 31N and simplifying, we get the equation 2N² – 34N + 32 = 0. The roots of this equation are N = 16 and 1. As 1 day is not viable, N must be 16. So the total number of cans bought initially = 31*16 = 496. And as it took (2N –1) = 31 days to finish the whole stock of food, only 1 cat was left unsold. — Anil Khanna, Ghaziabad *** Hi Kabir, Suppose the cat food was initially ordered for N days. Then, the number of cans ordered = 31N. Also, suppose K is the number of cats remaining unsold when the food stock got exhausted. On any day, the number of cans consumed is the equal to total number of unsold cats. Thus the total cans consumed = 31 + 30 + 29… + (K + 2) + (K + 1) + K = (31 + K)(31 – K + 1)/2 i.e. 31N = (31 + K)(31 – K + 1)/2 For the right-hand side to be a multiple of 31, K has to be 1. This means 31N = 32*31/2, or N = 16. The number of cans = 31 x 16 = 496. The food lasted for 31 days. If we add one more day, we get 32 days which is twice the original period of 16 days. — Professor Anshul Kumar, Delhi From Professor Kumar's approach, it emerges that the puzzle can be solved even without the information about the cans being exhausted in (2N – 1) days. 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