
Javi Gracia shares new Patrick Bamford injury news from Elland Road ahead of Leeds United vs Liverpool kick off
Javi Gracia has revealed that Patrick Bamford is missing from the Leeds United match day squad with a new calf injury picked up against Crystal Palace.
The striker has struggled with fitness issues throughout the past two seasons, and was withdrawn early in the shock 5-1 collapse to Roy Hodgson’s men last Sunday on 9 April.
Rodrigo has come in to start up front for the visit of Jurgen Klopp and Liverpool but it is not yet clear whether Bamford will be available going forward.

Speaking to Sky Sports from Elland Road ahead of kick off (17 April, 7.14pm) the Whites boss said: “[After Crystal Palace] We had some issues, some little problems, and one of them was Patrick.
“Patrick had problems in his calf and we’ll see his evolution, if he will be ready for the next [games].”
Not again
The same old story continues for Bamford in what has been a miserable couple of seasons for the one-time England man.
He has become the figure upon whom so many hopes are often pinned when absent, but when he comes back from one problem he is frequently soon struck down with another.
Dependent on the result on Monday night against the Reds the relegation battle could be looking very shaky again, in large part thanks to the unexpected thrashing to Palace last week, as Leeds go into this game just two points outside the drop zone.

Leeds United head to Fulham next Saturday (22 April) before a pair of meetings with relegation rivals, with Leicester and Bournemouth next up.
While the latter have suddenly shot up the table and may be out of trouble by then, the game against the Foxes looks incredibly crucial.
Having Bamford available, for the shape of the side as much as his goal-scoring ability, would be a key element to Gracia’s selection plans, so this is the latest in a number of damaging injury blows in recent weeks.
In other Leeds United news, “genuine fears” have arisen at Elland Road of a complete collapse in the 49ers takeover.